Chapter 3 Geometric Optics
Dr. Mohamed Saudy
1st. Year Biology 1
The Nature of Light
Before the beginning of the nineteenth century, light was considered to be a stream of particles The particles were either emitted by the object being viewed or emanated from the eyes of the viewer Newton was the chief architect of the particle theory of light
He believed the particles left the object and stimulated the sense of sight upon entering the eyes
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Nature of Light – Alternative View
Huygens argued that light might be some sort of a wave motion Thomas Young (in 1801) provided the first clear demonstration of the wave nature of light
He showed that light rays interfere with each other Such behavior could not be explained by particles
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Dual Nature of Light ď Ź
ď Ź
In view of these developments, light must be regarded as having a dual nature Light exhibits the characteristics of a wave in some situations and the characteristics of a particle in other situations
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The Ray Approximation in Geometric Optics
Geometric optics involves the study of the propagation of light It uses the assumption that light travels in a straight-line path in a uniform medium and changes its direction when it meets the surface of a different medium or if the optical properties of the medium are non-uniform The ray approximation is used to represent beams of light. 5
Ray Approximation ď Ź
ď Ź
The rays are straight lines perpendicular to the wave fronts With the ray approximation, we assume that a wave moving through a medium travels in a straight line in the direction of its rays
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Ray Approximation, cont.
If a wave meets a barrier, we will assume that λ<<d
d is the diameter of the opening
This approximation is good for the study of mirrors, lenses, prisms, etc. Other effects occur for openings of other sizes 7
Active Figure 35.4
Adjust the size of the opening Observe the effects on the waves passing through PLAY ACTIVE FIGURE 8
Reflection of Light
A ray of light, the incident ray, travels in a medium When it encounters a boundary with a second medium, part of the incident ray is reflected back into the first medium
This means it is directed backward into the first medium
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Specular Reflection
Specular reflection is reflection from a smooth surface The reflected rays are parallel to each other All reflection in this text is assumed to be specular
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Diffuse Reflection
Diffuse reflection is reflection from a rough surface The reflected rays travel in a variety of directions A surface behaves as a smooth surface as long as the surface variations are much smaller than the wavelength of the light
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Law of Reflection
The normal is a line perpendicular to the surface
It is at the point where the incident ray strikes the surface
The incident ray makes an angle of θ1 with the normal The reflected ray makes an angle of θ1’ with the normal
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Law of Reflection, cont.
The angle of reflection is equal to the angle of incidence θ1’= θ1
This relationship is called the Law of Reflection
The incident ray, the reflected ray and the normal are all in the same plane Because this situation happens often, an analysis model, wave under reflection, is identified 13
Active Figure 35.6
Use the active figure to vary the angle of incidence Observe the effect on the angle of reflection PLAY ACTIVE FIGURE 14
Refraction of Light
When a ray of light traveling through a transparent medium encounters a boundary leading into another transparent medium, part of the energy is reflected and part enters the second medium The ray that enters the second medium is bent at the boundary
This bending of the ray is called refraction 15
Refraction, 2
The incident ray, the reflected ray, the refracted ray, and the normal all lie on the same plane The angle of refraction depends upon the material and the angle of incidence
sin θ 2 v 2 sin θ1 v 1 v1 is the speed of the light in the first medium and v2 is its speed in the second 16
Refraction of Light, 3
The path of the light through the refracting surface is reversible For example, a ray that travels from A to B If the ray originated at B, it would follow the line AB to reach point A 17
Following the Reflected and Refracted Rays
Ray is the incident ray Ray is the reflected ray Ray is refracted into the lucite Ray is internally reflected in the lucite Ray is refracted as it enters the air from the lucite 18
Active Figure 35.10
Use the active figure to vary the incident angle Observe the effect on the reflected and refracted rays
PLAY ACTIVE FIGURE 19
Refraction Details, 1
Light may refract into a material where its speed is lower The angle of refraction is less than the angle of incidence
The ray bends toward the normal
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Refraction Details, 2
Light may refract into a material where its speed is higher The angle of refraction is greater than the angle of incidence
The ray bends away from the normal
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Active Figure 35.11
Use the active figure to observe the light passing through three layers of material Vary the incident angle and the materials Observe the effect on the refracted ray
PLAY ACTIVE FIGURE 22
Light in a Medium
The light enters from the left The light may encounter an electron The electron may absorb the light, oscillate, and reradiate the light The absorption and radiation cause the average speed of the light moving through the material to decrease 23
The Index of Refraction
The speed of light in any material is less than its speed in vacuum The index of refraction, n, of a medium can be defined as
speed of light in a vacuum c n speed of light in a medium v
24
Index of Refraction, cont.
For a vacuum, n = 1
We assume n = 1 for air also
For other media, n > 1 n is a dimensionless number greater than unity
n is not necessarily an integer
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Some Indices of Refraction
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Frequency Between Media
As light travels from one medium to another, its frequency does not change
Both the wave speed and the wavelength do change The wavefronts do not pile up, nor are created or destroyed at the boundary, so ƒ must stay the same 27
Index of Refraction Extended
The frequency stays the same as the wave travels from one medium to the other v = ƒλ
ƒ1 = ƒ2 but v1 v2 so λ1 λ2
The ratio of the indices of refraction of the two media can be expressed as various ratios c λ1 v1 n1 n2 λ2 v 2 c n1 n2
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More About Index of Refraction
The previous relationship can be simplified to compare wavelengths and indices: λ1n1 = λ2n2 In air, n1 = 1 and the index of refraction of the material can be defined in terms of the wavelengths λ n λn
λ in vacuum λ in a medium 29
Snell’s Law of Refraction
n1 sin θ1 = n2 sin θ2
θ1 is the angle of incidence θ2 is the angle of refraction
The experimental discovery of this relationship is usually credited to Willebrord Snell and is therefore known as Snell’s law of refraction Refraction is a commonplace occurrence, so identify an analysis model as a wave under refraction
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Snell’s Law – Example
Light is refracted into a crown glass slab θ1 = 30.0o, θ2 = ? n1 = 1.00 and n2 = 1.52
From Table 35.1
θ2 = sin-1(n1 / n2) sin θ1 = 19.2o
The ray bends toward the normal, as expected
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Example 1 Green light with a wavelength of 5Ă&#x2014;10-7 m in a vacuum enters a glass plate with refractive index 1.5.
A. What is the velocity of light in the glass? B. What is the wavelength of the light in the glass? Solution: A) From the definition n=c/v, the velocity in the glass is
B) In the vacuum, v=c and n=1, thus
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Example 2 A beam of light of wavelength 550 nm
traveling in air, is incident on a slab of transparent material. The incident beam makes an angle of 400 with the normal, and the refracted beam makes an angle of 260 with the
normal. Find the index of refraction of the material. Solution Using Snell's law of refraction, and taking n1=1 for air
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Example 3 A light ray of wavelength 589 nm traveling through air is incident on a smooth, flat slab of crown glass of refractive index 1.52 at an angle of 300 to the normal, as in figure. Find the angle of refraction. Solution We rearrange Snell's law of refraction with these data n1=1 for air and n2=1.52 for crown glass to obtain: sin 2
n1 sin 1 n2
1 sin 30 0.329 1.52 2 sin 1 (0.329) 19.20
Because this is less than the incident angle of 300, the refracted rays is bent toward the normal.
Its change in direction is called the angle of deviation and is given by 1 2 300 19.20 10.80 34
Example 4 A light beam passes from medium 1 to medium 2, with the later medium being a thick slab of material whose index of refraction is n2 as shown in figure. Show that the emerging beam is parallel to the incident beam. Solution Let us apply Snell's law of refraction to the upper surface:
n1 sin 2 sin 1 n2 Applying this law to the lower surface gives: n sin 3 2 sin 2 n1 n n sin 3 2 ( 1 sin 1 ) sin 1 n1 n2 Therefore, 3=1, and the slab does not alter the direction of the beam. The beam parallel to itself by a distance d. 35
Prism
A ray of singlewavelength light incident on the prism will emerge at angle from its original direction of travel
is called the angle of deviation F is the apex angle
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Dispersion
For a given material, the index of refraction varies with the wavelength of the light passing through the material This dependence of n on λ is called dispersion Snell’s law indicates light of different wavelengths is bent at different angles when incident on a refracting material 37
Variation of Index of Refraction with Wavelength ď Ź
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The index of refraction for a material generally decreases with increasing wavelength Violet light bends more than red light when passing into a refracting material
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Refraction in a Prism
Since all the colors have different angles of deviation, white light will spread out into a spectrum
Violet deviates the most Red deviates the least The remaining colors are in between
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The Rainbow
A ray of light strikes a drop of water in the atmosphere It undergoes both reflection and refraction
First refraction at the front of the drop
Violet light will deviate the most Red light will deviate the least
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The Rainbow, 2
At the back surface the light is reflected It is refracted again as it returns to the front surface and moves into the air The rays leave the drop at various angles
The angle between the white light and the most intense violet ray is 40° The angle between the white light and the most intense red ray is 42° 41
Active Figure 35.23
Use the active figure to vary the point at which the sunlight enters the raindrop Observe the angles and verify the maximum angles PLAY ACTIVE FIGURE 42
Observing the Rainbow
If a raindrop high in the sky is observed, the red ray is seen A drop lower in the sky would direct violet light to the observer The other colors of the spectra lie in between the red and the violet
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Total Internal Reflection ď Ź
A phenomenon called total internal reflection can occur when light is directed from a medium having a given index of refraction toward one having a lower index of refraction
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Possible Beam Directions ď Ź
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Possible directions of the beam are indicated by rays numbered 1 through 5 The refracted rays are bent away from the normal since n1 > n2
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Critical Angle
There is a particular angle of incidence that will result in an angle of refraction of 90°
This angle of incidence is called the critical angle, θC
n2 sin θC (for n1 n 2 ) n1 46
Active Figure 35.25
Use the active figure to vary the incident angle Observe the effect on the refracted ray PLAY ACTIVE FIGURE 47
Critical Angle, cont.
For angles of incidence greater than the critical angle, the beam is entirely reflected at the boundary
This ray obeys the law of reflection at the boundary
Total internal reflection occurs only when light is directed from a medium of a given index of refraction toward a medium of lower index of refraction 48
Example 5 What is the critical angle for light going from glass with refractive index 1.5 into air? Solution: Using n1=1.5 and n2=1
Thus, light incident from the glass on a glass-air boundary will be completely reflected if the angle of incidence exceeds 420. Example 6 Find the critical angle for an airâ&#x20AC;&#x201C;water boundary. (The index of refraction of water is 1.33) Solution: with the air above the water having index of refraction n2 and the water having index of refraction n1.
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Fiber Optics
An application of internal reflection Plastic or glass rods are used to “pipe” light from one place to another Applications include
Medical examination of internal organs Telecommunications 50
Fiber Optics, cont. ď Ź
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A flexible light pipe is called an optical fiber A bundle of parallel fibers (shown) can be used to construct an optical transmission line
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Construction of an Optical Fiber
The transparent core is surrounded by cladding
The cladding has a lower n than the core This allows the light in the core to experience total internal reflection
The combination is surrounded by the jacket
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Problems 1. The wavelength of red helium–neon laser light in air is 632.8 nm. (a) What is its frequency? (b) What is its wavelength in glass that has an index of refraction of 1.5? (c) What is its speed in the glass? 2. A ray of light is incident on a flat surface of a block of crown glass that is surrounded by water. The angle of refraction is 19.6°. Find the angle of reflection. 3. A laser in a compact disc player generates light that has a wavelength of 780 nm in air.
I.
Find the speed of this light once it enters the plastic of a compact disc (n=1.55).
II.
What is the wavelength of this light in the plastic? Answer
(v=1.94×108
m) and (n=503 nm)
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Notation for Mirrors and Lenses
The object distance is the distance from the object to the mirror or lens
The image distance is the distance from the image to the mirror or lens
Denoted by p
Denoted by q
The lateral magnification of the mirror or lens is the ratio of the image height to the object height
Denoted by M
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Images ď Ź
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Images are always located by extending diverging rays back to a point at which they intersect Images are located either at a point from which the rays of light actually diverge or at a point from which they appear to diverge
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Types of Images
A real image is formed when light rays pass through and diverge from the image point
Real images can be displayed on screens
A virtual image is formed when light rays do not pass through the image point but only appear to diverge from that point
Virtual images cannot be displayed on screens
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Images Formed by Flat Mirrors
Simplest possible mirror Light rays leave the source and are reflected from the mirror Point I is called the image of the object at point O The image is virtual 57
Images Formed by Flat Mirrors, 2
A flat mirror always produces a virtual image Geometry can be used to determine the properties of the image There are an infinite number of choices of direction in which light rays could leave each point on the object Two rays are needed to determine where an image is formed 58
Images Formed by Flat Mirrors, 3
One ray starts at point P, travels to Q and reflects back on itself Another ray follows the path PR and reflects according to the law of reflection The triangles PQR and P’QR are congruent 59
Active Figure 36.2
Use the active figure to move the object Observe the effect on the image
PLAY ACTIVE FIGURE 60
Images Formed by Flat Mirrors, 4
To observe the image, the observer would trace back the two reflected rays to P’ Point P’ is the point where the rays appear to have originated The image formed by an object placed in front of a flat mirror is as far behind the mirror as the object is in front of the mirror
|p| = |q| 61
Lateral Magnification
Lateral magnification, M, is defined as Image height h' M Object height h
This is the general magnification for any type of mirror It is also valid for images formed by lenses Magnification does not always mean bigger, the size can either increase or decrease
M can be less than or greater than 1 62
Lateral Magnification of a Flat Mirror
The lateral magnification of a flat mirror is +1 This means that h’ = h for all images The positive sign indicates the object is upright
Same orientation as the object
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Reversals in a Flat Mirror ď Ź
A flat mirror produces an image that has an apparent left-right reversal ď Ź
For example, if you raise your right hand the image you see raises its left hand
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Reversals, cont.
The reversal is not actually a left-right reversal The reversal is actually a front-back reversal
It is caused by the light rays going forward toward the mirror and then reflecting back from it
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Properties of the Image Formed by a Flat Mirror – Summary
The image is as far behind the mirror as the object is in front
The image is unmagnified
The image height is the same as the object height h’ = h and M = +1
The image is virtual The image is upright
|p| = |q|
It has the same orientation as the object
There is a front-back reversal in the image 66
Spherical Mirrors
A spherical mirror has the shape of a section of a sphere The mirror focuses incoming parallel rays to a point A concave spherical mirror has the silvered surface of the mirror on the inner, or concave, side of the curve A convex spherical mirror has the silvered surface of the mirror on the outer, or convex, side of the curve
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Concave Mirror, Notation
The mirror has a radius of curvature of R Its center of curvature is the point C Point V is the center of the spherical segment A line drawn from C to V is called the principal axis of the mirror
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Paraxial Rays
We use only rays that diverge from the object and make a small angle with the principal axis Such rays are called paraxial rays All paraxial rays reflect through the image point
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Spherical Aberration
Rays that are far from the principal axis converge to other points on the principal axis This produces a blurred image The effect is called spherical aberration
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Image Formed by a Concave Mirror
Geometry can be used to determine the magnification of the image
h' q M h p
h’ is negative when the image is inverted with respect to the object
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Image Formed by a Concave Mirror
Geometry also shows the relationship between the image and object distances
1 1 2 p q R This is called the mirror equation
If p is much greater than R, then the image point is half-way between the center of curvature and the center point of the mirror
p → ∞ , then 1/p 0 and q R/2 72
Focal Length
When the object is very far away, then p → ∞ and the incoming rays are essentially parallel In this special case, the image point is called the focal point The distance from the mirror to the focal point is called the focal length
The focal length is ½ the radius of curvature
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Focal Point, cont.
The colored beams are traveling parallel to the principal axis The mirror reflects all three beams to the focal point The focal point is where all the beams intersect
It is the white point
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Focal Point and Focal Length, cont.
The focal point is dependent solely on the curvature of the mirror, not on the location of the object
It also does not depend on the material from which the mirror is made
ƒ=R/2 The mirror equation can be expressed as 1 1 1 p q ƒ 75
Focal Length Shown by Parallel Rays
76
Example 1: Assume that a certain spherical mirror has a focal length of +10 cm. Locate and describe the image for object distances of (A) 25 cm, (B) 10 cm, and (C) 5 cm. Solution: A) Because the focal length is positive f=+10cm, we expect the image to be real. 1 1 1 p q f 1 1 1 25 cm q 10 cm
q 16.7 cm
The magnification of the image is given by M is less than unity tells us that
q 16.7 cm M 0.668 p 25 cm
the image is smaller than the object, and the negative sign for M tells us that the image is inverted
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B) When the object distance is 10 cm, the object is located at the focal point 1 1 1 p q f 1 1 1 10 cm q 10 cm
q
so that the image is formed at an infinite distance from the mirror; that is, the rays travel parallel to one another after reflection. This is the situation in a flashlight, where the bulb filament is placed at the focal point of a reflector, producing a parallel beam of light.
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C) When the object is at p=5 cm, it lies halfway between the focal point and the mirror surface 1 1 1 p q f 1 1 1 5 cm q 10 cm
q 10 cm
The image is virtual because it is located behind the mirror. The magnification of the image is
q 10 cm M ( ) 2 p 5 cm The image is twice as large as the object, and the positive sign for M indicates that the image is upright 79
Convex Mirrors
A convex mirror is sometimes called a diverging mirror
The light reflects from the outer, convex side
The rays from any point on the object diverge after reflection as though they were coming from some point behind the mirror The image is virtual because the reflected rays only appear to originate at the image point 80
Image Formed by a Convex Mirror
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In general, the image formed by a convex mirror is upright, virtual, and smaller than the object
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Sign Conventions ď Ź
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These sign conventions apply to both concave and convex mirrors The equations used for the concave mirror also apply to the convex mirror
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Sign Conventions, Summary Table
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Ray Diagrams
A ray diagram can be used to determine the position and size of an image They are graphical constructions which reveal the nature of the image They can also be used to check the parameters calculated from the mirror and magnification equations
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Drawing a Ray Diagram
To draw a ray diagram, you need to know:
Three rays are drawn
The position of the object The locations of the focal point and the center of curvature They all start from the same position on the object
The intersection of any two of the rays at a point locates the image
The third ray serves as a check of the construction 85
The Rays in a Ray Diagram – Concave Mirrors
Ray 1 is drawn from the top of the object parallel to the principal axis and is reflected through the focal point, F Ray 2 is drawn from the top of the object through the focal point and is reflected parallel to the principal axis Ray 3 is drawn through the center of curvature, C, and is reflected back on itself 86
Notes About the Rays
The rays actually go in all directions from the object The three rays were chosen for their ease of construction The image point obtained by the ray diagram must agree with the value of q calculated from the mirror equation
87
Ray Diagram for a Concave Mirror, p > R
The center of curvature is between the object and the concave mirror surface The image is real The image is inverted The image is smaller than the object (reduced) 88
Ray Diagram for a Concave Mirror, p < f
The object is between the mirror surface and the focal point The image is virtual The image is upright The image is larger than the object (enlarged) 89
The Rays in a Ray Diagram – Convex Mirrors
Ray 1 is drawn from the top of the object parallel to the principal axis and is reflected away from the focal point, F Ray 2 is drawn from the top of the object toward the focal point and is reflected parallel to the principal axis Ray 3 is drawn through the center of curvature, C, on the back side of the mirror and is reflected back on itself 90
Ray Diagram for a Convex Mirror
The object is in front of a convex mirror The image is virtual The image is upright The image is smaller than the object (reduced) 91
Active Figure 36.13
Use the active figure to
Move the object Change the focal length
Observe the effect on the images
PLAY ACTIVE FIGURE 92
Notes on Images
With a concave mirror, the image may be either real or virtual
When the object is outside the focal point, the image is real When the object is at the focal point, the image is infinitely far away When the object is between the mirror and the focal point, the image is virtual
With a convex mirror, the image is always virtual and upright
As the object distance decreases, the virtual image increases in size 93
The Image from a Convex Mirror Example 2: An anti-shoplifting mirror, as shown in Figure, shows an image of a woman who is located 3 m from the mirror. The focal length of the mirror is -0.25 m. Find (A) the position of her image and (B) the magnification of the image Solution: (A) We should expect to find an upright, reduced, virtual image 1 1 1 p q f 1 1 1 3m q 0.25 m
q 0.23 m
(B) The magnification of the image is
q 0.23m M ( ) 0.077 p 3m
The image is much smaller than the woman, and it is upright because M is positive 94
Images Formed by Refraction
Consider two transparent media having indices of refraction n1 and n2 The boundary between the two media is a spherical surface of radius R Rays originate from the object at point O in the medium with n = n1 95
Images Formed by Refraction, 2
We will consider the paraxial rays leaving O All such rays are refracted at the spherical surface and focus at the image point, I The relationship between object and image distances can be given by n1 n2 n2 n1 p q R
96
Images Formed by Refraction, 3
The side of the surface in which the light rays originate is defined as the front side The other side is called the back side Real images are formed by refraction in the back of the surface
Because of this, the sign conventions for q and R for refracting surfaces are opposite those for reflecting surfaces
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Sign Conventions for Refracting Surfaces
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Flat Refracting Surfaces
If a refracting surface is flat, then R is infinite Then q = -(n2 / n1)p
The image formed by a flat refracting surface is on the same side of the surface as the object
A virtual image is formed
99
Active Figure 36.18
Use the active figure to move the object Observe the effect on the location of the image PLAY ACTIVE FIGURE 100
Example 3: A set of coins is embedded in a spherical plastic paperweight having a radius of 3 cm. The index of refraction of the plastic is n1=1.5. One coin is located 2 cm from the edge of the sphere (Figure). Find the position of the image of the coin. Solution Because n1 > n 2, where n2=1 is the index of refraction for air, the rays originating from the coin are refracted away from the normal at the surface and diverge outward. Hence, the image is formed inside the paperweight and is virtual. Note that: (R=-3cm)
n1 n2 n2 n1 p q R 1.5 1 1 1.5 q 1.7cm 2 q 3
The negative sign for q indicates that the image is in front of the surface the image must be virtual. The coin appears to be closer to the paperweight surface than it actually is.
101
Example 4: A small fish is swimming at a depth d below the surface of a pond (Figure). What is the apparent depth of the fish, as viewed from directly overhead? Solution Because the refracting surface is flat, R is infinite. Hence, we can use the Equation to determine the location of the image with p =d
n2 1 q p d 0.752d n1 1.33 Because q is negative, the image is virtual, as indicated by the dashed lines in Figure. The apparent depth is approximately three-fourths the actual depth. 102
Lenses
Lenses are commonly used to form images by refraction Lenses are used in optical instruments
Cameras Telescopes Microscopes
103
Images from Lenses ď Ź
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Light passing through a lens experiences refraction at two surfaces The image formed by one refracting surface serves as the object for the second surface
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Locating the Image Formed by a Lens
The lens has an index of refraction n and two spherical surfaces with radii of R1 and R2 R1 is the radius of curvature of the lens surface that the light of the object reaches first R2 is the radius of curvature of the other surface The object is placed at point O at a distance of p1 in front of the first surface 105
Locating the Image Formed by a Lens, Image From Surface 1
There is an image formed by surface 1 Since the lens is surrounded by the air, n1 = 1 and n1 n2 n2 n1 1 n n 1 p q R p1 q1 R1
If the image due to surface 1 is virtual, q1 is negative, and it is positive if the image is real
106
Locating the Image Formed by a Lens, Image From Surface 2
For surface 2, n1 = n and n2 = 1
The light rays approaching surface 2 are in the lens and are refracted into air
Use p2 for the object distance for surface 2 and q2 for the image distance n1 n2 n2 n1 n 1 1 n p q R p2 q2 R2
107
Image Formed by a Thick Lens
If a virtual image is formed from surface 1, then p2 = -q1 + t
If a real image is formed from surface 1, then p2 = -q1 + t
q1 is negative t is the thickness of the lens
q1 is positive
Then 1 1 1 1 n 1 p1 q2 R1 R2
108
Image Formed by a Thin Lens
A thin lens is one whose thickness is small compared to the radii of curvature For a thin lens, the thickness, t, of the lens can be neglected In this case, p2 = -q1 for either type of image Then the subscripts on p1 and q2 can be omitted
109
Lens Makers’ Equation
The focal length of a thin lens is the image distance that corresponds to an infinite object distance
This is the same as for a mirror
The lens makers’ equation is
1 1 1 1 1 (n 1) p q R1 R2 ƒ 110
Thin Lens Equation
The relationship among the focal length, the object distance and the image distance is the same as for a mirror
1 1 1 p q ƒ
111
Notes on Focal Length and Focal Point of a Thin Lens
Because light can travel in either direction through a lens, each lens has two focal points
One focal point is for light passing in one direction through the lens and one is for light traveling in the opposite direction
However, there is only one focal length Each focal point is located the same distance from the lens 112
Focal Length of a Converging Lens
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The parallel rays pass through the lens and converge at the focal point The parallel rays can come from the left or right of the lens 113
Focal Length of a Diverging Lens
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The parallel rays diverge after passing through the diverging lens The focal point is the point where the rays appear to have originated 114
Determining Signs for Thin Lenses ď Ź
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The front side of the thin lens is the side of the incident light The light is refracted into the back side of the lens This is also valid for a refracting surface
115
Sign Conventions for Thin Lenses
116
Magnification of Images Through a Thin Lens
The lateral magnification of the image is h' q M h p When M is positive, the image is upright and on the same side of the lens as the object When M is negative, the image is inverted and on the side of the lens opposite the object 117
Thin Lens Shapes
These are examples of converging lenses They have positive focal lengths They are thickest in the middle
118
More Thin Lens Shapes
These are examples of diverging lenses They have negative focal lengths They are thickest at the edges
119
Ray Diagrams for Thin Lenses – Converging
Ray diagrams are convenient for locating the images formed by thin lenses or systems of lenses For a converging lens, the following three rays are drawn:
Ray 1 is drawn parallel to the principal axis and then passes through the focal point on the back side of the lens Ray 2 is drawn through the center of the lens and continues in a straight line Ray 3 is drawn through the focal point on the front of the lens (or as if coming from the focal point if p < ƒ) and emerges from the lens parallel to the principal axis
120
Ray Diagram for Converging Lens, p > f
The image is real The image is inverted The image is on the back side of the lens 121
Ray Diagram for Converging Lens, p < f
The image is virtual The image is upright The image is larger than the object The image is on the front side of the lens 122
Ray Diagrams for Thin Lenses – Diverging
For a diverging lens, the following three rays are drawn:
Ray 1 is drawn parallel to the principal axis and emerges directed away from the focal point on the front side of the lens Ray 2 is drawn through the center of the lens and continues in a straight line Ray 3 is drawn in the direction toward the focal point on the back side of the lens and emerges from the lens parallel to the principal axis
123
Ray Diagram for Diverging Lens
The image is virtual The image is upright The image is smaller The image is on the front side of the lens 124
Active Figure 36.26
Use the active figure to
Move the object Change the focal length of the lens
Observe the effect on the image PLAY ACTIVE FIGURE 125
Image Summary
For a converging lens, when the object distance is greater than the focal length, (p > ƒ)
For a converging lens, when the object is between the focal point and the lens, (p < ƒ)
The image is real and inverted
The image is virtual and upright
For a diverging lens, the image is always virtual and upright
This is regardless of where the object is placed
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Example 5: A converging lens of focal length 10 cm forms images of objects placed at (A) 30 cm, (B) 10 cm, and (C) 5 cm from the lens. In each case, construct a ray diagram, find the image distance and describe the image. Solution: (A) we construct a ray diagram as shown in Figure. The diagram shows that we should expect a real, inverted, smaller image to be formed on the back side of the lens.
1 1 1 p q f 1 1 1 q 15 cm 30 q 10 The positive sign for the image distance tells us that the image is indeed real and on the back side of the lens. The magnification of the image is q 15 M 0.5 p 30 Thus, the image is reduced in height by one half, and the negative sign for M tells us that the image is inverted
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(B) the object is placed at the focal point, the image is formed at infinity. This is readily verified by substituting p=10 cm into the thin lens equation. (C) The ray diagram in Figure shows that in this case the lens acts as a magnifying glass; that is, the image is magnified, upright, on the same side of the lens as the object, and virtual. Because the object distance is 5 cm.
1 1 1 p q f 1 1 1 q 10 cm 5 q 10 and the magnification of the image is
q 10 ( ) 2 p 5 The negative image distance tells us that the image is virtual and formed on the side of the lens from which the light is incident, the front side. The image is enlarged, and the positive sign for M tells us that the image is upright M
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Example 6: A diverging lens of focal length 10 cm forms images of objects placed at (A) 30 cm, (B) 10 cm, and (C) 5 cm from the lens. In each case, construct a ray diagram, find the image distance and describe the image. Solution: (A) constructing a ray diagram as in Figure taking the object distance to be 30 cm. The diagram shows that we should expect an image that is virtual, smaller than the object, and upright
1 1 1 p q ƒ 1 1 1 q 75 . cm 30 q 10 The magnification of the image is
M
q 7.5 ( ) 0.25 p 30
This result confirms that the image is virtual, smaller than the object, and upright 129
(B) When the object is at the focal point, the ray diagram appears as in Figure
1 1 1 p q f 1 1 1 q 5 cm 10 q 10 The magnification of the image is
M
q 5 ( ) 0.5 p 10
Notice the difference between this situation and that for a converging lens. For a diverging lens, an object at the focal point does not produce an image infinitely far away.
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(C) When the object is inside the focal point, at p =5 cm, the ray diagram in Figure, shows that we expect a virtual image that is smaller than the object and upright
1 1 1 p q f 1 1 1 q 3.33 cm 5 q 10 and the magnification of the image is
q 3.33 M ( ) 0.667 p 5 This confirms that the image is virtual, smaller than the object, and upright
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Example 7: A converging glass lens (n =1.52) has a focal length of 40 cm in air. Find its focal length when it is immersed in water, which has an index of refraction of 1.33. Solution We can use the lens makers’ equation in both cases, noting that R1 and R2 remain the same in air and water:
1 1 1 (n 1)( ) f air R1 R2 1 f water
(n 1)(
1 1 ) R1 R2
where is the ratio of the index of refraction of glass to that of water: =1.52/1.33=1.14. Dividing the first equation by the second gives
f water n 1 1.52 1 3.71 f air n 1 1.14 1
f water 3.71 f air 3.71 (40) 148 cm n 1
The focal length of any lens is increased by a factor n 1 when the lens is immersed in a fluid, where is the ratio of the index of refraction n of the lens material to that of the fluid.
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Fresnal Lens
Refraction occurs only at the surfaces of the lens A Fresnal lens is designed to take advantage of this fact It produces a powerful lens without great thickness
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Fresnal Lens, cont.
Only the surface curvature is important in the refracting qualities of the lens The material in the middle of the Fresnal lens is removed Because the edges of the curved segments cause some distortion, Fresnal lenses are usually used only in situations where image quality is less important than reduction of weight 134
Combinations of Thin Lenses
The image formed by the first lens is located as though the second lens were not present Then a ray diagram is drawn for the second lens The image of the first lens is treated as the object of the second lens The image formed by the second lens is the final image of the system 135
Combination of Thin Lenses, 2
If the image formed by the first lens lies on the back side of the second lens, then the image is treated as a virtual object for the second lens
p will be negative
The same procedure can be extended to a system of three or more lenses The overall magnification is the product of the magnification of the separate lenses 136
Two Lenses in Contact
Consider a case of two lenses in contact with each other
The lenses have focal lengths of ƒ1 and ƒ2
For the first lens, 1 1 1 p q1 ƒ1
Since the lenses are in contact, p2 = -q1 137
Two Lenses in Contact, cont.
For the second lens, 1 1 1 1 1 p2 q2 ƒ2 q1 q
For the combination of the two lenses 1 1 1 ƒ ƒ1 ƒ 2
Two thin lenses in contact with each other are equivalent to a single thin lens having a focal length given by the above equation 138
Combination of Thin Lenses, example
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Combination of Thin Lenses, example
Find the location of the image formed by lens 1 Find the magnification of the image due to lens 1 Find the object distance for the second lens Find the location of the image formed by lens 2 Find the magnification of the image due to lens 2 Find the overall magnification of the system 140
Example 8: Two thin converging lenses of focal lengths f1=10 cm and f2=20 cm are separated by 20 cm, as illustrated in Figure. An object is placed 30 cm to the left of lens 1. Find the position and the magnification of the final image. Solution: To analyze the problem, we first draw a ray diagram showing where the image from the first lens falls and how it acts as the object for the second lens. The location of the image formed by lens 1 is found from the thin lens equation:
1 1 1 p1 q1 f1 1 1 1 q1 15 cm 30 q1 10 The magnification of this image is
q1 15 M1 0.5 p1 30
The image formed by this lens acts as the object for the second lens. Thus, the object distance for the second lens is 20 cm - 15 cm =5 cm. We again apply the thin lens equation to find the location of the final image: 141
1 1 1 p2 q2 f 2 1 1 1 q2 6.67 cm 5 q2 20 The magnification of the second image is M 2 Thus, the overall magnification of the system is
q2 6.67 ( ) 1.33 p2 5
M M1M 2 (0.5) (1.33) 0.667 Note that the negative sign on the overall magnification indicates that the final image is inverted with respect to the initial object. The fact that the absolute value of the magnification is less than one tells us that the final image is smaller than the object. The fact that q2 is142 negative tells us that the final image is on the front, or left, side of lens 2.
Lens Aberrations
Assumptions have been:
The rays from a point object do not focus at a single point
Rays make small angles with the principal axis The lenses are thin
The result is a blurred image This is a situation where the approximations used in the analysis do not hold
The departures of actual images from the ideal predicted by our model are called aberrations 143
Spherical Aberration
This results from the focal points of light rays far from the principal axis being different from the focal points of rays passing near the axis For a camera, a small aperture allows a greater percentage of the rays to be paraxial For a mirror, parabolic shapes can be used to correct for spherical aberration 144
Chromatic Aberration
Different wavelengths of light refracted by a lens focus at different points Violet rays are refracted more than red rays The focal length for red light is greater than the focal length for violet light Chromatic aberration can be minimized by the use of a combination of converging and diverging lenses made of different materials 145
The Camera
The photographic camera is a simple optical instrument Components
Light-tight chamber Converging lens Produces a real image Film behind the lens Receives the image
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Camera Operation
Proper focusing will result in sharp images The camera is focused by varying the distance between the lens and the film
The lens-to-film distance will depend on the object distance and on the focal length of the lens
The shutter is a mechanical device that is opened for selected time intervals
The time interval that the shutter is opened is called the exposure time 147
Camera Operation, Intensity
Light intensity is a measure of the rate at which energy is received by the film per unit area of the image
The intensity of the light reaching the film is proportional to the area of the lens
The brightness of the image formed on the film depends on the light intensity
Depends on both the focal length and the diameter of the lens 148
Digital Camera
Digital cameras are similar in operation The image does not form on photographic film The image does form on a charge-coupled device (CCD)
This digitizes the image and turns it into a binary code The digital information can then be stored on a memory chip for later retrieval 149
The Eye
The normal eye focuses light and produces a sharp image Essential parts of the eye:
Cornea – light passes through this transparent structure Aqueous Humor – clear liquid behind the cornea
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The Eye – Parts, cont.
The pupil
A variable aperture An opening in the iris
The crystalline lens Most of the refraction takes place at the outer surface of the eye
Where the cornea is covered with a film of tears
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The Eye â&#x20AC;&#x201C; Close-up of the Cornea
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The Eye – Parts, final
The iris is the colored portion of the eye
It is a muscular diaphragm that controls pupil size The iris regulates the amount of light entering the eye
It dilates the pupil in low light conditions It contracts the pupil in high-light conditions
The f-number of the eye is from about 2.8 to 16
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The Eye – Operation
The cornea-lens system focuses light onto the back surface of the eye
This back surface is called the retina The retina contains sensitive receptors called rods and cones These structures send impulses via the optic nerve to the brain
This is where the image is perceived
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The Eye – Operation, cont.
Accommodation
The eye focuses on an object by varying the shape of the pliable crystalline lens through this process Takes place very quickly Limited in that objects very close to the eye produce blurred images
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The Eye – Near and Far Points
The near point is the closest distance for which the lens can accommodate to focus light on the retina
Typically at age 10, this is about 18 cm The average value is about 25 cm It increases with age Up to 500 cm or greater at age 60
The far point of the eye represents the largest distance for which the lens of the relaxed eye can focus light on the retina
Normal vision has a far point of infinity 156
The Eye – Seeing Colors
Only three types of color-sensitive cells are present in the retina
They are called red, green and blue cones
What color is seen depends on which cones are stimulated
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Conditions of the Eye
Eyes may suffer a mismatch between the focusing power of the lens-cornea system and the length of the eye Eyes may be:
Farsighted
Light rays reach the retina before they converge to form an image
Nearsighted
Person can focus on nearby objects but not those far away 158
Farsightedness
Also called hyperopia The near point of the farsighted person is much farther away than that of the normal eye The image focuses behind the retina Can usually see far away objects clearly, but not nearby objects 159
Correcting Farsightedness
A converging lens placed in front of the eye can correct the condition The lens refracts the incoming rays more toward the principal axis before entering the eye
This allows the rays to converge and focus on the retina 160
Nearsightedness
Also called myopia The far point of the nearsighted person is not infinity and may be less than one meter The nearsighted person can focus on nearby objects but not those far away 161
Correcting Nearsightedness
A diverging lens can be used to correct the condition The lens refracts the rays away from the principal axis before they enter the eye
This allows the rays to focus on the retina 162
Presbyopia and Astigmatism
Presbyopia (literally, “old-age vision”) is due to a reduction in accommodation ability
The cornea and lens do not have sufficient focusing power to bring nearby objects into focus on the retina Condition can be corrected with converging lenses
In astigmatism, light from a point source produces a line image on the retina
Produced when either the cornea or the lens or both are not perfectly symmetric Can be corrected with lenses with different curvatures in two mutually perpendicular directions 163
Diopters
Optometrists and ophthalmologists usually prescribe lenses measured in diopters
The power P of a lens in diopters equals the inverse of the focal length in meters
P = 1/ƒ
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Example 9: A particular nearsighted person is unable to see objects clearly when they are beyond 2.5 m away (the far point of this particular eye). What should the focal length be in a lens prescribed to correct this problem? Solution The purpose of the lens in this instance is to “move” an object from infinity to a distance where it can be seen clearly. This is accomplished by having the lens produce an image at the far point. From the thin lens equation, we have
1 1 1 p q f 1 1 1 2.5 m f f 2.5 m
We use a negative sign for the image distance because the image is virtual and in front of the eye. As you should have suspected, the lens must be a diverging lens (one with a negative focal length) to correct nearsightedness. 165
Simple Magnifier ď Ź
ď Ź
ď Ź
A simple magnifier consists of a single converging lens This device is used to increase the apparent size of an object The size of an image formed on the retina depends on the angle subtended by the eye
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The Size of a Magnified Image
When an object is placed at the near point, the angle subtended is a maximum
The near point is about 25 cm
When the object is placed near the focal point of a converging lens, the lens forms a virtual, upright, and enlarged image
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Angular Magnification
Angular magnification is defined as θ angle with lens m θo angle without lens
The angular magnification is at a maximum when the image formed by the lens is at the near point of the eye
q = - 25 cm 25 cm Calculated by m max 1
f
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Angular Magnification, cont.
The eye is most relaxed when the image is at infinity
Although the eye can focus on an object anywhere between the near point and infinity
For the image formed by a magnifying glass to appear at infinity, the object has to be at the focal point of the lens θ 25 cm The angular magnification is mmin θo
ƒ
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Example 10: What is the maximum magnification that is possible with a lens having a focal length of 10 cm, and what is the magnification of this lens when the eye is relaxed? Solution The maximum magnification occurs when the image is located at the near point of the eye
mmax
25 cm 25 cm 1 1 3.5 f 10 cm
When the eye is relaxed, the image is at infinity
mmin
25 cm 25 cm 2.5 f 10 cm 170
Magnification by a Lens
With a single lens, it is possible to achieve angular magnification up to about 4 without serious aberrations With multiple lenses, magnifications of up to about 20 can be achieved
The multiple lenses can correct for aberrations
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Compound Microscope
A compound microscope consists of two lenses
Gives greater magnification than a single lens The objective lens has a short focal length, ƒo< 1 cm The eyepiece has a focal length, ƒe of a few cm
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Compound Microscope, cont.
The lenses are separated by a distance L
The object is placed just outside the focal point of the objective
L is much greater than either focal length
This forms a real, inverted image This image is located at or close to the focal point of the eyepiece
This image acts as the object for the eyepiece
The image seen by the eye, I2, is virtual, inverted and very much enlarged 173
Active Figure 36.41
Use the active figure to adjust the focal lengths of the objective and eyepiece lenses Observe the effect on the final image
PLAY ACTIVE FIGURE 174
Magnifications of the Compound Microscope
The lateral magnification by the objective is
The angular magnification by the eyepiece of the microscope is
Mo = - L / ƒ o
me = 25 cm / ƒe
The overall magnification of the microscope is the product of the individual magnifications L M M o me ƒo
25 cm ƒ e 175
Other Considerations with a Microscope
The ability of an optical microscope to view an object depends on the size of the object relative to the wavelength of the light used to observe it
For example, you could not observe an atom (d 0.1 nm) with visible light (λ 500 nm)
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Telescopes
Telescopes are designed to aid in viewing distant objects Two fundamental types of telescopes
Refracting telescopes use a combination of lenses to form an image Reflecting telescopes use a curved mirror and a lens to form an image
Telescopes can be analyzed by considering them to be two optical elements in a row
The image of the first element becomes the object of the second element 177
Refracting Telescope
The two lenses are arranged so that the objective forms a real, inverted image of a distant object The image is formed at the focal point of the eyepiece
p is essentially infinity
The two lenses are separated by the distance ƒo + ƒe which corresponds to the length of the tube The eyepiece forms an enlarged, inverted image of the first image 178
Active Figure 36.42
Use the active figure to adjust the focal lengths of the objective and eyepiece lenses Observe the effects on the image
PLAY ACTIVE FIGURE 179
Angular Magnification of a Telescope
The angular magnification depends on the focal lengths of the objective and eyepiece
θ ƒo m θo ƒe
The negative sign indicates the image is inverted
Angular magnification is particularly important for observing nearby objects
Nearby objects would include the sun or the moon Very distant objects still appear as a small point of light
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Disadvantages of Refracting Telescopes
Large diameters are needed to study distant objects Large lenses are difficult and expensive to manufacture The weight of large lenses leads to sagging which produces aberrations
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Reflecting Telescope
Helps overcome some of the disadvantages of refracting telescopes
Replaces the objective lens with a mirror The mirror is often parabolic to overcome spherical aberrations
In addition, the light never passes through glass
Except the eyepiece Reduced chromatic aberrations Allows for support and eliminates sagging 182
Reflecting Telescope, Newtonian Focus
The incoming rays are reflected from the mirror and converge toward point A
At A, an image would be formed
A small flat mirror, M, reflects the light toward an opening in the side and it passes into an eyepiece
This occurs before the image is formed at A
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Examples of Telescopes
Reflecting Telescopes
Largest in the world are the 10-m diameter Keck telescopes on Mauna Kea in Hawaii
Each contains 36 hexagonally shaped, computercontrolled mirrors that work together to form a large reflecting surface
Refracting Telescopes
Largest in the world is Yerkes Observatory in Williams Bay, Wisconsin
Has a diameter of 1 m 184