Refraction of Light at Plane Surfaces — Question 2

(a) As we know,

$_aμ_g = \dfrac{1}{\text {sin C}} \\[0.5em] = \text {cosec C}$

Refractive index is

$_aμ_g = \dfrac{3}{2}$

Substituting the values in the formula we get,

$\dfrac{3}{2} = \dfrac{1}{\text {sin C}} \\[1em] \text {sin C} = \dfrac{1}{\dfrac{3}{2}} \\[1em] \text {sin C} = \dfrac{2}{3}$

As,

$\text {sin 42°} = \dfrac {2}{3}$

Hence, critical angle for glass air surface is 42°.

(b) As we know,

$_aμ_w = \dfrac{1}{\text {sin C}} = \text {cosec C} \\[0.5em]$

Refractive index is

$_aμ_w = \dfrac {4}{3} \\[0.5em]$

Substituting the values in the formula we get,

$\dfrac{4}{3} = \dfrac{1}{\text {sin C}} \\[1em] \text {sin C} = \dfrac{1}{\dfrac{4}{3}} \\[1em] \text {sin C} = \dfrac{3}{4} \\[1em] \text {sin 49°} = \dfrac{3}{4} \\[0.5em]$

Hence, critical angle for water air surface is 49°.