Handbook p5 refraction

Agastya International Foundation

Refraction Handbook P5

“All truths are easy to understand once they are discovered; the point is to discover them.�

-Galileo Galilei (1564-1642)

1

Handbook P5 Refraction OVERVIEW OF HANDBOOK ABL

CONCEPT

NO OF ACTIVITIES

TIME (min)

PAGE NO

ABL 1 ABL 2 ABL 3 ABL 4 ABL 5 ABL 6 ABL 7

Introduction to Refraction Refraction and Normal Shift Refraction through prism Total Internal Reflection Refraction through curved surfaces Convex lens Magnification of convex lens; focal length of combination of lenses Optical instruments (Do it yourself)

3 2 4 2 4 2 2

45 65 60 50 45 60 60

4 15 28 34 44 61 71

6

-

79

ABL 8

ABLs WITH REFERENCE TO STANDARD

S.No. 1 2 3 4

STANDARD 9 and 10 9 and 10 9 and 10 9 and 10

PREFERRED ABL ABL 1 ABL 2 ABL 3 ABL 4

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LIST OF FIGURES, CHARTS AND WORKSHEETS S. No

Name

Page No

Fig 1

Refraction

8

Fig 2

Looking at a coin through water

8

Fig 3

Lateral shift model

9

Fig 4

Refraction through a glass slab

17

Fig 5

Calculating Refractive Index using real depth and apparent depth

25

Fig 6

Total internal reflection through semi-circular slab

35

Fig 7

Total internal reflection in prism

40

Fig 8

Optical fibre

41

Fig 9

Refraction through curved surface

51

Fig 10

Set up for ABL 3.2b

61

Chart 1

Fermat’s Principle

11

Chart 2

Normal shift

23

Chart 3

Refraction through prism

29

Chart 4

Critical angle and Total internal reflection

37

Chart 5

Total internal reflection in right angled prism

39

Chart 6

Parallel beam passing through prism

39

Chart 7

Refraction through a curved surface

52

Chart 8

Image formation by convex and concave lens when object is placed at infinity

55

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Chart 9

Image formation by convex and concave lens when object is placed beyond 2F

55

Chart 10

Image formation by convex and concave lens when object is placed at 2F

55

Chart 11

Image formation by convex and concave lens when object is placed between F and 2F

56

Chart 12

Image formation by convex and concave lens when object is placed at F

56

Chart 13

Image formation by convex and concave lens when object is placed between F and O

57

Chart 14

Image formation by concave lens

57

Chart 15

Nature of image formed by convex lens when object is at different positions

62

Worksheet 1

Observations for image formation by convex lens

60

Worksheet 2

Observations for calculating focal length of a convex lens

65

Worksheet 3

Observations for calculating magnification of a convex lens

69

Worksheet 4

Observations for determining focal length of a combination of two lenses

73

Note to Instructor: All the figures in this handbook are for the Instructor’s reference only. The Charts need to be printed and shown to the learners during the course of the activity. Worksheets need to be printed out in advance for the learners. The number of worksheets required is mentioned in the Material List.

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4

ABL 1 Introduction to Refraction Activity

Learning objective

1.1

Studying a few illustrations that demonstrate the refraction of light.

1.2

Arriving at Snell’s law from Fermat’s principle.

Key messages

The path of a light ray and its speed change whenever a light ray is incident obliquely on the boundary separating the media and only speed changes when incident normally. 

Time (min)

25

Light ray has to choose a path for 20 which sin i/sin r = constant for a given pair of media in accordance with Fermat’s principle.

Total Time

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45 mins

5

Introduction (Note to Instructor: This introduction is for you. Please read it before you perform the activities with your class) Let us recall some of our experiences in our daily life:         

Very minute particles that cannot be seen with naked eye become visible when viewed through a lens.

The images of objects appear to be of different size and nature when viewed through a thick curved transparent medium.

Searchlight transmitting light to very large distances.

You seem to misjudge to catch a coin placed at the base of water tank.

The base of a water tank appears to be shifted towards you.

The path of light rays bent conveniently using a prism or a lens.

The intensity of light rays being changed using a lens.

The formation of VIBGYOR when which light is passed through a prism.

The formation of a rainbow, dazzling of diamonds.

An understanding of all these phenomenon is based on the ability of an optical medium to change the path of a light ray that tries to pass through – i.e., refraction.

What is refraction?

Explanation: Light ray ( ) in any homogeneous optical medium has a definite velocity and follows a linear path i.e., a straight line. If the medium is non-homogeneous the path becomes curved because the velocity is different at different regions. Hence it is to be expected for a light ray to change its path whenever it travels from one optical medium to another. The change in the path is sudden and occurs only at the boundary. Such bending of a ray of light whenever it travels from one optical medium to another is called refraction of light.

When a ray of light is incident normally on the boundary there is only a change in the magnitude of velocity without change in direction. On the other hand when a ray of light is incident obliquely at the boundary there will be a change in magnitude as well as direction. Hence refraction is due to change in velocity when light travels from one optical medium to another.

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Normal incidence oblique incidence

Bending towards normal and away from the normal It is observed that 1. A ray of light is bent towards the normal when it travels from a rarer to a denser optical medium.

Deviation, d = i - r 2. A ray of light bends away from the normal when it travels from denser to rarer medium. This bending of light either towards or away from the normal follows Fermat’s principle. The path of the light is that for which the time of travel is minimum. d=i-r 3. It is known that the velocity of light decreases when it travels from rarer to denser medium and increases when it travels from denser to rarer medium.

4. When a ray of light is incident normally on a boundary it travels without change in direction. However there is a change in speed of light.

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min

7

ABL 1.1 LEARNING OBJECTIVE: Studying a few illustrations that demonstrate the refraction of light.

ADVANCE PREPARATION Material List S.no 1

Material Glass beaker

Quantity 1 per group

2

Pencil

2

per group

3

Water

4

Coin

1

per group

Things to do: Keep the materials ready before the session Safety precautions: Not Applicable

SESSION Link to known information/previous activity Not Applicable Procedure Divide the learners into five groups. Illustration 1 is a group activity and Illustration 2 is demo.

Illustration 1 Pour water in to a beaker. Take a pencil or metal rod and dip partially into water. View the pencil full from the side. Look at the pencil from the top. What do you observe when you look from sides? What do you observe when you look from the top?

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Fig 1: Refraction The part of the pencil under water appears to be bent when viewed from the sides. When viewed from the top surface, the part of the pencil under water is not seen to be in line with the part in air.

Illustration 2

Fig 2: Looking at a coin through water Place a coin at the bottom of an empty beaker. Cover the sides with paper. Looking at the coin from the top slowly move your eye towards the edge of the beaker. At one position you stop seeing the coin. Do not go too far from the beaker. Ask one of your friends to pour water into beaker. At some level of water the coin again is in your field of view though, the direct light is blocked by the sides.

UNDERSTANDING THE ACTIVITY Leading questions 1. Why was the coin observable when viewed directly or to some extent moving towards sides?

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2. Why coin was not observable when viewed from the side at some angle?

3. Why do you think the coin was observable when water was poured up to a certain minimum height?

Discussion and explanation 1. The coin could be seen when viewed directly through air. When viewed at some angle from the sides at some point the light from the air was blocked on the way by the walls of the beaker.

2. When water is poured into beaker up to a certain level, some of the light rays incident directly on water boundary seem to have bent away from the normal and reached the eye.

Illustration 3 Place the given set of colored rods on one side of prism as shown in the figure. Ask one of the students to hold the prism on hand and look at colored rods through the prism. Let him hold the rod of same color, remove the prism. What do you observe?

Fig 3: Lateral shift model

UNDERSTANDING THE ACTIVITY: Leading Questions 1. Did he hold the rod of the expected colour?

2. Was the rod he held in the line of sight when the prism was moved?

3. Why did he miss to hold the rod of expected colour?

Discussion and explanation When you try to pick the rod on the other side the hand holding the chosen rod is not in the line of sight. The light ray from the said rod enters the prism in some direction and you start seeing the image in a different direction. The light ray on passing the prism seems to have changed path on reaching your eye.

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All these experiments point to the fact that a light ray while travelling from one optical medium to another seems to change its path or bend. This effect that arises at the boundary separating media is called refraction.

As we study later, the bending is due to change in velocity of light from one medium to another. Different media bend the light ray to different extent. Since light is a vector, the change in velocity may mean a change in magnitude or change in magnitude and direction.

KEY MESSAGES 

The path of a light ray and its speed change whenever a light ray is incident obliquely on the boundary separating the media and only speed changes when incident normally.

LEARNING CHECK: Ask learners to list the key things they have learnt. Guide them to the key messages listed and then put up the chart of key messages. If you have time during the class, make up a small game, quiz or match the following as a learning check. This may have to be done as part of advance preparation.

Time: 20 min

ABL1.2

LEARNING OBJECTIVE: Arriving at Snell’s law from Fermat’s principle.

ADVANCE PREPARATION Material List S.no 1

Material Scale

Required Quantity 1 per group

2

Pencil

1 per group

3

Eraser

1 per group

4

Sharpener

1 per group

Things to do: Agastya International Foundation. For Internal Circulation only.

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Not Applicable Safety precautions: Not Applicable

SESSION Link to known information/previous activity In previous activities you have seen the bending of light rays and in this activity we learn why the light rays bend.

Procedure: Let us consider a ray of light travelling from a point ‘A’ in medium1 to a point ‘B’ in another optical medium2.

While moving from A to B the light ray chooses the path of minimum time ‘in accordance with Fermat’s principle’.

Chart 1: Fermat’s Principle Let v1 be the velocity of light in medium1 (rarer) and v 2be the velocity in medium2 (denser). If it chooses a straight path the time of journey is t1 = AC/v1 + CB/v2 Agastya International Foundation. For Internal Circulation only.

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If the light ray chooses a path (2) AEB, the time is t2 = AE/v1 + EB/v2v2< v1 The light is travelling a longer path in denser medium and shorter path in rarer medium (AC > AE and

EB > CB)

t2 > t1 i.e., bending along AEB is not allowed.

Consider a ray choosing the path ADB t3 = AD/v1 + DB/v2 Since DB/v2 is smaller than BC/v2it can be seen t3< t1 the path of the light is on the other side (to the right) of ACB and not to left. In travelling from point A to point B which is in different media, the light chooses only the path AMB because the time of joining is minimum along this path only.

If D is moved away from C it can be shown that t3 generally decreases and at some position becomes minimum and then begins to increases. That position says M is the correct position for minimum time of journey. Therefore AMB is the right path across the boundary. If we measure angle of incidence and angle of refraction then it is seen that Sin I / sin r = constant for the given pair of media.

This is the essence of Snellâ&#x20AC;&#x2122;s law.

There are two laws of refraction

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i)

The ratio of sin of angle of incidence to the sin of angle of refraction is always a constant and is called refractive index of the second medium with respect to the first medium. μ2 = Sin i/ Sin r

Choose two points P and Q equidistant from M. draw PK and QL perpendicular on The normal.

Sin i = PK/PM , Sin r = QL/LM Sin i/Sin r = PK/QL x QM/PM PK/QL = μ2

ii)

= PK/QL

The incident ray and the refracted ray are always in a plane containing the normal at that point.

The laws of refraction help us to quantify the effect of refraction when the media are known Since the ray is incident at a point on the boundary the laws remain the same for all kinds of surfaces.

Sin i/ Sin r = refractive index of medium 2 (μ2) with respect to the medium 1. If the ray travels from medium 2 to medium 1 then the refracted ray is in medium 1. Then paths are interchange ‘r’ is angle of incidence and ‘i’ is angle of refraction. The sin r/sin i =2μ1 is the refractive index of medium (with respect to medium2).

KEY MESSAGES: 

Light ray has to choose a path for which sin i/sin r = constant for a given pair of media in accordance with Fermat’s principle.

LEARNING CHECK: Ask learners to list the key things they have learnt. Guide them to the key messages listed and then put up the chart of key messages. If you have time during the class, make up a small game, quiz or match the following as a learning check. This may have to be done as part of advance preparation.

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14

ABL 2 Refraction and Normal Shift Activity

Learning Objective

2.1

Understanding the phenomenon of refraction of light through a rectangular glass slab.



A ray of light incident obliquely and passing through the rectangular slab emerges parallel to itself.



There is no deviation but there is a lateral shift.

To understand the phenomenon of normal refraction and normal shift associated with it.



Object at a certain distance from 35 our eye when viewed along normal through a denser medium appears to be nearer. This is due to normal refraction.

2.2

Key message

Total Time

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Time 30

65 mins

15

ABL2.1 LEARNING OBJECTIVE: Understanding the phenomenon of refraction of light through a rectangular glass slab.

ADVANCE PREPARATION Material List S.no 1

Material Rectangular Glass slab

Required Quantity 1 per group

2

Scale

1 per group

3

Pencil

1 per group

4

Eraser

1 per group

5

Bell pins

10 per group

Things to do: Not Applicable Safety precautions: Not Applicable

SESSION Link to known information/previous activity You have learnt about phenomenon of refraction, now let us see how it happens when light travels from rarer (air) to denser (glass) medium and vice versa Procedure: Spread a white sheet on a drawing board. Fix the white sheet with bell pins. Place the glass slab with its edges parallel to the edges of the paper. Mark the boundary with a sharp pencil. Remove the slab and draw a Agastya International Foundation. For Internal Circulation only.

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straight line (PO) touching the earlier boundary line of slab at a point Replace the slab into the positions marked earlier, fix two or three bell pins (L1 and L2) vertically on the line PO. Looking through the opposite face fix pin L3 such that images of L1, L2 and the pin L3 are collinear. Likewise fix another pin L4that is collinear with images of L1, L2 and pin L3. Mark the position of L3 and L4. Remove the slab.

Draw a straight line through positions of L3 and L4 to touch the opposite face at O2. Call this line as O2Q. Join O1 and O2 between opposite faces. Draw normal lines at O 1 and O2.

Measure the angle subtended by PO with normal (I) i.e. angle of incidence.

Measure the angle between QO2 and normal (e) i.e. the angle of emergence.

Measure the angle between normal and O1O2 inside (r). Produce PO1 and measure the perpendicular distance between incident ray and emergent ray (S).

Fig 4: Refraction through glass slab Note to Instructor: Please draw Fig 4 on the board after all the learners have completed the activity. Ask them to compare their work with the figure on the board.

UNDERSTANDING THE ACTIVITY Leading questions: 1. Why do we need bell pins for this experiment?

2. What do i, r and e represent?

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3.

4.

5.

6.

7.

How are the incident and emergent rays oriented with respect to each other?

What is the deviation at the first boundary?

What is the deviation at the second boundary?

What is the total deviation produced by the slab for the ray passing through it?

What is the separation between incident and emergent rays called? And what factors does it depend upon?

8. Is it possible to calculate refractive index of the medium of the slab in this experiment?

9. Summarize the total behavior of light ray on passing through the slab?

Discussion and explanation Rectangular slab is an optical medium bound by two opposite refracting surfaces, which are parallel. The bell pins help to fix a ray of light. i) PO1 passing through the pins L1 and L2 acts as light ray. The ray is incident on the boundary 1 at an angle (i = 500). When viewed from the other bounding surface the ray appears to emerge along O 2Q , which passes through the pins L3 and L4. O1 is the point of incidence and O2 is the point of emergence. Therefore O1 O2 is the path of refracted ray inside the slab.

For boundary 1, PO1 is the incident ray and O1 O2 is refracted ray. The ray has bent towards normal (rarer to denser). i is angle of incidence and r is angle of refraction.

Refractive index of glass with respective to air aμg = sin i/ sin r.

The deviation of the ray at first boundary d 1= i - r. ii) The ray O1 O2 is incident on boundary2 and emerges along O 2Q. Hence O1O2 is incident ray and O2Q is emergent ray.

Angle of incidence = r Angle of refraction = e Angle of deviation = e-r = d2 Therefore refractive index of air with respect to glass gμa = sin r / sin e But aμg = μg/μa and gμa = μa/μg Agastya International Foundation. For Internal Circulation only.

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18 aÎźg=

1/gÎźa

Sin i / sin r = sin e / sin r Sin i = sin e i = e Inclinations of incident ray and emergent ray are same. The two rays are parallel. Also total deviation of slab = deviation at first boundary + deviation at second boundary d = d1+d2 = ( i - r) + ( r - e ) = i - e but i = e d = 0 The incident and emergent rays are parallel. iii) The incident ray PO1 and emergent ray QO2 are parallel but displaced laterally through a distance called lateral shift (S).

Lateral shift S = perpendicular distance between PO 1 and QO2 A ray of light on passing through the rectangular slab has only undergone lateral shift without deviation.

Replace the slab on the marked boundaries, let the pins be in place L1, L2 on PO1 and L3, L4 on QO2.

Now look through the face1 and observe images of L 3 and L4.

The images of L3,L4 are collinear with L1and L2.

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It means, if PO1 is incident ray then QO2 is emergent ray.

If QO2 is incident ray then PO1 is emergent ray.

The two rays are interchangeable. The paths are reversible.

The lateral shift is larger, i) If i increases ii) If refractive index of slab with respect to surrounding medium is larger. iii) If wave length decreases

KEY MESSAGES  

A ray of light incident obliquely and passing through the rectangular slab emerges parallel to itself.

There is no deviation but there is a lateral shift.

LEARNING CHECK: 1. Refractive index of glass with respect to air is 1.5. What is the refractive index of air with respect to glass?

Given aμg = 1.5 , but

gμa

= 1/ aμg

= 1/1.5

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INTERESTING INFORMATION: When are the incident and emergent rays not parallel?

Ans: When the media on either side of the slab are not the same, the two rays are not parallel.

When incident ray is close to the boundary ( i = 900) the emergent ray grazes the second boundary.

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Time: 30 min

21

ABL2.2 LEARNING OBJECTIVE: To understand the phenomenon of normal refraction and normal shift associated with it.

ADVANCE PREPARATION Material List S.no 1

Material Glass beaker 400 ml

Required Quantity 1 per class

2

Microscope

1 per class

3

Bell pins

10 per class

4

Cello tape

1 per class

5

Scale

1 per class

6

Water

7

Saw dust

1 cup per class

8

Pencil/Sketch

2 per class

9

1 rupee coin

1 per class

Things to do: Not Applicable Safety precautions: Not Applicable

SESSION Link to known information/previous activity: A few interesting things like lateral shift, deviation of a ray at boundary, give rise to very interesting effects. In this activity we try to understand another interesting situation arising out of refraction.

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Procedure: Take an empty beaker and fix a bell pen at the base with a cello tape. Fix a compound microscope to an adjustable stand and focus it vertically to see the pin clearly through the microscope. Measure the distance of the objective of the microscope from the table top (R 1) Ask one of the students to pour water into beaker till the brim. The image of the pin is lost out of the field of view. Catch the image again by raising the microscope (do not operate the focusing screw). Measure the distance of the microscope from the tabletop again (R2).

Now sprinkle some saw dust on water surface and raise the microscope to see clearly the saw dust on water surface. Measure the distance of microscope from tabletop (R 3).

Chart 2: Normal Shift

UNDERSTANDING THE ACTIVITY Leading questions 1. What is the significance of the initial adjustment R 1?

2. The image of the object went out of field of view on pouring water into beaker. In which direction did you move the microscope to catch the image?

3. Why did you raise or lower the microscope to regain the image into field of view?

4. How do you explain the observed effect?

5. What is normal shift?

6. Is normal shift related to refractive index of the medium?

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Discussion and explanation 





  

When the object is viewed through the microscope the image is distinct and clear only when the object is in the focal plane of the eyepiece. This distance for a given microscope is fixed. R1 gives the distance between the microscope objective and the object for clear image.

When water is poured into beaker the image is out of field of view indicating that the light rays passing out of water have gone out of focus due to refraction. The microscope is raised through a certain distance to see the image. It only means that the position of the image seems to have shifted upward in the water medium. In other wards the image is nearer to the microscope (than the object) and therefore the microscope has to be raised further to maintain the right focusing distance. The distance by which the image appears to be nearer when viewed along the normal through a denser medium is called normal shift. It is due to refraction of light rays from the pin at the water boundary.

In the figure O is a luminous point object in a denser medium of refractive index μ w (water). It is at a depth h0 from the water air boundary. Two rays of light from O, incident very close to the normal are refracted at the boundary, bending away from it. When produced backwards, they meet at I. For a person viewing from air along the normal the refracted rays appear to come from I. therefore I is the image of O as seen by the observer. Due to normal refraction, an object at O appears to be shifted from O to I and appears nearer. This distance through which the object appears to be nearer when viewed through a denser medium is called normal shift.

In the figure, Real depth =R3 – R1 Apparent depth = R3 – R2 Normal shift = Real depth – apparent depth. = R2 – R 1 .

This shift increases as the water level in the beaker increases and refractive index of the medium increases.

When different liquids are poured to same height the normal shift is different. Since refractive index is different for different colours, normal shift also depends on .

It is possible to calculate the refractive index of the light from the formula, w = real depth from water surface/apparent depth from water surface. = R3 – R1/R3 – R2 = h0/hi μ0 = h0/hi

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Fig 5: Calculating the refractive index using real depth and apparent depth

Note: If a person in water looks at an object above water surface the bending of light from air to water is opposite (towards normal). The object appears to be farther.

KEY MESSAGES ď&#x201A;ˇ

An object at a certain distance from our eye, when viewed along normal through a denser medium appears to be nearer. This is due to normal refraction.

LEARNING CHECK: o A coin placed at the base of a beaker appears to be at a depth of 4.5 cm under water when viewed normally. If water stands to a height of 6 cm in beaker, calculate the refractive index of water. = real depth/ apparent depth = 6/4.5 = 4/3 o An object is viewed through a diamond slab of 12cm thickness. What is the distance by which it appears nearer? d = 2.4 Apparent depth = real depth/

d

= 12/2.4 = 5cm shift = 12 â&#x20AC;&#x201C; 5 = 7cm o A swimmer under water looks at a building 24m tall. What is the height as observed by him? = apparent depth /real depth Agastya International Foundation. For Internal Circulation only.

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4/3 =apparent depth/24 Apparent depth = 24X 4/3 = 32 The building appears to 32m tall.

INTERESTING INFORMATION: ď&#x201A;ˇ

Let us take a bright and point source of light and fix it at the base of a tank. The light from it diverges and spreads in all directions. If we pour water up to certain level, all the light diverging from the bulb will not come out of water surface. Those in the middle come out along a cone and other rays incident outside this zone are totally internally reflected. The water surface appears bright within a circular patch.

ď&#x201A;ˇ

It is for the same reason explained above, for a fish looking up from water everything on land appears to be within a cone.

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ABL 3 Refraction through prism Activity

3.1

Learning objective

To study refraction of light through a glass prism.

Key messages

ď&#x201A;ˇ

A ray of monochromatic light on passing through a prism is deviated from its path and for a given prism this deviation changes with angle of incidence, wavelength and refractive index (RI).

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Time (min)

45

27

ABL3.1

Time: 45 min

LEARNING OBJECTIVE: To study refraction of light through a glass prism.

ADVANCE PREPARATION Material List S.no 1

Material Drawing board

Required Quantity 1 per group

2

A4 sheet

1 per group

3

Bell pens

10 per group

4

Fixing pins

1 per group

5

Prism

1 per group

Things to do: Not Applicable Safety precautions: Not Applicable

SESSION Link to known information/previous activity In the earlier activities we have under stood the behavior of light ray when passed through an optical medium bound by two parallel bounding surfaces. In this activity we learn how the behavior of light changes if the two bounding surfaces are inclined instead of being parallel.

Procedure: Place the drawing board on the table, spread a sheet of white paper on the board and fix it with pins. The given prism is placed on the paper and the boundaries are marked (ABC) with a sharpened pencil. Let BC be the grounded surface and AC, AB the refracting surfaces. Remove the prism; draw a straight line (PO1) inclined at some angle 400 with AB. Draw a normal to AB at O1.

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Replace the prism into the marked boundary. Fix two pins L 1 and L2 on the line PO1. Look through the face AC and fix two pins L3 and L4 one after another such that images of L1, L2 are collinear with the pins L3, L4. Mark the positions of L3 and L4. Remove the prism and draw a straight line (QO 2) through the positions of L3 and L4.

Chart 3: Refraction through Prism Join O1and O2Produce incident ray and emergent ray. They meet at N. Measure the angle between them (d) as deviation produced by the prism.

UNDERSTANDING THE ACTIVITY Leading questions 1. How is a prism different from a rectangular slab?

2. Do you observe any similarities in the way light ray passing through a prism and a ray passing through a slab behave?

3. What are ‘i’ and ‘r’ at each boundary? And what is the deviation at each boundary?

4. Is the total deviation produced by the prism related to the deviations at the two boundaries?

5. Is it possible to calculate the refractive index of the material of the prism applying Snell’s law?

6. On what factors does the deviation produced by the prism depend upon?

7. Will the deviation be different when different colours of light are incident at the same angle?

8. Is it possible to calculate the refractive index of the material of the prism?

Discussion and explanation

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Refraction through prism is treated as sum of refractions at the two surfaces. A ray of light along L1, L2 is incident at O1 on AB. It is refracted along O1 O2 in the prism and emerges along O2Q as emergent ray after two refractions at O1 and O2.

Refraction at O1: PO1 is a ray of light along L1,L2 and incident at angle i1 on the refracting surface AB. It bends towards the normal in the prism and follows the path O 1O2.

PO1R = i1 and O2O1M = r1 Deviation of the ray at the face AB is,

d 1 = i 1 – r1

From Snell’s law, aμg = Sin i1 / sin r1 Refraction at O2 : The ray O1O2 is incident at an angle r2 on the second boundary inside the prism. The ray travelling from glass to air (denser → rarer) bends away from the normal.

Angle of incidence = MO2O1 = r2 Angle of refraction = angle of emergence = QO 2S = i2 Deviation at second boundary, d2 = r2 – i2 Refractive index, gμa= Sin r2 / sin i2 aμg

=

Sin i2 / sin r2

aμg

calculated from refraction at the two surfaces is same.

Deviation: The deviation produced by the prism = angle between incident ray and emergent ray d =----This is same as the sum of deviations at the two boundaries is d 1 +d2. d  

= d1 + d2.

Prism and slab are both the optical media bound by two plane surfaces. A ray of light is refracted on entering the prism or slab and is refracted again while coming out.

The two refracting faces are parallel for a slab and the two refracting faces are inclined for a prism.

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 

 

In a prism the deviation of the ray is in the same direction at the two boundaries. d = d 1 + d2 In a slab the deviation of the ray at the first boundary is opposite to the deviation at second boundary. d = d1 – d2 and d1 = d2 d = 0 The incident ray, refracted ray and emergent ray are retraceable or reversible.

The refractive index Sin i1 / r1 aμg = r1 = sin i1/ aμg

For a given i1 the angle of refraction (r1) depends on refractive index.

Therefore, deviation depends on i) Angle of the prism ii) Angle of incidence at the first boundary iii) Nature of material of prism iv) Nature of the surrounding medium v) Wavelength of light ray.

KEY MESSAGES 

A ray of monochromatic light on passing through a prism is deviated from its path and for a given prism this deviation changes with angle of incidence, wave length and refractive index (RI)

LEARNING CHECK: 1. A ray incident at say, 500 on one face deviates through 220 on emergence. Will the deviation be same if the prism is immersed under water?

No. it will be lesser in water than in air.

INTERESTING INFORMATION 

What do you observe if a narrow beam of white light is passed through the prism?

White light is a composite light, a mixture of several colours or wavelengths. When they fall at the boundary AB, light rays of different colours suffer different deviation after refraction. Red suffers least and violet suffers maximum deviation. The emergent light from AC forms a spectrum of colours. This is dispersion of lights. The dispersion is due to different colours of light travelling with different velocities in the prism (V R> VG>VV).

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ABL 4 Total Internal Reflection ACtivity

Learning Objective

4.1

To understand the phenomenon of total internal reflection.

ď&#x201A;ˇ

4.2

Understanding the working of few devices based on total internal reflection.

ď&#x201A;ˇ

Key message A light ray incident at a certain angle greater than critical angle on the boundary in a denser medium does not show refraction. It suffers total reflection inside.

The optical devices discussed make use of total internal reflection where reflection is 100%.

Total Time

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Time 25

25

50 min

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ABL4.1 LEARNING OBJECTIVE: To understand the phenomenon of total internal reflection .

ADVANCE PREPARATION Material List S.no 1

Material Critical angle model

Required Quantity 1 per class

2

Laser light

2 per class

3

Semicircular glass slab

1 per class

Things to do: Not Applicable Safety precautions: Not Applicable

SESSION Link to known information/previous activity: We have learnt earlier that a ray of light incident obliquely on a boundary separating two optical media is refracted or bent while it travels into second medium. It is shown here that under certain circumstances the light ray may get reflected without being refracted.

Procedure: Arrange the critical angle model on the table. Place the semicircular glass slab on the rotating table disc. The centre of the straight edge is at the centre of the table and the edges along 900 - 900line of circular scale.

Rotate the laser source arm to a convenient position and note the angle, which the light ray from laser source makes with normal (0 â&#x20AC;&#x201C; 0 line). On entering the curved surface along the radial line the light ray travels without deviation and falls on glass-air boundary. The refracted ray emerging from plane surface produces a bright spot on a screen. Notice a small part of light is reflected at the boundary.

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Fig 6: Total Internal Reflection through semicircular slab Gradually rotate the turntable and increase the angle of incidence. The refracted ray bends more towards plane boundary of the slab. Continue till the refracted ray just grazes the plane surface and not get into air. To confirm the exact position, give a slight rotation back words the refracted ray spot appears in air and if rotated other way the entire light is totally reflected.

Measure the angle between incident/reflected light ray and normal (0 -0 line)

UNDERSTANDING THE ACTIVITY Leading questions 1.

2.

3.

4.

5.

6.

7.

8.

9.

What is the optical phenomenon proposed for demonstration?

Why is it related to refraction?

The boundary of the slab where the light gains entry into it is made semicircular. Why? What do you observe when the angle of incidence at glass-air boundary is continuously varied from 00 ?

What is critical angle?

Why is the reflection of light at (i>c) called total internal reflection?

What are the conditions for total internal reflection?

Is the critical angle dependent on the nature of light and the kind of the media on either side?

Does total internal reflection obey the usual laws of reflection? Then how is this reflection different from the reflection from mirror?

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Discussion and explanation 

  



 





The phenomenon observed here is called total internal reflection. It occurs when ever i) light travels from denser to rarer medium and ii) When light is incident at an angle greater than critical angle at the boundary.

The light from air incident along the radial line ( i = 900 )on semicircular surface is not deviated and travels without being bent towards the glass-air boundary.

Initially when the incident ray is set for an angle ( say20 0 or 250 ) one can see strong refracted ray and weak reflected ray. This is the usual reflection at any smooth boundary.

On gradually increasing the angle of incidence you can notice the refracted ray bending more and more towards the boundary of separation. At some point when the angle of incidence is just equal to (a limiting angle ) critical angle the refracted ray is seen travelling along or grazing the boundary.

Mathematically, when i = c the angle of refraction r = 900. On further increasing the angle of incidence by rotating the turn table in the same direction there is no refraction because angle r cannot be greater than 900 in air. The light ray that is incident at i>c, is totally reflected without refraction. All the incident light is reflected into the medium inside. We say that the incident light is totally internally reflected.

Total internal reflection is the phenomenon in which a ray of light travelling from denser to rarer medium is totally reflected into the medium when incident at an angle greater than critical angle.

The behavior of light at the boundary depends on the nature of the light used and the kind of media on either side of the boundary. In other words the critical angle depends on refractive index of the pair of media and the wavelength of light.

It is naturally expected that ‘c’ is different i) Even when one or both media are changed and ii) When wavelength of light is changed. It is found that Sin C= 1/μ ‘c’ decreases as μ increases or decreases.

Total internal reflection for all purposes is normal reflection (when refraction is absent) and light reflected is 100%. Even a best plane mirror has 90 – 95% reflection. So totally reflection prisms are superior to plane mirrors.

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KEY MESSAGES 

A light ray incident at a certain angle greater than critical angle on the boundary in a denser medium does not show refraction. It suffers total reflection inside.

LEARNING CHECK: Ask learners to list the key things they have learnt. Guide them to the key messages listed and then put up the chart of key messages. If you have time during the class, make up a small game, quiz or match the following as a learning check. This may have to be done as part of advance preparation.

INTERESTING INFORMATION: 

In the figure ABCD is a non-homogenous optical medium. The refractive index decreases from CD to BA. A ray of light enters the medium from the lower surface. It is continuously going from region of larger to smaller refractive index region (denser to rarer)and hence keeps bending away all the while.

At some place when i > c it gets internally reflected. Such effects seen near Polar Regions during winter are the cause for ‘looming’.



Let us reverse the situation where the refractive index increases on moving towards AB. A ray of light coming down is totally internally reflected and moves up. This is the principle of mirages.

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Time: 20 min

ABL4.2

LEARNING OBJECTIVE: Understanding the working of few devices based on total internal reflection. a) Total reflection prisms b) Optical fibers.

ADVANCE PREPARATION Material List S.no 1

Required Quantity 1 per group

2 3

Material Right angular equilateral prism Laser light A4 sheets

4

Pencil

1 per student

5

Equilateral prism

1 per group

6

Optical fiber toy model

1 per class

2 per group 1 per student

Things to do: Not Applicable Safety precautions: Not Applicable

SESSION Activity 1 a) Total reflection prisms A right-angled isosceles glass prism (Îź = 1.5) is called a totally reflection prism. Its base angles are each 45 0 . It can be used to deviate a ray of light like a plane mirror.

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i)

ii)

0

A ray of light through (ray simply turned through 90 ) two pins (L1, L2) incident normally on PQ travels without deviation and falls on PR at angle i = 45 0 . The critical angle for glass is 420. Since i>c, the ray is internally reflected and falls normally on QR. You can fix pins L3 and L4 to be collinear with images of L1 and L2. The straight line through L1 and L2 is normal to straight line through L3 and L4.

Chart 5: Total Internal Reflection in right angled prism Parallel beam passes through the prism as parallel but inverted.

Chart 6: Parallel beams passing through prism In Chart 6,1 and 2 are two rays of parallel beam entering a prism through PQ. After refraction they are incident on the base at an angle greater than the critical angle. They are reflected totally, incident on boundary PR and emerge from PR as passed beam. The rays 1 and 2 come out as 2 and 1. The prism thus produces inversion without change in size. iii)

Reversal of direction of light:

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The light rays 1 and 2 from an object are incident normal on the face QR. They enter the prism without deviation and are incident at 450on the face QP ( i>c). The rays are totally internally reflected from PR. They fall normally on QR and come out parallel to incident beam.

The incident beam is reflected back along the same direction. Image of an object is reversed.

Such prisms are used in binoculars, periscopes, view finders, cameras.

Fig 7: Total Internal Reflection in Prism Reflection prisms find applications in Binoculars, periscopes . . . . . . . . . ..

Activity 2b) Optical fibers: It is a thin hair like transparent medium, which is flexible covered by a coating/ cladding of smaller refractive index. Hundreds of such fibers bundled form a cable.

A ray of light passing close to the axis after refraction at one end is incident on the boundary at angle greater than critical angle. It is totally internally reflected and falls on the opposite boundary. Since the medium has same diameter it undergoes series of total internal reflections and emerges at the other end. Such long cables can be used to transmit light that carries information.

Optical fibers have revolutionized the field of communications. They are also used in medical diagnosis to view the interior of intestine, throat etc.

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Fig 8: Optical Fibre

KEY MESSAGES  The optical devices discussed make use of total internal reflection where reflection is 100%.

LEARNING CHECK: Ask learners to list the key things they have learnt. Guide them to the key messages listed and then put up the chart of key messages. If you have time during the class, make up a small game, quiz or match the following as a learning check. This may have to be done as part of advance preparation.

INTERESTING INFORMATION: 

On a rainy day with sun behind, if you look into the sky sometimes you get a chance to see rainbow. It is due to dispersion and total internal reflection of light entering water drop the viewing angle is around 420.



A diamond sparkles brilliantly because of the multiple total internal reflections and dispersion of light.

If we partially dip the diamond in water the brilliant sparkling is very much reduced.

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ABL 5 Refraction through curved surfaces Activity

Learning objective

5.1

Demonstration of refraction at curved surfaces

5.2

Drawing ray diagrams for images formed by convex lens.

Key messages

ď&#x201A;ˇ

Refraction at the curved boundary produces images, which are magnified, real or virtual and erect or inverted, only in the plane of curvature.

Time (min)

30

15

Total Time

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45 mins

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Note to Instructor: Please read the following information before performing the following ABLâ&#x20AC;&#x2122;s.

REFRACTION AT A SPHERICAL SURFACE When a parallel beam of light is incident on a plane boundary, it emerges as a parallel beam in the other medium (boundary) with each ray incident at the same angle being bent through the same angle. If a parallel beam of light is incident on a curved surface, different rays incident at different angles suffer different deviations. As a result the rays either bend inwards or outwards. We then speak of focusing action of the surface â&#x20AC;&#x201C; convergence or divergence. Nevertheless, the laws of refraction, which holds good for a plane boundary, remain the same even for a curved boundary because every ray is incident only at a point on the boundary. This focusing action can be noticed in human eye and is also the basis of working of several optical instruments. Let us limit our study only to the spherical boundaries.

Defining the parameters, characteristics for a spherical boundary Spherical boundary is a part of a sphere and separates two optical media.

Centre of curvature (C): The center of the sphere of which the spherical boundary is a part is called the center of curvature.

Pole (P): It is the Centre of the spherical surface.

Principal axis: It is a straight line passing through the centre and pole of the spherical surface.

Radius of curvature (R): It is the distance between the pole and the center of curvature.

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INTRODUCTION: Imagine now a situation where a parallel beam of light is incident on a spherical boundary (convex) parallel to the principle axis. The ray incident close to the principle axis (I is small) suffers a smaller deviation. On moving away from the principle axis, the angle of incidence for each ray increases resulting in the increase in angle of refraction. The refracted rays thus come to common focus F on the principle axis.

If the surface is concave each ray bends towards normal. Again the bending increases as we move out ward.

For a convex surface is, all the rays diverge after refraction and appear to originate from a focal point F on the principle axis in the same medium. This is true when medium 1 is rarer and medium 2 is denser. The situations get interchanged when medium 1 is denser and medium 2 is rarer. Thus a spherical or curved surface has ability to render the incident rays convergent or divergent.

This ability to converge or diverge a beam of light incident on a boundary separating two media is called the power of the surface. If R is the radius of curvature and μ 1 and μ2 the refractive indices of separated media.

Power

= μ2 – μ1/ R

It is expressed in unit diopter, where R is in meters. *If an object is passed at a distance u from the pole in one medium (μ 1) and image is formed in the other(μ2) at a distance v, then (μ2/v) – (μ1 /u) = (μ2 - μ1)/R Agastya International Foundation. For Internal Circulation only.

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Where R is the radius of curvature of the boundary

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LENSES: A lens is a transparent medium bounded by two surfaces of which at least one is spherical. If lens is thick in the middle and thin at the edges it is called a convex lens. If the lens is thin at the middle and thick at the edges, it is called concave lens. There are several variants of these lenses.

1. Biconvex lens; Lens whose both the surfaces are convex.

2. Biconcave lens; Lens whose both the surfaces are concave

3. Plano-convex lens; Lens whose one surface is convex and the other is plane.

4. Plano-concave lens; Lens whose one surface is concave and the other is plane.

5. Concavo-convex lens; Lens whose one surface is concave and the other is convex.

A beam of light passing through a lens is refracted twice, once at each of the two boundaries. The power of the lens (P) is sum of power of the two bounding surfaces.

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Definitions related to lenses; Centre of curvature: The center of the sphere of which the spherical boundary is a part is called the center of curvature. There are two centers of curvature (C1,C2).

Radius of curvature: It is the distance between the pole and the center of curvature.

Principal axis: It is a straight line passing through the centre and pole of the spherical surface.

Optic centre : The geometrical center of a lens (on along of the optic axis) is called optic center.

Principal focus: The point of common focus on the principal axis is called the principal focus F Focal length:

The distance between the optic centre and principal focus (f).

Convex lens 1. Consider a beam of light incident parallel to principal axis on surface.

After refraction through the two surfaces the rays meet at a common point on the principal axis on the other side. This point of common focus is called the principle focus F 1.

2. Similarly a parallel beam of light

Concave lens 1. Consider a beam of light incident parallel to principal axis on the first surface. After refraction through the two boundaries the rays form a divergent beam that appears to originate from a common point on the principal focus on the same side. This point of common divergence is called the principle focus (F1).

2. If the beam enters from the other side

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incident on the second boundary from right comes to focus in front of the first boundary on the principle axis (F 2).

3. The 2 points of focus on either side are at equal equidistance from the lens.

The distance of principle focus from the optic centre is called focal length of the lens (f).

there is another principal focus (F2) on that side of the lens.

3. The 2 focal points are equidistance from the lens. The distance of either of principal focus from the lens (optic centre) is called focal length (f).

f is +ve for convex and –ve for concave lenses.

Focal length of a lens depends upon i) Radius of curvature of two bounding surfaces and ii) The refractive indices of the material of the lens and that of surroundings.

1/f = (1μ2- 1)(1/R1 + 1/R2)- Lens maker’s formula 1/v – 1/u = 1/f P

= 1/f

------------ relation between u, v and f

------------ Power

Sign conventions and rules for formation of images.

Sign convention: i) All distances are measured from the optic centre. ii) The distances measured along the directions of incident light are +ve. iii) The heights measured above principal axis are +ve The heights measured below principal axis are –ve.

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Consider rays of light emanating from the object and passing through the lens. The point where the refracted rays converge gives the position of real image. If the refracted rays appear to diverge from a point, it gives the position of virtual image. To fix the position of image consider the point of convergence or divergence of any two of these following rays after refraction. i) ii) iii)

A ray from the object travelling parallel to the principal axis after passing through the lens either passes through F or diverges from F. A ray of light from the object passing through F after refraction through the lens emerges parallel to the axis.

A ray from the object passing through, optic centre emerges through the lens without deviation.

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ABL5.1

Time: 30 min Time: 25

LEARNING OBJECTIVE: Demonstration of refraction at curved surfaces ADVANCE PREPARATION Material List S.no 1

Material A tank with cylindrical shape, one side plane and another side cylindrical

2

Water

3 A small Wooden or metal strip on which an arrow crosses is painted.

Required Quantity 1 per class

1 per class

Things to do The experiment can be demonstrated using a 400 ml glass beaker Safety precautions Not Applicable

SESSION Link to known information/previous activity: In ABL 1, 2 you have seen refraction at plane surfaces, now let us see, refraction through curved surfaces.

Procedure Place the empty tank on the table so that it is in proper view of all the students in the class. Ask one student to hold the object (crossed arrow) right behind the plane surface. Call another student and ask him to look at the object through the tank and describe how the object is seen by him.

Possibly the student may say, the sharp tips of the cross are, one pointed to the left and another upwards.

Call one more student to the table and instruct him to fill the tank with water without spilling on the working table.

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Call the second student to come to the model again and view the cross-mark through water. Let him describe as to how the cross-mark appears. What does he observe if the cross is gradually moved away from the tank without disturbing the line of sight? Let him record his observations on a work sheet.

Carry on this exercise for the benefit of other students in the class. When all the students complete observations instruct them to sit in groups and lead the discussion to understand the science behind the observations.

Fig 9: Refraction through curved surface

UNDERSTANDING THE ACTIVITY Leading questions 1.

2.

3.

4.

How does the object appear when viewed directly in air? Or through the empty tank?

What have you created after you have filled the tank with water?

How does the object appear when viewed through the water medium?

What changes do you notice in the appearance of the image on moving the object farther from the tank?

5. What changes do you expect in the nature and size of the image, if the tank were to have curved boundary (convex out wards) on both the sides?

6. What would be the situation if the surface has uniform curvature in all the directions?

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Chart 7: Refraction through a curved surface at different positions Discussion and explanation 1. The object, which is a double cross-arrow looks the same when viewed directly in air or through the empty tank. Though the walls of tank are made of glass, the thickness being very small, the light rays from the object most of the time travel in air. The effect of refraction through the thin glass walls is safely ignored.

2. On filling the tank with water we have created a water medium that has a plane boundary on one side and cylindrical boundary at the other. The light rays first pass through plane boundary and get refracted. The rays are just bent into water medium. The plane boundary neither has converging or diverging power. If the other surface were also to be plane the refracted rays in water would simply be deviated at the second boundary on emergence into air. It is like viewing the object through a rectangular water medium where there is only a shift in the direction of light. The image looks same as the object without a change in size or nature of appearance.

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3. When the second boundary is curved (in horizontal plane in case of cylindrical wall) the rays after refraction at the boundary either converge or diverge. The image formed may have size and nature different from that of the object. These changes are noticed only in the plane of curvature.

4. When the object is held close to the tank the image appears only enlarged but erect. The enlargement is only in the horizontal plane. On gradually moving the object away the image becomes larger and larger in size but still remains erect. After crossing certain position the image suddenly becomes inverted (horizontal cross is reversed) without any change in the vertical direction. On moving the object farther, the nature of the image remains the same but only (horizontally) becomes smaller and smaller in size.

5. If the boundary is curved on both sides, the effects of refraction at the two boundaries get compounded. The change over from erect to inverted image occurs at a much nearer distance than earlier and the magnification would be much larger.

6. When the curvature is uniform in all directions, the changes in size and inversion would take place in horizontal as well as vertical directions. The water medium behaves as a double convex lens.

KEY MESSAGES ď&#x201A;ˇ

Refraction at the curved boundary produces images, which are magnified, real or virtual and erect or inverted, only in the plane of curvature.

LEARNING CHECK: Ask learners to list the key things they have learnt. Guide them to the key messages listed and then put up the chart of key messages. If you have time during the class, make up a small game, quiz or match the following as a learning check. This may have to be done as part of advance preparation.

ABL 5.2 Time: 15 min

LEARNING OBJECTIVE: Drawing ray diagrams for images formed by convex lens.

ADVANCE PREPARATION Material List Agastya International Foundation. For Internal Circulation only.

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S.no 1

Material Compass

Required Quantity 1 per group

2

Scale

1 per group

3

Pencil

1 per group

4

Eraser

1 per group

Things to do: The diagrams have to be drawn by each student in their notebooks. Initially for the first couple of diagrams, the instructor will have to go round the classroom to check if the students have done it correctly. Once they get the expertise of making the basic diagrams, it will be easier for them. Insist that they practice the diagrams a few times on their own at home.

Safety precautions: Not Applicable

SESSION Link to known information/previous activity In the previous activity we saw how and where the images are formed. Now we will learn how to draw the ray diagrams.

Procedure: Draw the convex lens and principle axis on the board. With inputs from the students, mark 2F, focus (F) and optic centre of the lens. Now take position of the object one at a time. To find the position of the image, trace the paths of at least two incident rays and find their points of intersection or divergence after reflection from the lens.

Case I: (object at infinity) 1. Draw the object (in the shape of an arrow) facing upwards at infinity. Let the object be represented by A B. Here A is the head and B is bottom of the object.

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2. Draw two parallel lines above and below the principle axis. These two rays after refraction from the lens will meet at focus (F). Draw arrow marks to the incident and refracted rays to indicate the

direction of travel.

Chart 8- Image formation by convex and concave lens when object is placed at infinity Case II: (object is beyond at 2F) 1. From a point A draw an incident ray parallel to principle axis. After refraction it passes through F on the other side. Draw another incident ray from A. Passing through O. this ray is undeviated after refraction.

The two refracted rays intersect at A’. Draw A’B’. This is the real image. It is real inverted, diminished and between F and 2F.

Chart 9- Image formation convex lens when object is placed beyond 2F Case III: (object is at 2F) 1. Draw a ray parallel to the principle axis from A. This ray when refracted will pass through focus (F) of the lens. Now draw another incident ray from A through the optical centre of the lens. This ray travels un deviated and meets the first refracted ray. Draw A’B’. this is the image. The image is real, inverted, of the same size as object and formed at 2F.

Chart 10- Image formation by convex lens when object is placed at 2F Agastya International Foundation. For Internal Circulation only.

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Case IV: (object is between F and 2F) 1. Draw a ray parallel to the principle axis from A. This ray when refracted will pass through focus (F) of the lens. Now draw another incident ray from the top of the object through the optical centre of the lens. This ray travels un deviated and meets the first refracted ray. Draw A’B’ perpendicular to principal axis. This is the image of AB. The image is real, inverted, enlarged and formed beyond 2F.

Chart 11- Image formation by convex lens when object is between F and 2F Case V: (object is at F) 1. Draw a ray parallel to the principle axis from A this ray when refracted will pass through focus (F) of the lens. Now draw another incident ray from through the optical centre (O) of the lens. This ray travels un deviated and two refracted rays do not meet. They travel parallel and form the image at infinity.

Chart 12- Image formation by convex lens when object is placed at F Case VI: (object is between F and O) 1. Draw a ray parallel to the principle axis from A, this ray when refracted will pass through focus (F) of the lens. Now draw another incident ray from the top of the object through the optic centre (O) of the lens. The two refracted rays diverge and when produced backwards appear to be coming from the point A’ behind the lens. Draw A’B’. This is the image position. The image is virtual, enlarged, erect and behind the object.

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Chart 13- Image formation by convex lens when object is laced between F and O Concave lens Draw a ray (1) from the point A, parallel to principal axis. After refraction it appears to diverge from F. draw another ray (2) through incident along optic centre (O). It passes through un-deviated. The two refracted rays appear to diverge from A’. Draw A’B’. This is the image of AB. It is erect, virtual, diminished and always between O and F on the same side.

Chart 14- Image formation by concave lens

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ABL 6Convex Lens ACtivity

Learning Objective

6.1

To study the nature of the image formed by a convex lens for different positions of the object.



6.2

Measurement of focal length of a convex lens by u–v method.



Time

Key Message Like a spherical mirror, the image formed by a lens can be magnified or diminished, erect or inverted and real or virtual, depending on the object position.

The focal length of a lens is measured by measuring the object distance and real image distance from the lens.

Total Time

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30

60 mins

Time: 30 min

ABL 6.1

58

LEARNING OBJECTIVE: To study the nature of the image formed by a convex lens for different positions of the object.

ADVANCE PREPARATION Material List S.no 1

Material Candle

Quantity 1 per group

2

Match box

1 per group

3

Screen

1 per group

4

Lens stand

1 per group

5

Convex lens

1 per group

6

Chart paper

1 per group

7

Scale

1 per group

8

Sketch pen

1 per group

9

Pencil

1 per group

10

Eraser

1 per group

Things to do: This activity will have to be demonstrated by the instructor first and then the students can be allowed to do it on their own in groups of 6 to 10. The instructor should insist that the students take down the readings. This will help them to understand the concept better.

Safety precautions: The mirror, candle and screen should be handled with care.

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Link to known information/previous activity We know that the light is refracted at spherical surfaces. Now we will see how and where the images are formed when light is refracted at two boundaries, i.e., a lens.

Procedure: Secure the white chart paper on the table. Draw a straight line in the middle of the paper along the length.

This line will be the principle axis of the lens. Mark a point ‘O’ on the line in the middle. Mount a convex lens on the lens stand and keep the lens stand exactly at the point ‘O’. Now keep a screen on one side of the lens and adjust the screen such that a clear image of the distant object is formed on the screen.

Measure the distance between the lens and the screen. This approximately gives the focal length of the lens (distant object method). Now mark the positions of F and 2F on either side of the lens placed at ‘O’.

Light a candle and place it on the principle axis at a large distance. Adjust the position of the screen to get a sharp image. Now observe position of the image formed, nature and relative size of the image. Note down the characteristics of the image in the tabular column.

S.no

Position of object

Position of image

Size of image

Nature of image

Worksheet 1- Observations for image formation by convex lens Place the object beyond 2F i.e. at a distance more than 2F from the lens. Mark this point. Now move the screen on the other side of the lens along the principle axis such that a clear image is formed on the screen.

Note down the position, size and nature of the image in the tabular column.

Now, place the object at the point 2F. Note down the position, nature and relative size of the image in the tabular column.

Repeat this activity by placing object between F and 2F, at F and between F and O. Note down the size and nature of the image in the tabular column in each case. When the object is at F the image is formed at infinity ( ) and cannot be caught on the screen.

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Fig 10- Set up for ABL 3.2b

UNDERSTANDING THE ACTIVITY Leading questions 1. Where is the image seen when the object is kept beyond 2F? 2. What kind of image is this?

3. Where is the image seen when the object is kept at 2F? 4. What kind of image is this?

5. Where is the image seen when the object is kept between 2F and F? 6. What kind of image is this?

7. Where is the image seen when the object is kept at F? 8. What kind of image is this?

9. Where is the image seen when the object is kept between the O and F? 10. What kind of image is this?

Discussion and explanation Let us take each of these positions of the object one at a time.

1. When the object is placed beyond 2F, the image is formed between F and 2F. The image is captured on the screen. So this is a real image (only real images can be captured on screen). This image is inverted, and smaller than the object.

2. When the object is placed at 2F, the image is formed at 2F. The image is a real image. The image is inverted, and is of the same size as the object.

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3. When the object is placed between F and 2F, the image is formed beyond 2F. The image is real, inverted and larger than the object.

4. When the object is placed at F, an image is formed at infinity. The image is real, inverted and larger than the object.

5. When the object is placed between F and the lens, the image cannot be caught on screen. So it is a virtual image. The image is seen to be formed on the same side of the lens. This image is erect and larger than the object.

This information can be referred in the following table: S.no

Position of object

Position of image

Size of image

Nature of image

1

At infinity

At F

Very much diminished

Real and inverted

2

Beyond 2F

Between F and 2F

Diminished

Real and inverted

3

At 2F

At 2F

Same size

Real and inverted

4

Between F and 2F

Beyond 2F

Magnified

Real and inverted

5

At F

At infinity

Highly magnified

Real and inverted

6

Between lens and F

Same side of the object

Magnified

Virtual and erect

Chart 15- Nature of image formed by convex lens when object is at different positions

KEY MESSAGES: ď&#x201A;ˇ

Like a spherical mirror, the image formed by a lens can be magnified or diminished, erect or inverted and real or virtual, depending on the object position.

LEARNING CHECK: Ask learners to list the key things they have learnt. Guide them to the key messages listed and then put up the chart of key messages. If you have time during the class, make up a small game, quiz or match the following as a learning check. This may have to be done as part of advance preparation.

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Time: 30 min

ABL 6.2 LEARNING OBJECTIVE: Measurement of focal length of a convex lens by uâ&#x20AC;&#x201C;v method ADVANCE PREPARATION Material List S.no 1

Material Ray pad or optic box

Quantity 1 per group

2

Convex lens

1 per group

3

Concave lens

1 per group

4

LED torch

1 per group

5

Junction box

1 per group

6

Screen

1 per group

Things to do Not Applicable Safety Precautions Not Applicable

SESSION Link to known information/previous activity In the earlier activities we have seen the nature of images formed by a convex lens. Now we are going to actually measure the focal length of the lens.

Procedure

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Stretch a drawing sheet on the table, draw a straight line and mark a point ‘O’ in the middle. Mount the given convex lens on the lens stand and place the stand at ‘O’. Place a screen on one side and try to catch a distinct image of a distant object formed by the lens. The distance between lens and image screen gives roughly the focal length (f).

Mark ‘F’ and 2F on either side of ‘O’ on the straight line drawn. Place the lighted candle at a position beyond 2F. Measure its distance from the lens as u. Adjust the position of the screen on the other side to get a distinct, bright image on the screen. Measure the distance of image screen as v. calculate the focal length.

Next place the object at 2F and adjust the position of the screen to obtain bright and distinct image on the screen. Measure u and v. calculate f the focal length again. Next place the object between F and 2F and adjust the screen to obtain an enlarged distinct image. Measure u and v. calculate focal length f. f = (u v)/u + v Calculate the mean of the three measurements of focal length.

UNDERSTANDING THE ACTIVITY Leading questions 1. Why do we measure the approximate value of the focal length of the convex lens by distant object method before starting the experiment?

2. What is the position and nature of the image in each case?

3. Why did you not try to measure f by keeping the object between P and F? 4. Is focal length the same when calculated for different positions?

5. On what factors does focal length depend upon?

6. How does the size of the image vary on moving the object nearer and nearer to the lens?

Discussion and explanation 1. A convex lens can form real or virtual images for different positions of the object. u>f real image u<f virtual image.

2. Since virtual image cannot be caught on a screen it is advised to measure f approximately so that you can mark F, 2F and keep u always greater than f. The image in all the cases is real and inverted.

When i) u > 2f, the image is diminished ii) u > f and u < 2f, the image is enlarged and inverted. u = 2f, the image and the object are of same size.

3. In all, the size of the image increases as the object is brought nearer to the lens. It is infinity when u = f.

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4. When the object is placed between O and F the image is virtual and cannot be caught on a screen.

But if we look into the lens from the other side one can only see an enlarged and erect image of the object behind the screen.

5. The focal length of the lens is the same for all the settings because it is a constant and characteristic of the lens. The value of f depends on the radii of curvature of the two surfaces and the relative refractive index of the material of the lens with respect to that of the surroundings.

S.no

Distance of candle from lens ‘u’

Distance of the image from lens ’v’

Focal length f = uv/u+v

Mean f = ……………….m Worksheet 2- Observations for calculating focal length of convex lens

KEY MESSAGES 

The focal length of a lens is measured by measuring the object distance and real image distance from the lens.

LEARNING CHECK: 1. From the lens maker’s formula 1/f = (μ – 1)(1/R1 +1/ R2) calculate f if the material of the lens is glass ( μ = 1.5) and the two surfaces are of radius 40cm each.

1/f = ( 1.5 – 1) 2/40 = 0.5/20 = 1/40 f = 40cm 2. What is the power of convex lens of f= 0.25 m. The power, P = 1/f = 1/0.25 = 4diopter.

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INTERESTING INFORMATION 1. Suppose a lens has a focal length 20cm in air. It means a distant object can be focused to a point 20cm distant from it.

If the surrounding medium is water (any other medium optically less dense than glass) then relative refractive index of glass with respect to water (if water) is 1.125. Then f = 80cm. The distant object is focused now at a very large distance. Same lens has focal length 20cm in air and 80cm in water.

If we increase μ of surrounding medium f increases its power decreases. When μ of lens is equal to that of surroundings, f = . The rays pass through without bending.

2. If the lens material has refractive index lesser than that of the surroundings, f becomes –ve. The convex lens in a denser optical medium behaves as divergent lens.

A concave lens surrounded by medium of larger μ acts as convex lens.

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ABL 7 Magnification of convex lens; focal length of combination of lenses ABL

Learning Objective

Key Messages

7.1

Measurement of magnification produced by a convex lens.

o

It is possible to get desired magnification by fixing the object distance (m = v/u).

30

7.2

To determine the focal length of a combination of two lenses.

ď&#x201A;ˇ

It is possible to combine two or more thin lenses to get a lens system of required focal length.

30

Total Time

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Time

60 mins

Time: 30 min

ABL 7.1 LEARNING OBJECTIVE: Measurement of magnification produced by a convex lens ADVANCE PREPARATION Material List S.no 1

Material LED torch

Required Quantity 1 per group

2

Graph sheet

1 per student

3

A4 sheet

1 per group

4

Sketch pens

4 per group

5

Cello tape

1 per group

6

Scissor

1 per group

7

Scale

1 per group

8

Pencil

1 per student

9

Eraser

1 per group

10

Convex lens f=10 cm

1 per group

11

Torch stand

1 per group

12

Lens stand

1 per group

13

Screen

1 per group

Things to do: Take a powerful LED torch. Cover it with a paper cut in the form of an arrow and stick it. This torch with the arrow is our object instead of the candle flame. Measure the height of the arrow mark (h 0).

Fix a graph sheet on the image screen to help you measure the height of the image.

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Safety precautions: Not Applicable

SESSION Link to known information/previous activity In the previous activity, we found the focal length of a convex lens. Now let us measure the magnification.

Procedure: Measure the focal length of the lens by distant object method, mark F and 2F on either side of the lens holder, on a straight line drawn (explained in 3.3a).

Mount the torch on a stand and place it on one side of the lens at a distance u > 2f. Adjust the position of the image screen till a distinct and bright image is formed on the screen. Measure the distance of the image screen (v) as well as the height of the image (h 0).

Repeat the trial for other distances u = 2F, u between f and 2f. In each case measure the height of the image.

Enter your measurements in the tabular column.

Calculate magnification using formula m = v/u and m = hi/h0.

S.no

Distance of object from lens ‘u’

Distance of the image from lens ’v’

Height of the image ‘hi‘

Height of the object ‘h0‘

Magnification m = v/u or hi/h0 m = v/u

m = hi/h0

1 2 3 4 5 Worksheet 3: Observations for calculating magnification of a convex lens

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UNDERSTANDING THE ACTIVITY Leading questions 1.

2.

3.

4.

5.

What is magnification?

Does the magnification remain the same or change?

For which position is magnification ‘1’? What does it mean?

For which positions is magnification greater than 1 and less than 1?

Are the relations hi/h0and v/u the same for magnification?

Discussion and explanation 





 

When light rays from an object are passed through a different medium there is refraction. If the bounding surfaces are curved, the image formed is real or virtual, enlarged or diminished depending on the object distance from the lens.

If hi is the height of the image formed then h i/h0 is magnification (m).

If hi>h0, m > 1 for magnified image, hi< h0, m < 1 for diminished image, hi = h0 m = 1 for image and object of same size.

It is geometrically established that v/u = hi/h0= m magnification is calculated by both the formulae.

On bringing the object nearer, v/u increases. Therefore magnification gradually increases, becomes 1 (at 2F) and then becomes greater than 1.

If u is very close to F, the real image becomes so large that it is difficult to focus.

KEY MESSAGES 

It is possible to get desired magnification by fixing the object distance (m = v/u).

LEARNING CHECK: Ask learners to list the key things they have learnt. Guide them to the key messages listed and then put up the chart of key messages. If you have time during the class, make up a small game, quiz or match the following as a learning check. This may have to be done as part of advance preparation.

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Simple microscope: A convex lens because of the magnifying property is used as a reading lens or magnifier, that is a simple microscope.

Compound microscope: It has two convex lenses. One lens called objective gives a magnified image of the object. This image becomes object of another convex lens that forms a further magnified final image.

Telescope: It also has two lenses of suitable focal length. Here also image formed by one lens becomes object for the second lens. But a telescope helps to bring the distant object nearer.

ABL 7.2

Time: 30 min

LEARNING OBJECTIVE: To determine the focal length of a combination of two lenses.

ADVANCE PREPARATION Material List S.no 1

Material Concave lens f 40 cm

Required Quantity 1 per group

2

Convex lens f 10 cm

1 per group

3

Chart

1 per group

4

Sketch pens

4 per group

5

Cello tape

1 per group

6

Scissor

1 per group

7

Scale

1 per group

8

Pencil

1 per student

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9

Eraser

1 per group

10

Lens stand

1 per group

11

Screen

1 per group

12

Torch stand

1 per group

13

LED torch

1 per group

Things to do: Choose the two lenses such that their focal lengths differ appreciably. Otherwise it may become difficult to focus the images at distances within the size of the table. Focal length of the convex lens should be much lesser than that of the concave lens. ( f1 – the focal length of convex lens around 10 cm and f2 – the focal length of concave lens around 40 cm) Safety precautions: Not Applicable

SESSION Procedure: Keep the given two thin lenses in contact and if necessary secure them together with an a adhesive tape.

Arrange the lens system on the lens holder.( convex f1 and concave f2) Stretch a white sheet on the table and draw a line. Mark a point ‘O’ on the line and place the lens system at ‘O’.

Mount a torch with arrow (as object) on the holder and keep it at a distance of about 40cm to 50 cm from the lens. Bring the screen on the other side and adjust its position till a distinct and bright image is formed.

Measure the distance of torch from lens as u and that of image screen as v. Repeat this trial for another two positions (as explained in 3.4). Enter the data into tabular column. Calculate the focal length of the system using the relation f = uv/u + v.

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Leading questions 1. Does your lens system behave as converging lens or diverging lens?

2. Are you sure the system behaves as a ‘single lens’?

3. Does your result agree with the focal length calculated from theoretical formula Fe = f1f2/f1 – f2?

4. Is it possible to combine more than 2 lenses?

5. Why did we are not use concave lens of focal length smaller than the focal length of convex lens?

Discussion and explanation  





When two thin lenses are combined the power of the combination is sum of powers of each lens. P = P1 + P2 The power of the convex lens is P1 = + i/f1 and that of concave lens is P2 = -1/f2 i) If f1<f2 then P1 + P2= +ve, the system behaves as converging lens and ii) If f1> f2then P1 - P2= -ve, the system behaves as diverging lens.

In our experiment f1<<f2, or P1>>P2 P =P1 – P2 is +ve, the system is converging fe = 1/P -------------(1) In our experiment, the effective focal length is determined experimentally using fm = uv/u+v -------(2) used for single lens in 3.4.

For all the measurements the experimental value (2) is nearly same as the theoretical value (1). It only means that the two lens system behaves as a single lens. fe= fm

It is possible to combine more than two lenses to get required power provided they are thin. If in this experiment f1 is concave and f2 is convex and f1<f2 then P1 is -ve ; P2 is +ve and P1 – P2 = P is – ve.

The system acts as diverging lens and it is not possible to catch virtual images.

P1 = 1/f1,

S.no

P2 = 1/f2 and P = P1 – P2

Distance of object u

Distance of image v

fm = uv/u+v

fe = 1/P

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fe = fm Worksheet 4- Observations for determining focal length of a combination of two lenses

KEY MESSAGES ď&#x201A;ˇ

It is possible to combine two or more thin lenses to get a lens system of required focal length.

LEARNING CHECK: Ask learners to list the key things they have learnt. Guide them to the key messages listed and then put up the chart of key messages. If you have time during the class, make up a small game, quiz or match the following as a learning check. This may have to be done as part of advance preparation.

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ABL 8– Optical Instruments Do it yourself Note to Instructor The following section is on Optical instruments and it includes working principle of camera, simple microscope, compound microscope, projector and telescope. Also there are simple procedures, which can help us construct these instruments. These sections can be given as a ‘do it yourself’ activity/1 week summer/winter project/construction of low cost optical instruments etc. All these require some concave and convex lenses and stationery. Lenses are available at almost all optical shops. .

So far we discussed about the phenomena of reflection, refraction and dispersion of light. We shall now see how the basic ideas developed in the formation of images by lenses and spherical mirrors will help us to understand the working of various optical instruments like microscope, telescope, camera and binocular.

Optical instruments have extended the range of our vision from very tiny, nearby objects to very large distant objects.

The normal human eye cannot see very small objects like bacteria, cells, viruses etc. Similarly, it cannot provide finer details of distant objects like surface of the moon or the individual leaves of a distant tree etc.

However, the range of vision of the human eye can be extended with the help of optical instruments.

Optical instruments that form real images are Camera, Eye and the Film Projector. Optical instruments that form virtual images are microscopes, Telescopes and Prism Binoculars.

The lens Camera A camera consists of a light-tight box with a convex lens at one end and the film at the other end.

In camera object is outside2F of a lens.

Therefore image is real,

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inverted and diminished.

Activity – 1 Materials: Pinhole camera, Lens Procedure:      Think

During one of the previous activities you have made a pinhole camera.

Take the pinhole camera Enlarge the hole at the front of the box.

Fix the lens over the hole.

Adjust the position of the lens for either near or far objects to make a sharp image on the screen.

 Is the image erect or inverted?  If the object is coloured, is the image is coloured?  Is the image brighter or dimmer than in pinhole camera?

The controls on camera 1. Focusing A camera id focused by moving the lens. If the object moves nearer the lens its image moves farther away.

2. The shutter The amount of light entering the camera can be controlled by the length of time that the shutter is open. Fast moving objects will appear blurred unless the exposure time is very short.

3. The aperture or diaphragm The amount of light entering the camera can be controlled by varying the size of the hole in the diaphragm, which is just behind the lens. To take a photograph in dim light, a large hole is needed.

The size of the hole also controls the depth of focus. A large depth of focus means that both near and far objects will appear to bi in focus at the same time. This is obtained by having small hole in the diaphragm.

The control, which varies the size of hole, is marked in f-numbers A value of f/8 means that the diameter of the hole is 1/8th of the focal length.

The shutter-time and f-number which should be used, depend upon i.

Brightness of the object

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ii. iii.

Sensitivity or â&#x20AC;&#x153;speedâ&#x20AC;? of the film The kind of effect that is wanted.

Digital camera Digital cameras need no film. They convert light images into electrical signals (digits). These signals are stored in the memory of the camera. A special screen enables viewing of photographs immediately. The photographs from these cameras can be transferred to a computer and can be viewed on its screen. Then they can be sent to any person anywhere using a computer network within a few seconds. They can also be printed using a printer connected to the computer.

The Projector A projector contains a lamp and a concave mirror to make the image brighter.

The lamp is placed at the center of curvature of the mirror so that light are reflected back along their own path.

the rays of

If the film is placed just after the lamp, the image on the screen would be poorly illuminated because the rays of light are diverging. To give a bright picture, a condenser is included. It is usually made of 2 Plano-convex lenses.

Now the light is converging towards the screen and the film is illuminated brightly and evenly.

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The light is then scattered by the film and focused by the convex projection lens on to the screen.

The film O is placed between F and 2F of the projection lens so that the image I is real, inverted and magnified.

The film is put in the projector upside-down so that the picture is seen the right way up.

Activity – 2 Making a Projector Materials: 1.

2.

3.

4.

Plano-convex lenses (f=5 cm) – 2 Convex lens (f=10 cm) – 1 Incandescent bulb (100 W) – 1 A piece of cine film or a slide.

Procedure: Take a ray box and set the plane-convex lenses in place of condensers and convex lens in place of projection lens. Your projector is ready.

The Simple Microscope A convex lens of short focal length held close to the eye acts as a simple microscope.

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Watchmakers, jewelers, often use simple microscope to get a magnified view of the fine parts of a watch or jewelry. It is sometimes used in magnifying printed letters in a book, texture of fibers or threads of a cloth, engravings, size of the particles of different soils etc.

A convex lens is held near the object at a distance less than its focal length and the eye is placed just behind the lens. An erect virtual and magnified image of the objects is formed.

The Compound Microscope We have seen how a simple converging lens (or simple microscope) can be used to magnify objects.

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To give a higher magnifying power we need two lenses. Therefore compound microscope contains two lenses.

1. An objective lens, which has short focal length.

2. An eye lens, which has long focal length.

The ray diagram: I. II.

III.

The object is placed just outside F 0, the principal focus of the objective lens and so a real inverted magnified image I1 is formed.

Then the magnified image I1 acts as an object for the eye-lens, which is used as magnifying glass. F e is the principal focus of the eye-lens.

The dotted lines are construction lines used to find the position of final image I 2 The final image I2 is virtual and magnified still further. It is inverted compared with the object.

I2 may appear 1000 times larger than the object.

Activity – 3 Making a compound microscope Materials: 1.

2.

3.

4.

Convex lens (f=5 cm) – 1 Convex lens (f-10 cm) – 1 Half meter scale Plasticine

Procedure:  Affix two lenses on the scale as shown in the diagram, with the help of Plasticine  Move the object to and fro until it appears in the sharp focus.

Think  What do you notice about the image?  Is it distorted?  Is it coloured differently in any way?

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Telescopes 1. The Astronomical Telescopes A simple refracting telescope contains two lenses i. ii.

An objective lens which has long focal length An eye lens, which has short focal length.

The ray diagram: i. ii.

Parallel rays of light from the distant object are converged to form a real, inverted image I1 at F0, the principal focus of objective lens.

The image I1 acts as an object for the eye lens since I1 is at Fe , the principal focus of the eye-lens, the magnified final image I2 is at infinity.

Astronomers use telescope to look at very faint stars. Why is it an advantage to have a large objective lens?

The final image is inverted. Is this a disadvantage to an astronomer?

Activity – 4 Making a refracting Telescope Materials: 1. Convex lens (f=20 cm) – 1 (Objective) 2. Convex lens (f=50 cm) – 1 (eye lens) Agastya International Foundation. For Internal Circulation only.

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3. Half meter scale 4. Plasticine Procedure;  Affix two lenses on half meter scale using Plasticine as shown in the diagram  Focus the system to a distance object  Adjust the distance between the lenses until the mage is in sharp focus.

Think  Is the image magnified?  Is the image erect or inverted?  What happens to the magnifying power if you change the objective lens to one with longer focal length?

2. Galilean Telescopes Galileo constructed a telescope in 1610. Hence it is called Galilean telescope. It consists of two lenses, mounted coaxially on the outer ends of two hollow metallic tubes. Convex lens acts as objective and concave lens as eye piece.

Light rays coming from distant object, after passing through objective will be converged at its focal plane, and forms the image of an object. If the eyepiece is brought in between image and the objective a virtual image is formed.

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Galilean Telescope has two advantages over astronomical telescope.

1. It produces erect image. Therefore, it is convenient to view the terrestrial objects.

2. Tube length of Galilean telescope is less than astronomical telescope. In astronomical telescope tube length is equal to fo+fe where as in Galilean telescope, it is equal to fo-fe 3. Major disadvantage of Galilean telescope is its field of view is very limited.

Activity – 5 Making Galilean Telescope Materials: 1.

2.

3.

4.

Convex lens (f=50 cm) – 1 (Objective) Concave lens (f=10 cm) – 1 (eye lens) Half meter scale Plasticine

Procedure;  Affix two lenses on half meter scale using plasticine as shown in the diagram  Focus the system to a distance object  Adjust the distance between the lenses until the image is in sharp focus.

Think  Is the image erect or inverted?  Is the image magnified?

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Prism Binoculars Binoculars are made of two refracting telescopes side by side, one for each eye. In each telescope two prisms are added as shown in the figure.

While discussing about total internal reflection, we saw that 45 0 - 450 - 900 prisms can use total internal reflection to invert rays of light.

By including two of these prisms in each telescope, the final image can be turned the right way-up and right way round.

The prisms do two things 1. They invert the rays of light so that the final image is seen the correct way.

2. By reflecting light up and down the inside of the binoculars, they shorten the length of the instrument so it is more compact.

ADDITIONAL INFORMATION Magnification and Resolution Activity â&#x20AC;&#x201C; 6 Materials: Torch having two LEDs, Convex lens of f=10 cm Agastya International Foundation. For Internal Circulation only.

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Procedure: A) Magnification    

Take a newspaper Look at the letters with bare eye Now look at the same letters through a convex lens What difference do you find?

When you look through convex lens, letters appear magnified.

B) Resolution     

Keep a white paper on the table Illuminate the paper with the help a torch having 2 LEDs holding torch very near to the paper.

Now light emitted by two LEDs forms two separate images.

Now take torch at some distance above the paper and allow the light to fall on the paper.

The two images formed by the LEDs will merge and form a single image and this image is called an unresolved image.  Now again bring torch closer to the paper, at some distance two images are neither fully merged nor separated. This situation is called just resolved.  Now bring the torch very close to the paper. You see that the two images are clearly separated. This image is fully resolved.

Thus, greater useful magnification with adequate resolution forms the essential component in the design of optical instruments for viewing an object distinctly.

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Magnification of Optical Instruments

1. Magnification of a simple microscope is Size of the image

Near Point

D

M = ----------------------- = --------------- = ---Size of the object

Focal length

f

Near Point – The nearest point to the entrance of the pupil of the normal eye at which focus is attained without strain, 25 cm.

2. Magnification of the compound microscope is given by Size of the final image m= ----------------------------Size of the object

A2B2

L

D

= --------- = ------ -----AB

fofe

Here fo – focal length of the objective fe– focal length of the eyepiece Agastya International Foundation. For Internal Circulation only.

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L â&#x20AC;&#x201C; Length of tube D â&#x20AC;&#x201C; Least distance of distinct vision.

To achieve a larger magnification of small object, both the eyepiece and objective of a compound microscope should have smaller focal lengths. Magnification of the order of 1000 is possible with a very good design of the microscopes.

3. Magnification of the telescope M= fo/fe Thus, in order to increase the magnification of a telescope, objective must have a large focal length and eyepiece have a small focal length.

4. Magnification of Galilean Telescope fo

Focal length of objective M = -- = ---------------- ----------------

fe

Focal length of eye piece

where fe and fo are focal length of eye piece and objective respectively.

Magnification shall be increased either by increasing focal length of the objective or decreasing focal length of the eye piece.

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