Question

A prism of cross section ABC and refractive index under root of 3/2 and angle CAB=8° is given. A ray of light incident on surface AB at an angle of incidence 60°. The angle of emergence will be

Answer 1

Given that,

Refractive index $\mu=\sqrt{\dfrac{3}{2}}$

Angle CAB =8°

Incidence = 60°

We need to calculate the minimum deviation

Using formula of deviation

$\mu=\dfrac{\sin\dfrac{A+\delta m}{2}}{\sin\dfrac{A}{2}}$

Put the value into the formula

$\dfrac{\sqrt{3}}{2}=\dfrac{\sin\dfrac{8+\delta m}{2}}{\sin\dfrac{8}{2}}$

$\sin\dfrac{8+\delta m}{2}=\dfrac{\sqrt{3}}{2}\times0.0697$

$\dfrac{8+\delta m}{2}=\sin^{-1}(0.0604)$

$8+\delta m=3.462\times2$

$\delta m=3.462\times2-8$

$\delta m = -1.07$

The deviation angle is very small.

So, When the minimum deviation is very small then the emergence angle will be equal to the incidence angle

$i_{i}=i_{e}$

Put the value into the formula

$i_{e}=60^{\circ}$

Hence, The angle of emergence will be 60°.