Question

A ray of light passes through an equilateral prism such that the angle of incidence is equal of emergence and the later is equal to $\frac{3}{4}$ th of the angle of the prism. Calculate the angle of deviation.

Answer 1

Given: A ray of light passes through an equilateral prism such that angle of incidence is equal of emergence and the later is equal to ${\left(\frac{3}{4}\right)}^{th}$ of the angle of prism.

To find the angle of deviation.

Solution:

If angle of incidence be $i$, angle of emergence be $e$ and angle of prism be $A$.

According to the given criteria,

$i=e=\frac{3}{4}A.......\left(i\right)$

Angle of deviation for the light in prism is,

$\delta =i+e-A$

$⟹\delta =2i-A$ (as given $i=e$)

$⟹\delta =2\left(\frac{3}{4}A\right)-A$ (from eqn(i))

$⟹\delta =\frac{3-2}{2}A=\frac{1}{2}A$

As the prism is an equilateral triangle, all the angles of the prism is equal to

${60}^{\circ }$ each, i.e., $A={60}^{\circ }$

$\delta =\frac{1}{2}\times 60={30}^{\circ }$

is the angle of deviation.