Refraction of Light from Plane Surface, Critical Angle & Total Internal Reflection

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]

$1 : \sqrt{2}$
$1 : 2$
$\sqrt{2} : 1$
$2 : 1$

1

2

3

4

(3)

We know

$\sin c = \frac{n_{2}}{n_{1}} \Rightarrow \frac{n_{1}}{n_{2}} = \frac{1}{\sin c}$

$\Rightarrow \frac{n_{1}}{n_{2}} = \frac{1}{\sin 45} = \frac{\sqrt{2}}{1}$

$n_{1} : n_{2} = \sqrt{2} : 1$

Q 2 :
A light ray is incident on a glass slab of thickness $4 \sqrt{3}$ cm and refractive index $\sqrt{2}$ . 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 sin $15^{o}$ = 0.25) [2024]

0%

(2)

$i = θ_{c} \Rightarrow i = \sin^{- 1} (\frac{1}{μ})$

$\Rightarrow i = 45 °$

According to Snell’s law,

$1 \sin 45 ° = \sqrt{2} \sin r \Rightarrow \frac{1}{\sqrt{2}} = \sqrt{2} \sin r \Rightarrow \frac{1}{\sqrt{2}} = \sqrt{2} \sin r$

$\Rightarrow r = 30 °$

Now, $Δ x = t \sec 30 ° \sin 15 ° = 4 \sqrt{3} \times \frac{2}{\sqrt{3}} \times \frac{1}{4} = 2 cm$

Q 3 :
Two immiscible liquids of refractive indices $\frac{8}{5}$ and $\frac{3}{2}$ 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 $\frac{α}{4}$ cm. The value of $α$ is ____. [2024]

0%

(31)

$h_{app} = \frac{h_{1}}{μ_{1}} + \frac{h_{2}}{μ_{2}} = \frac{6}{8 / 5} + \frac{6}{3 / 2} = 4 + \frac{15}{4} = \frac{31}{4} cm$

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]

$\frac{h \tan (i - r)}{\tan r}$
$\frac{h \cos (i - r)}{\sin r}$
h
$\frac{h \sin (i - r)}{\cos r}$

1

2

3

4

(4)

AB = h sec r

BC = h sec r sin (i – r)

BC = $\frac{h \sin (i - r)}{\cos r}$

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]

$\sqrt{3}$
$\frac{3}{2}$
$\sqrt{2}$
$\frac{\sqrt{3}}{2}$

1

2

3

4

(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 $θ_{c}$ inside the liquid.

$μ = \frac{1}{\sin θ_{c}}$

$μ$ is minimum when $θ_{c}$ is maximum.

Maximum value of $θ_{c} = 45 °$

$\Rightarrow μ$ has a minimum value of $\sqrt{2}$ .

Q 6 :
At the interface between two materials having refractive indices $n_{1}$ and $n_{2}$ , the critical angle for reflection of an em wave is $θ_{1 C}$ . The $n_{2}$ material is replaced by another material having refractive index $n_{3}$ , such that the critical angle at the interface between $n_{1}$ and $n_{3}$ material is $θ_{2 C}$ . If $n_{3} > n_{2} > n_{1}$ ; $\frac{n_{2}}{n_{3}} = \frac{2}{5}$ and $\sin θ_{2 C} - \sin θ_{1 C} = \frac{1}{2}$ , then $θ_{1 C}$ is [2025]

$\sin^{- 1} (\frac{1}{6 n_{1}})$
$\sin^{- 1} (\frac{2}{3 n_{1}})$
$\sin^{- 1} (\frac{5}{6 n_{1}})$
$\sin^{- 1} (\frac{1}{3 n_{1}})$

1

2

3

4

(none)

$n_{2} \sin (θ_{1 C}) = n_{1} \Rightarrow s in (θ_{1 C}) = \frac{n_{1}}{n_{2}}$

$n_{3} \sin (θ_{2 C}) = n_{1} \Rightarrow s in (θ_{2 C}) = \frac{n_{1}}{n_{3}}$

Also, $n_{3} = \frac{5 n_{2}}{2} \Rightarrow s in (θ_{2 C}) = \frac{2 n_{1}}{5 n_{2}}$

$\Rightarrow \frac{n_{1}}{n_{2}} = \frac{5}{2} \cdot \sin (θ_{2 C}) \Rightarrow s in (θ_{2 C}) - \sin (θ_{1 C}) = \frac{1}{2}$

Given, $\frac{2 n_{1}}{5 n_{2}} - \frac{n_{1}}{n_{2}} = \frac{1}{2} \Rightarrow \frac{n_{1}}{n_{2}} (\frac{- 3}{5}) = \frac{1}{2}$

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]

$\tan^{- 1} (4 / 3)$
$\tan^{- 1} (3 / 4)$
$\sin^{- 1} (3 / 4)$
$\cos^{- 1} (3 / 4)$

1

2

3

4

(3)

$r + θ_{C} = 90 ° \Rightarrow r = 90 ° - θ_{C}$

At $1^{st}$ refraction : $1 \times \sin θ = \frac{5}{4} \sin r$

$\Rightarrow \sin r = \frac{4}{5} \sin θ$ ... (i)

At point TIR:

$\frac{5}{4} \sin θ_{C} = 1 . \sin 90 ° \Rightarrow \frac{5}{4} \sin θ_{C} = 1$

$\sin (90 - r) = \frac{4}{5} \Rightarrow \cos r = \frac{4}{5}$

$\Rightarrow \sin r = \frac{3}{5}$ ... (ii)

From (i) and (ii), $θ = \sin^{- 1} (\frac{3}{4})$

Q 8 :
Two light beams fall on a transparent material block at point 1 and 2 with angle $θ_{1}$ and $θ_{2}$ , 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, $d = 4 \sqrt{3}$ cm and $θ_{1} = θ_{2} = \cos^{- 1} (\frac{n_{2}}{2 n_{1}})$ , where refractive index of the block $n_{2}$ > refractive index of the outside medium $n_{1}$ , then the thickness of the block is __________ cm. [2025]

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(6)

$n_{1} \sin (90 - θ_{1}) = n_{2} \sin r$

$n_{1} \times \frac{n_{2}}{2 n_{1}} = n_{2} \sin r$

$\sin r = \frac{1}{2} \Rightarrow r = 30 °$

$\tan r = (\frac{d / 2}{t})$

$t = \frac{d}{2 \tan r} = \frac{d \sqrt{3}}{2} = \frac{(4 \sqrt{3}) \sqrt{3}}{2} = 6 cm$

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]

0%

(80)

Shift $△ t = 40 cm$

$△ t = 60 (1 - \frac{1}{1.2}) + H (1 - \frac{1}{1.6})$

$\Rightarrow 40 = (\frac{1}{6}) 60 + H (\frac{3}{8})$

$\Rightarrow 30 = H \times \frac{3}{8} \Rightarrow H = 80 cm$

Topic Question Set

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]

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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]

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Topic Question Set

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]

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]

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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]

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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]

0%

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Want Us to Email you About Special Offers & Updates?

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]

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]