Question

Determine angle of deviation due to refraction at Air-Water interface as shown in the figure below $(\text{take}\sin {37}^{\circ }=\frac{3}{5})$

• A.
  ${53}^{\circ }$
• B.
  ${30}^{\circ }$
• C.
  ${16}^{\circ }$
• D.
  ${15}^{\circ }$

Answer 1

The correct option is C

${16}^{\circ }$
Given,
Refractive index of water, ${\mu }_1=1.33$

Refractive index of air, ${\mu }_1=1$

Incident angle, $\angle i={37}^{\circ }$

Let, $r$ be the angle of refraction.


Using snell's law we get,

${\mu }_1\sin i={\mu }_2\sin r$

$\Rightarrow 1.33\times \sin {37}^{\circ }=1\times \sin r$

$\Rightarrow \sin r=1.33\times \frac{3}{5}$

$\Rightarrow \sin r=\frac{3.99}{5}\simeq \frac{4}{5}$

$\therefore \angle r={53}^{\circ }$

So,
Angle of deviation, $\delta =\angle r-\angle i$
$={53}^{\circ }-{37}^{\circ }$
$={16}^{\circ }$

Hence, option (c) is the correct answer.

Why this question? To understand the application of snell's law in oblique incidence.