Question

A water film is formed on a glass-block. A light ray is incident on water film from air at an angle of ${60^\circ }$ with the normal. The angle of incidence on glass slab is:
( ${\mu _g} = 1.5,{\mu _w} = \dfrac{4}{3}$ ).
A. ${\sin ^{ - 1}}(\dfrac{{3\sqrt 3 }}{8})$
B. ${\sin ^{ - 1}}(\dfrac{1}{{\sqrt 3 }})$
C. ${\sin ^{ - 1}}(\dfrac{{4\sqrt 3 }}{9})$
D. ${\sin ^{ - 1}}(\dfrac{{9\sqrt 3 }}{{16}})$

Answer 1

To solve this question, first we will rewrite the given facts about the question and then apply Snell’s Law to find the angle of refraction which is equal to the angle of incidence in this case.

Complete step by step solution:
Firstly, rewrite the given information about the question:
Refractive Index of the glass, ${\mu _g} = 1.5$ ;
Refractive Index of the water, ${\mu _w} = \dfrac{4}{3}$ .
The angle of incidence on the glass slab is ${60^\circ }$ with the normal.
So, the angle of incidence on glass slab will be equal to the angle of refraction on the air water interface, i.e.. $r = i$ .
Applying Snell’s Law on air water interface, we get:
$\begin{gathered} \sin {60^\circ } = ({\mu _w}).\sin r \\\ \Rightarrow \dfrac{{\sqrt 3 }}{2} = \dfrac{4}{3}.\sin r \\\ \Rightarrow \sin r = \dfrac{{\sqrt 3 }}{2} \times \dfrac{3}{4} \\\ \Rightarrow \sin r = \dfrac{{3\sqrt 3 }}{8} \\\ \end{gathered}$
As we proved above, $r = i$ :
$\sin r = \sin i = \dfrac{{3\sqrt 3 }}{8}$
Therefore, $i = {\sin ^{ - 1}}\dfrac{{3\sqrt 3 }}{8}$
The angle of incidence on glass slab, $i = {\sin ^{ - 1}}\dfrac{{3\sqrt 3 }}{8}$ .
Hence, the correct option is (A.) ${\sin ^{ - 1}}(\dfrac{{3\sqrt 3 }}{8})$ .

Note:
Snell’s law tells us the degree of refraction and relation between the angle of incidence, the angle of refraction and refractive indices of a given pair of media. We know that light experiences the refraction or bending when it travels from one medium to another medium. Snell’s law predicts the degree of the bend.