Refraction of Light (Plain and Curved Surfaces) Refraction (The spoken/unspoken rules) As all angles are to be measured from Normal, so DRAW ‘normal’ first, All distances are to be measured from optical centre C (Treat as Origin) Put arrows on rays (else get zero for diagram) Dashed lines – imaginary, solid lines – actual rays (with arrow) [“Diagram” – GOOD / SENSIBLE Centre: For curved surface draw full circle, for lens draw two intersecting circles]
Mostly there are 3 things that you keep doing in ray optics in refraction – 1) Diagrams, 2) Snell’s law and 3) for small angles sin i = i = tan i [When you come to curved surface there is just one more addition – the formula for refraction at curved surfaces] Refractive Index
µ1
i1
µ2
W
µG =
i2
c Absolute = vm , Relative
µG µW
1
µ2 =
1 2 µ1
Principal of reversibility of light. Its consequence Laws of refraction Law 1 - Incident ray, refracted ray, normal lie in same plane Law 2 - Snell’s Law use this form; µ1 sin i1 = µ 2 sin i2
[AI 2005C] Monochromatic light is refracted from air into glass of refractive index n. Find the ratio of wavelengths of the incident and refracted light. Question on Snell’s Law [IIT 1999] The X-Y plane is the boundary between two transparent media. Medium 1 with z 0 has refractive index 2 and the medium 2 with z 0 has refractive index r ˆ ˆ ˆ 3. A ray of light in medium 1 given by the vector A = 6 3i + 8 3 j − 10k is incident on the plane of separation. Find the unit vector in the direction of the refracted ray in medium 2. REFRACTION by Dr Rajeev Tyagi
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Deviation of ray due to refraction Condition for deviation – Either 1) The mediums must be different (If there are more than one interfaces, then the first and last mediums must be different, OR 2) the interfaces must not be parallel (Prism). Condition for ‘No Bending’ – For a single interface – Either a) the angle of incidence must be zero (normal incidence), OR b) the two mediums have same refractive index. (for more than 2 mediums – the first and the last medium are same). For more than one interface, additionally the interfaces must be parallel. Lateral shift / lateral displacement – better derive! Q. On what factors does the lateral shift depend?
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Apparent depth / Apparent Shift / Apparent height d actual App. depth = µdenser or =
d actual μ relative
App. height = µdenser × hactual If more than one medium, then treat each one to be of d/ thickness. If other than air, take to be relative.
[NCERT] A diver under water, looks obliquely at a fisherman standing on the bank of a lake. Would the fisherman appear taller or shorter? [NCERT Similar] A tank of water is looked upon vertically and then obliquely. In which case does it appear deeper? Q. As shown in the fig., a point source of light is placed at the bottom of a tank filled with water ( μ = 4/3) to a depth of 4m. Find the apparent depth of the source below the surface as viewed by a) observer 1(vertical), and b) observer 2 (larger angle). Q. A container completely filled with water ( μ = 4/3) has a scratch at its bottom. An observer looks at the scratch from a height 1m above the surface of water. A small hole is now made in the container such that water level reduces to half. It appears to the observer that the scratch has moved AWAY by a distance of 25 cm. Find the initial height of water. [h = 2m] Q. A fish rising vertically to the surface of water in a lake uniformly at the rate of 3 m/s observes a bird diving vertically towards the water at a rate of 9 m/s vertically above it. Find the actual velocity of the bird. [4.5 m/s] Q. A piece of plane glass is placed on a word with letters of different colours. The letters which appear minimum raised are ____ [Red] Application – Travelling microscope expt [NCERT] A beaker is filled with water to a height of 12.5 cm. The apparent depth of a needle lying at the bottom of the beaker is measured by a microscope to be 9.4 cm. What is the refractive index of water? If water is replaced by a liquid of refractive index 1.63 upto the same height, by what distance would the microscope have to be moved to focus on the needle again? [1.33, 1.7 cm] Dealing with questions where position of image is asked and there are other mediums in front of glass slab etc. followed by mirror (You have to first find how far do “you see the mirror”. That much away from the mirror will you see the image)
Q. A vessel having perfectly reflecting plane bottom is filled with water ( μ = 4/3) to a depth d. A point source of light is placed at a height h above the surface of water. Find the distance of final image from the water surface. [h + (3d/2)]
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Dealing with situations involving curved mirrors FOUR cases:
Image Dist in Med. will be more
Object dist will appear to be more
Dist in air will be less
Object dist will appear to be less
Q. A concave mirror of radius of curvature 1m is placed at the bottom of a tank of water. The mirror forms an image of the sun when it is directly overhead. Calculate the distance of the image from the mirror for a)80 cm and b) 40 cm of water in the tank. [a) 50 cm, b) 47.5 cm] Q. An object is placed 21 cm in front of a concave mirror of R = 10 cm. A glass slab of thickness 3 cm and Îź = 1.5 is then placed close to the mirror in the space between object and mirror. Find the position of the final image formed. Take distance of near surface of slab from mirror to be 1 cm. [7.67 cm] Îź [IIT 2005] A container is filled with water ( = 1.33) upto a height of 33.25 cm. A concave mirror is placed 15 cm above the water level and the image of an object placed at the bottom is formed 25 cm below the water level. The focal length of the mirror is ___ [out of four options (10, 15, 20, 25), 20 cm is nearest]
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Critical angle & Total Internal Reflection When light goes from denser to rarer medium (Java Applet on TIR) µ1 sin i1 = µ 2 sin i2 Put i1 = iC and i2 = 90o In case of air, sin (iC) = 1/μ Also, According to Cauchy’s relation B µ = A+ 2
λ
Which colour shows TIR first? Result – Violet bends the most Red bends the least If Red is showing TIR, then all other colours are definitely showing TIR. In general, if the higher wavelength is showing TIR, then lower wavelengths will surely be going through TIR. [AI 2009] (i) What is the relation between critical angle and refractive index of a material? (ii) Does critical angle depend on the colour of light? Explain. [AI 2009] (i) State the principle on which the working of an optical fibre is based. (ii) What are the necessary conditions for this phenomenon to occur? [Delhi 2011] part (b) Explain briefly how the phenomenon of total internal reflection is used in fibre optics. [NCERT] A small bulb is placed at the bottom of a tank containing water to a depth of 80 cm. What is the area of the surface of water through which light from the bulb can emerge out? [2.6 m2] [NCERT Exemp] The minimum area of a disc is asked which if placed symmetrically will make a dot at bottom invisible, when seen from above. (h and μ are given) [Foreign 2009] State the condition under which total internal reflection occurs. [EAMCET 2003] A ray of light is incident on the hypotenuse of a right angled prism after travelling parallel to the base inside the prism. The max value of base angle for which light is totally reflected from the hypotenuse is _____ [cos-1(1/μ)] Q. Two media having speeds of light 2x108 m/s and 2.4 x 108 m/s are separated by a plane surface. What are the possible angle for a ray going from medium I to medium II? OPTICAL FIBRES as Application of TIR
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Questions on TIR in Entrance Exams [IIT 2004] While light is incident on the interface of glass and air as shown, if green light is just totally internally reflected then the emerging ray in air contains ______ [Yellow, Orange, Red] [IIT 2010] A ray OP of monochromatic light is incident on the face AB of prism ABCD near vertex B at an incident angle of 60° (see figure). If the refractive index of the material of the prism is √3 , which of the following is (are) correct? A) The ray gets totally internally reflected at face CD B) The ray comes out through face AD C) The angle between the incident ray and the emergent ray is 90° D) The angle between the incident ray and the emergent ray is 120° (A, B, C) Using snell’s law Sin-1(1/√3) < sin-1 (1/√2) Net deviation is 90°
[IIT 2005, 2M] AB and CD are two slabs. The medium between the slabs has refractive index 2. Find the minimum angle of incidence at Q, so that the ray is totally reflected by both the slabs. A
Q
= 2
B
= 3
D
=2 C
[IIT 1986, 6M] Monochromatic light is incident on a plane interface AB between two media of refractive indices n1 and n2 (n2 > n1) at an angle of incidence as shown in the figure. The angle is infinitesimally greater than the critical angle for the two media so that total internal reflection takes place. Now if a transparent slab DEFG of uniform thickness and of refractive index n3 is introduced on the interface (as shown), show that for any value of n3 all light will ultimately be reflected back again into medium II. Consider separately the cases: a) n3 < n1, b) n3 > n1.
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Medium 1 n1
D
A
E
Medium 3 n3
G
θ
F
B
Medium 2 n2
[IIT 1992, 8M] Light is incident at angle on one planar end of a transparent cylindrical rod of refractive index n. Determine the least value of n so that the light entering the rod does not emerge from the curved surface of the rod irrespective of the value of . θ
[Ans. 2 ]
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Prism : Overall picture : You need to – a) Draw diagram, b) Apply Snell’s law (on left and right face of the prism), c) Remember expressions r1 + r2 = A and i1 + i2 = A + Terms – a) Angle of the Prism or refracting angle of the prism b) Refracting surfaces / faces of the prism [The diagram below is not the typical diagram. It is valid only at MINIMUM DEVIATION] A
M1
i2 – r2
i 1 – r1
M2
i1
i2 = e r1
r2 N
In M1M2N r1 + r2 + N = 180 In Quad. AM1NM2 A + 90 + N + 90 = 360 So r1 + r2 = A = (i1 – r1) + (i2 – r2) => i1 + i2 = A +
…1 …2
From these two, we get the following two results – a) For small angle Prism (thin prism), from r1 + r2 = A, if A is small then both r1, r2 and also i1, i2 are small. Using Snell’s law, = ( - 1)A A + δm Sin( ) 2 µ= Sin( A / 2) b) At minimum deviation, i1 = i2 = i, r1 = r2 = r, so [Delhi 2010] Graph of δ vs I [Delhi 2009] Condition for minimum deviation – a) i1 = i2 b) r1 = r2 c) ray inside the prism is parallel to the base [Delhi 2011, Foreign 2011] Draw ray diagram to show refraction of a ray of monochromatic light passing through a glass prism. Deduce expression for the refractive index of glass in terms of angle of prism and angle of minimum deviation. Q. A ray of light incident at 49o on the face of an equilateral prism passes symmetrically. Calculate the refractive index of the material of the prism. [ = 1.5] REFRACTION by Dr Rajeev Tyagi
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[NCERT Exemp] For a glass prism ( = √3) the angle of minimum deviation is equal to the angle of prism. Find the angle of the prism. [Delhi 2008] A ray of light passing through an equilateral triangular glass from air undergoes minimum deviation when angle of incidence is 3/4th of the angle of prism. Calculate the speed of light in the prism. Condition for maximum deviation Either i = 90 OR e = 90
θC Condition for No Emergence: A > 2 [DCP62] Q. A ray of light is incident on a prism ( = 1.5) at an angle of 600. The refracting angle of the prism is also 600. Find the angle of emergence and the angle of deviation. Is there any other angle of incidence which will produce the same deviation? Q. A ray of light makes an angle of 600 on a prism and suffers a deviation of 300 on emergence from the other surface. If A = 300 show that the emergent ray is perpendicular to the other surface. Also calculate the refractive index of the prism. Q. A ray of light falls on one side of a prism whose refracting angle is 600. Find the angle of incidence in order that the emergent ray may just graze the other face ( = 1.5) [ i = 27o55’] Questions on small angle prism [AI 2005] A right-angled (isosceles) crown glass prism with critical angle 41o is placed before an object PQ in two positions as shown in fig (i) and (ii). Trace the path of the rays from P and Q passing through the prisms in the two cases.
[NCERT] At what angle should a ray of light be incident on the face of a prism of refracting angle 60o so that it suffers total internal reflection at the other face? Refractive index of material of prism is 1.524 [30o] [Foreign 2009] One face of a prism with refracting angle 30o is coated with silver. A ray incident on another face at an angle of 45o is refracted, and reflected from the silver coated face and retraces its path. Find the refractive index of the material of the prism. IIT questions on Prism [IIT 1991; 2+2+4M] Two parallel beam of light P and Q (separation d) containing radiations of wavelengths 4000 A0 and 5000 A0 are incident normally on a prism as shown. The refractive index of the prism as a function of wavelength is given by the b µ (λ ) = 1.20 + 2 λ . The value of b is such that that condition for TIR at the relation face AC is just satisfied for one wavelength and is not satisfied for the other. a) Find the value of b, b) Find the deviation of the beams, REFRACTION by Dr Rajeev Tyagi
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c) A convergent lens is used to bring these transmitted beams into focus. If the intensities of upper and lower beam after transmission are 4I and I respectively, find the resultant intensity at the focus. A sin = 0.8 P Q
Ans: (a) b = 8x10
-15
B C 0 0 m , (b) (4000) = 37 , (5000) = 27.13 , (c) 9I 2
[Foreign 2011] part (b) (Just like above diagram) – Three light rays of red, green, and blue are incident on the right angled isoceles prism at face AB. The refractive index for red, green and blue wavelengths are respectively 1.39, 1.44, 1.47. Trace the paths of these rays reasoning out the difference in their behaviour.
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[IIT 1996, 3M] A right angle prism of refractive index n has a plane of refractive index n1 (n1 < n) cemented to its diagonal face. The assembly is in air. The ray is incident on AB. a) Calculate the angle of incidence at AB for which the ray strikes the diagonal face at the critical angle, b) Assuming n = 1.352, calculate the angle of incidence at AB for which refracted ray passes through the diagonal face undeviated. A
n
n1 450
B { Ans: i1 = sin-1
C
1 ( n 2 − n12 − n1 )} 2 , (b) 730
[IIT 2003, 4M] (The second part is on thin films) A prism of refracting angle 300 is coated with a thin film of = 2.2 on face AC. A light of wavelength 6600 A0 is incident on face AB such that angle of incidence is 600. Find a) the angle of emergence, b) the minimum value of thickness of the coated film on AC for which light emerging from the face has maximum intensity. (Given = 3 for prism). A 300 600 = 2.2 = 3 for part (b) x = 2t = nλ Ans: a) zero, b) 1500 A0 [IIT 2005, 4M] A ray of light is incident on a prism ABC of refractive index 3 as shown in the fig. a) Find the angle of incidence for which the deviation of light ray by the prism ABC is minimum, b) By what angle the second prism must be rotated so that the final ray suffer net minimum deviation.
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B
D
600
600
600 A Ans: a) 600, b) 600
C
E
[IIT 1998, 8M] A prism of refractive index n1 and another of n2 are struck together with a gap as shown. The angles of the prism are shown. n1 and n2 depend on λ according to D C
700 n2
n1 200 600 A
400 B
n1 = 1.20 +
10.8 ×10 4 λ2 and
1.80 ×10 4 λ2 (λ is in nm). a) Calculate the wavelength λ0 for which rays incident at any angle on the interface BC pass through without bending at that interface. b) For light of wavelength λ0, find the angle of incidence i on the face AC such that the deviation produced by the combination of prisms is minimum.
n2 = 1.45 +
[Ans: a) 600 nm, b) sin-1(3/4) ] [IIT 1987, 7M] A right prism is to be made by selecting a proper material and the angles A and B (B A), as shown in the figure. It is desired that a ray of light incident on the face AB emerges parallel to the incident direction after two internal reflections. a) What should be the minimum refractive index n for this to be possible? b) for n = 5/3 is it possible to achieve this with the angle B equal to 300?
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[Ans. a) ď&#x192; 2, b) No ]
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Dispersion – = ( - 1)A [true only if A is small] b µ = a+ 2 λ , So As according to Cauchy’s relation (deviation) depends on [Remember – Violet bends the most, red the least] [Delhi 2010] Angular dispersion - = V - R Dispersive power – δV − δR µV − µ R θ δY = δ mean = = µY − 1 If for Blue and Red is given (Or for any two colours) then B - R will give angular dispersion for these two colours.
µY =
If Y is not given then we take the mean of V and R i.e. Also if we need for a prism, then it is found the same way.
µV + µ R 2
Q. Calculate the dispersive power of crown and flint glass prism from given data – For crown glass V = 1.522, R = 1.514 For flint glass V = 1.662, R = 1.644 For crown glass µ − µ R 1.522 − 1.514 µ + µR ω= V = µY = V µY − 1 1.518 − 1 = 0.01544 2 = 1.518, Dispersive power Similarly for flint glass = 1.653, ω = 0.276 Dispersion without Deviation (Direct vision spectroscope)
δ = ( µ − 1)A
δ ' = ( µ '− 1) A '
As the combination produces no deviation, so A' µ −1 =− µ '− 1 δ + δ ' = 0 => A ' ' ' Net Angular dispersion = 1 + 2 = [( µV − µ R ) A] + [( µ V − µ R ) A ] Substituting for A’/A, we get Net angular dispersion = ( - 1)A[ω – ω’]
Q. Find the angle of the flint glass prism which should be combined with a crown glass prism of 5o so as to give dispersion but no deviation. For crown glass, V = 1.523, R = 1.515 For flint glass, V = 1.688, R = 1.650 A' µ −1 =− µ '− 1 For no deviation, A REFRACTION by Dr Rajeev Tyagi
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µ=
µV + µ R 2 = 1.519
µ'=
µ 'V + µ 'R 2 = 1.659
So A’ = -3.94o Deviation without Dispersion (Achromatic combination of Prism) –
θ1 = δV − δ R = ( µV − µ R ) A . Similarly θ 2 = δ 'V − δ ' R = ( µ 'V − µ ' R ) A' As there is no dispersion, so 1 + 2 = 0 which gives µ − µR A' = −[ V ] A µ 'V − µ 'R This is the condition for Achromatism. Another form – from [( µV − µ R ) A] + [( µ 'V − µ 'R ) A '] = 0
µV − µ R µ ' − µ 'R ( µ − 1) A] + [ V ( µ '− 1) A '] = 0 µ '− 1 we have µ − 1 ωδ + ω ' δ ' = 0 [
Net deviation produced by the combination is = δ + δ ' =
( µ − 1) A[1 −
ω ] ω'
[IIT 2001, 5M] The refractive indices of the crown glass for blue and red light are 1.51 and 1.49, and those of flint glass are 1.77 and 1.73. An isosceles prism of angle 60 is made of crown glass. A beam of white light is incident at a small angle on this prism. The other flint glass isosceles prism is combined with the crown glass prism such that there is no deviation of incident light. a) Determine the angle of flint glass prism, b) Calculate the net dispersion of the combined system. (For no deviation put 1 + 2 = 0. Being a small angle prism = ( - 1). The result A2 = 40 is –ve, means the second prism is kept reversed. Net dispersion = 1 + 2 = (1B - 1R)(6) + (2B - 2R)(-4) = -0.040)
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