Question

A light ray suffers minimum deviation while passing through a prism of refractive index $1.5$and refracting angle $60°$. Calculate the angle of deviation and angle of incidence.

Answer 1

Step 1: Given data

The angle of the prism, $A=60°$

Refractive index $n_{glass}=1.5$

For the minimum angle of deviation $\delta$, the angle of refraction $\left(r\right)$ should be half of the angle of the prism

$r=\frac{A}{2}=30°$

Step 2: To find

The angle of deviation $\delta$ and the angle of incidence $i$.

Step 3: Solution

Using snell's law

$$\begin{gathered} n_{air}\sin i=n_{glasses}\sin r \\ \Rightarrow \sin i=1.5\times \sin 30° \\ \Rightarrow i=48.6° \end{gathered}$$

For the minimum angle of deviation

$\delta =2i-A$

$$\begin{gathered} \delta =2(48.6)-60° \\ \delta =37.9° \end{gathered}$$

Hence, the angle of deviation is $37.9°$.