Question

Angle of refraction of a ray of light travelling from air into glass is 30 $^{\circ }$ . Then the angle between incident ray and surface of separation of air and glass if refractive index of glass is 1.5 is:
$\begin{align}
  & (A){{\sin }^{-1}}(0.75) \\
 & (B){{90}^{\circ }} \\
 & (C){{180}^{\circ }} \\
 & (D){{60}^{\circ }} \\
\end{align}$

Answer 1

Hint: Refraction occurs when light falls from one medium to another. Snell's law governs refraction of light. It gives us the relation between angle of incidence, angle of refraction and the refractive index. In the above question, the angle between the incident ray and the surface separating air and glass is the same as the incident angle.
Formula used:
Snell’s law is given by
$\eta =\dfrac{\sin (i)}{\sin (r)}$

Complete answer:
Given,
The refractive index is given as 1.5
Angle of refraction is given as=30$^{\circ }$
To find: the incident angle
The Snell’s law is given by,
$\eta =\dfrac{\sin (i)}{\sin (r)}$ ………… (1)
Here, $\eta $is the refractive index.
i is the angle of incidence.
r is the angle of refraction.
Plugging the values of r and $\eta $ in equation (1),
$\begin{align}
  & 1.5=\dfrac{\sin (i)}{\sin ({{30}^{^{\circ }}})} \\
 & \Rightarrow \sin ({{30}^{\circ }})\times 1.5=\sin (i) \\
 & \Rightarrow \dfrac{3}{4}=\sin (i) \\
 & \Rightarrow i={{\sin }^{-1}}(0.75) \\
\end{align}$
The angle of incidence is given by the Snell’s law and is calculated as ${{\sin }^{-1}}(0.75)$

So, the correct answer is “Option A”.

Additional Information:
With the help of Snell’s law, a quantitative description of refraction is made clear. When a ray of light is travelling from a medium of higher refraction index to a medium of lower refractive index, the ray of light bends away from the normal. On the other hand, when the ray of light is travelling from a medium of lower refractive index to a medium of higher refractive index, the ray of light moves towards the normal.

Note:
While solving this problem, we must be able to point out the angle of refraction and the angle of incidence and then proceed with the Snell’s law. The trigonometric value of certain angles must be kept in mind. In this problem, the final answer is expressed in inverse trigonometric ratio.