Question

A ray of light is travelling from air into a medium in which its velocity is reduced to $\frac{3}{4}$ of its velocity in air. If it is incident at a very small angle on the surface of separation, and if i and r are the angles of incidence and reflection respectively, then the deviation of the ray is given by?

• A. $\delta =i-{\sin }^{-1}\left(\frac{4}{3}\sin i\right)$
• B. $\delta =\frac{i}{2}$
• C. $\delta =\frac{i}{4}$
• D. $\delta =2r$

Answer 1

The correct option is A $\delta =i-{\sin }^{-1}\left(\frac{4}{3}\sin i\right)$

We know that deviation of the ray $\delta$ = $angleofincidence\left(i\right)-angleofrefraction\left(r\right)$ -------(1)

Also from Snell's law we get

$\frac{\sin i}{\sin r}=\mu =\frac{v_m}{v_a}$,where $v_{m}and{v}_a$ are the velocity of light in medium and air respectively.

Given when light travels from air to medium its velocity decreases by a factor of $\frac{3}{4}$ times the speed of light in air.

Thus reducing the above values in Snell's law we get,

$\frac{\sin i}{\sin r}=\frac{3v_a}{4v_a}\sin r=\frac{4}{3}\sin ir={\sin }^{-1}\left(\frac{4}{3}\sin i\right)$

Thus equation 1 becomes

$\delta =i-{\sin }^{-1}\left(\frac{4}{3}\sin i\right)$