Q 1 :

Critical angle of incidence for a pair of optical media is 45°. The refractive indices of first and second media are in the ratio:           [2024]

  • 1:2

     

  • 1:2

     

  • 2:1

     

  • 2:1

     

(3)             

                 We know

                  sinc=n2n1n1n2=1sinc

                  n1n2=1sin45=21

                   n1:n2=2:1

 



Q 2 :

A light ray is incident on a glass slab of thickness 43 cm and refractive index 2. The angle of incidence is equal to the critical angle for the glass slab with air. The lateral displacement of ray after passing through glass slab is ____ cm.

(Given sin15o = 0.25)                               [2024]



(2)

i=θci=sin-1(1μ)

i=45°

According to Snell’s law,

1sin45°=2sinr12=2sinr12=2sinr

r=30°

Now, Δx=tsec30°sin15°=43×23×14=2 cm



Q 3 :

Two immiscible liquids of refractive indices 85 and 32 respectively are put in a beaker as shown in the figure. The height of each column is 6 cm. A coin is placed at the bottom of the beaker. For near normal vision, the apparent depth of the coin is α4cm. The value of α is ____.         [2024]



(31)

happ=h1μ1+h2μ2=68/5+63/2=4+154=314 cm

 



Q 4 :

What is the lateral shift of a ray refracted through a parallel sided glass slab of thickness 'h' in terms of the angle of incidence 'i' and angle of refraction 'r', if the glass slab is placed in air medium?          [2025]

  • h tan (ir)tan r

     

  • h cos (ir)sin r

     

  • h

     

  • h sin (ir)cos r

     

(4)

AB = h sec r

BC = h sec r sin (ir)

BC = h sin (ir)cos r



Q 5 :

A hemispherical vessel is completely filled with a liquid of refractive index μ. A small coin is kept at the lowest point (O) of the vessel as shown in figure. The minimum value of the refractive index of the liquid so that a person can see the coin from point E (at the level of the vessel) is ________.          [2025]

  • 3

     

  • 32

     

  • 2

     

  • 32

     

(3)

For the rays from coin to reach the point E, the refracted rays must graze the surface, i.e., they must be incident at critical angle θc inside the liquid.

μ=1sin θc

μ is minimum when θc is maximum.

Maximum value of θc=45°

 μ has a minimum value of 2.



Q 6 :

At the interface between two materials having refractive indices n1 and n2, the critical angle for reflection of an em wave is θ1C. The n2 material is replaced by another material having refractive index n3, such that the critical angle at the interface between n1 and n3 material is θ2C. If n3>n2>n1n2n3=25 and sin θ2Csin θ1C=12, then θ1C is           [2025]

  • sin1(16n1)

     

     

  • sin1(23n1)

     

  • sin1(56n1)

     

  • sin1(13n1)

     

(none)

n2 sin(θ1C)=n1  sin(θ1C)=n1n2

n3 sin(θ2C)=n1  sin(θ2C)=n1n3

Also, n3 =5n22  sin(θ2C)=2n15n2

 n1n2=52·sin(θ2C)  sin(θ2C)sin(θ1C)=12

Given, 2n15n2n1n2=12  n1n2(35)=12

Coming out to be (–ve).



Q 7 :

A transparent block A having refractive index μ = 1.25 is surrounded by another medium of refractive index μ = 1.0 as shown in figure. A light ray is incident on the flat face of the block with incident angle θ as shown in figure. What is the maximum value of θ for which light suffers total internal reflection at top surface of the block?          [2025]

  • tan1(4/3)

     

  • tan1(3/4)

     

  • sin1(3/4)

     

  • cos1(3/4)

     

(3)

 

r+θC=90°  r=90°θC

At 1st refraction : 1×sin θ=54sin r

 sin r=45sin θ          ... (i)

At point TIR:

54sin θC=1.sin 90°  54sin θC=1

sin(90r)=45  cos r=45

 sin r=35          ... (ii)

From (i) and (ii), θ=sin1(34)



Q 8 :

Two light beams fall on a transparent material block at point 1 and 2 with angle θ1 and θ2, respectively, as shown in figure. After refraction, the beams intersect at point 3 which is exactly on the interface at other end of the block.

Given: the distance between 1 and 2, d=43 cm and θ1=θ2=cos1(n22n1), where refractive index of the block n2 > refractive index of the outside medium n1, then the thickness of the block is __________ cm.          [2025]



(6)

n1 sin (90θ1)=n2 sin r

n1×n22n1=n2 sin r

sin r=12  r=30°

tan r=(d/2t)

t=d2 tan r=d32=(43)32=6 cm



Q 9 :

A container contains a liquid with refractive index of 1.2 up to a height of 60 cm and another liquid having refractive index 1.6 is added to height H above first liquid. If viewed from above, the apparent shift in the position of bottom of container is 40 cm. The value of H is __________ cm. (Consider liquids are immisible)          [2025]



(80)

Shift t=40 cm

t=60(111.2)+H(111.6)

 40=(16)60+H(38)

30=H×38  H=80 cm



Q 10 :

A monochromatic light wave with wavelength λ1 and frequency ν1 in air enters another medium. If the angle of incidence and angle of refraction at the interface are 45° and 30° respectively, then the wavelength λ2 and frequency ν2 of the refracted wave are                         [2023]

  • λ2=12λ1, ν2=ν1

     

  • λ2=λ1, ν2=12ν1

     

  • λ2=2λ1, ν2=ν1

     

  • λ2=λ1, ν2=2ν1

     

(1)

Snell's law μ1sin45°=μ2sin30°

μ1μ2=12

 μ1μ2=λ2λ1=12

 λ2=λ12

Frequency doesn't change on change in medium.



Q 11 :

The critical angle for a denser-rarer interface is 45°. The speed of light in rarer medium is 3×108 ms. The speed of light in the denser medium is             [2023]

  • 5×107 m/s

     

  • 2.12×108 m/s

     

  • 3.12×107 m/s

     

  • 2×108 m/s

     

(2)

iC = Critical angle

vC=1μ=siniC=sin45°=12

 v=C2=3×1082m/s=2.12×108 m/s



Q 12 :

An ice cube has a bubble inside. When viewed from one side the apparent distance of the bubble is 12 cm. When viewed from the opposite side, the apparent distance of the bubble is observed as 4 cm. If the side of the ice cube is 24 cm, the refractive index of the ice cube is                    [2023]

  • 43

     

  • 32

     

  • 23

     

  • 65

     

(2)

dapparent=dactualμrel

12=xμ    (i)

d=24-xμ    (ii)

On solving we get μ=1.5



Q 13 :

A vessel of depth 'd' is half filled with oil of refractive index n1 and the other half is filled with water of refractive index n2. The apparent depth of this vessel when viewed from above will be                        [2023]

  • dn1n2(n1+n2)

     

  • d(n1+n2)2n1n2

     

  • dn1n22(n1+n2)

     

  • 2d(n1+n2)n1n2

     

(2)

Formula used dapp=d1n1+d2n2

dapp=d2[n1+n2n1n2]



Q 14 :

A ray of light is incident from air on a glass plate having thickness 3 cm and refractive index 2. The angle of incidence of a ray is equal to the critical angle for glass-air interface. The lateral displacement of the ray when it passes through the plate is _______ ×10-2 cm (given sin15°=0.26)                  [2023]



(52)

sini=2sinr

sinr=12r=30°

μ=1sinc

c=45°

Lateral displacement,    x=tsin(i-r)cosr

x=3sin(45-30)cos30°

=3×sin15°(32)=2×0.26=0.52 cm

=52×10-2 cm



Q 15 :

A pole is vertically submerged in swimming pool, such that it gives a length of shadow 2.15 m within water when sunlight is incident at an angle of 30° with the surface of water. If swimming pool is filled to a height of 1.5 m then the height of the pole above the water surface in centimeters is (nw=4/3) _______.              [2023]



(50)

By Snell's law

    1sin60°=43sinr  sinr=338  tanr=3337

By the diagram

    x3+1.5tanr=2.15

     0.50 meter  x=50 cm



Q 16 :

A fish rising vertically upward with a uniform velocity of 8 ms-1 observes that a bird is diving vertically downward towards the fish with the velocity of 12 ms-1. If the refractive index of water is 43, then the actual velocity of the diving bird to pick the fish, will be _____ ms-1.                   [2023]



(3)

Vb/f(43)=-8(43)+(-v)1  -12(43)=-8(43)+(-v)1

  v=3 m/s



Q 17 :

Consider light travelling from a medium A to medium B separated by a plane interface. If the light undergoes total internal reflection during its travel from medium A to B and the speed of light in medium A and B are 2.4×108 m/s and 2.7×108 m/s respectively, then the value of critical angle is:            [2026]

  • tan-1(817)

     

  • cot-1(313)

     

  • sin-1(98)

     

  • cos-1(89)

     

(1)

μAsinc=μBsin90

sinc=μBμA=vAvB

   sinc=2.4×1082.7×108=89

tanc=881-64=817

c=tan-1(817)



Q 18 :

The wavelength of light, while it is passing through water is 540 nm. The refractive index of water is 43. The wavelength of the same light when it is passing through a transparent medium having refractive index of 32 is ______ nm.       [2026]

  • 840

     

  • 380

     

  • 480

     

  • 540

     

(3)

μ1μ2-v2v1=λ2λ1|,    v=fλ

μ1μ2=λ2λ1

μB3/2=λ540

λ=(4×2×5403×3)

λ=480 nm