Question

A ray of light undergoes refraction and loses $0.25$ times its velocity. If 'i' is the angle of incidence and is very small, the angle of deviation for the ray of light is______

Answer 1

Step 1: Given data

1. The angle of incidence $=i$
2. $v_2=v_1-0.25v_1=0.75v_1$

Step 2: Formula used

1. ${\mu }_1\sin i={\mu }_2\sin r$
2. $\frac{{\mu }_2}{{\mu }_1}=\frac{{\nu }_1}{v_2}$
3. $\delta =i-r$

Step 3: Solution

Let $\delta$ be the angle of deviation

By Snell's law

${\mu }_1\sin i={\mu }_2\sin r......\left(1\right)$

Also,

$\frac{{\mu }_2}{{\mu }_1}=\frac{{\nu }_1}{v_2}......\left(2\right)\left[\because Refractiveindex=\frac{speedofthelightinvaccum}{speedoflightinthegivenmedium}\right]$

From $\left(2\right)$

$\frac{{\mu }_2}{{\mu }_1}=\frac{1}{0.75}$

Putting the above value in $\left(1\right)$

$\sin r=0.75·\sin i$

$\Rightarrow r=0.75i$ $\left[\because if\theta issmall,\sin \left(\theta \right)\approx \theta \right]$

So, the angle of deviation is, $\delta =i-r$

$$\begin{gathered} \Rightarrow \delta =i-0.75i \\ \Rightarrow \delta =0.25i \end{gathered}$$

Hence, the angle of deviation is $0.25i$.