Question

A torch submerged in water sends a light beam towards the surface of water at an angle of incidence of 30°. The angle of refraction in air is: (a) greater than 30° but less than 45º (b) less than 30° (c) equal to 30° (d) equal to 0° 55.​

Answer 1

The angle of refraction in air is (a) greater than 30° but less than 45º

Reason :

Using Snell's Law,

                    sin i × $u_{1}$ = sin r × $u_{2}$

(where i = angle of incidence; u$_{1}$ = refractive index of medium 1 i.e, water; u$_{2}$ = refractive index of medium 2; r = angle of refraction)

        or,      sin 30° × $\frac{4}{3}$ = sin r x 1

        or,      $\frac{1}{2}$ × $\frac{4}{3}$           = sin r

        or,         $\frac{2}{3}$             = sin r

        or,         r            =  41.8°

Angle of refraction(r) = 41.8°

Here the refractive angle is coming  41.8° which is greater than 30° but less than 45º.

Answer 2

Given:

• Torch light ray goes from water to air through an interface.

• Angle of incidence = 30°.

To find:

• Angle of refraction?

Calculation:

First of all, we know that :

• Refractive index of water is higher than refractive index of air.

So, applying SNELL'S LAW:

$\mu_{water} \sin(i) = \mu_{air} \sin(r)$

$\implies \sin(r) = \dfrac{\mu_{water}}{ \mu_{air} } \times \sin(i)$

• Now, the fraction will be greater than 1.

$\implies \sin(r) > \sin(i)$

$\boxed{ \implies \angle r > \angle i}$

Now, if we put correct values of refractive indices:

$\implies \sin(r) = \dfrac{ \frac{4}{3} }{1} \times \sin( {30}^{ \circ} )$

$\implies \sin(r) = \dfrac{ 4 }{3} \times \sin( {30}^{ \circ} )$

$\implies \sin(r) = \dfrac{ 4 }{3} \times \dfrac{1}{2}$

$\implies \sin(r) = \dfrac{2 }{3}$

$\boxed{ \implies \angle r = {41.8}^{ \circ} }$

So, OPTION a) IS CORRECT ✔️