Question 6The refractive index of glass is 1.5. From a point P inside a glass slab, draw rays PA, PB and PC incident on the glass-air surface at an angle of incidence 30°, 42° and 60° respectively.(a) In the diagram show the approximate direction of these rays as they emerge out of the slab.(b) What is the angle of refraction for the ray PB?(Take sin 42° = 2 / 3)

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Accepted Answer

(a) Diagram with rays PA, PB and PC is shown below:

The refractive index of glass is 1.5. From a point P inside a glass slab, draw rays PA, PB and PC incident on the glass-air surface at an angle of 30°, 42° and 60° respectively. In the diagram show the approximate direction of these rays as they emerge out of the slab. What is the angle of refraction for the ray PB? Refraction of light at plane surfaces, Concise Physics Class 10 Solutions.

(b) Given,

μ_g = 1.5

As we know,

$\text {sin i}_c = \dfrac {1}{μ} \\[0.5em]$

Substituting the values in the formula we get,

$\text {sin i}_c = \dfrac{1}{1.5} \\[0.5em] \Rightarrow \text {sin i}_c = 0.667 \\[0.5em] \Rightarrow i_c = 41.8$

Hence, we can round off i_c = 42°

Applying,

$\dfrac{\text {sin r}}{\text {sin i}} = _a\mu_g \\[0.5em] \text {sin r} = _a\mu_g \times \text {sin i} \\[0.5em] \text {sin r} = _a\mu_g × \text {sin 42°} \\[0.5em]$

Given,

$\text {sin 42°} = \dfrac{2}{3}$

and

$_aμ_g = \dfrac{3}{2}$

Substituting the values in the formula we get,

$\text{sin r} = \dfrac{3}{2} \times \dfrac{2}{3} \\[0.5em] \Rightarrow \text {sin r} = 1 \\[0.5em]$

As sin r = 1 , so angle of refraction = 90°.

Refraction of Light at Plane Surfaces — Question 6

Question 6