Question

A ray of light passing through an equilateral triangular glass prism from air undergoes minimum deviation when angle of incidence is 3/4th of the angle of prism. Calculate the speed of light in the prism. (ans. 2.13 x 108m/s)

Answer 1

Answer:

A ray of light passing through an equilateral triangular glass prism from air undergoes minimum deviation when angle of incidence is 34th of the angle of prism. Calculate speed of light in prism. As μ=cv,v=cμ=3×108√2=2.12×108m/s.

Answer 2

Given:

A ray of light passing through an equilateral triangular glass prism from air undergoes minimum deviation when angle of incidence is 3/4th of the angle of prism.

To find:

Speed of light in the prism ?

Calculation:

Angle of incidence be "i" :

$\therefore \: \angle i = \dfrac{3}{4} \times \angle A$

$\implies\: \angle i = \dfrac{3}{4} \times {60}^{ \circ}$

$\implies\: \angle i = {45}^{ \circ}$

Now , minimum deviation occurs when the refracted ray travels parallel to the base of prism:

$\therefore \: \angle r = {90}^{ \circ} - {60}^{ \circ} = {30}^{ \circ}$

Now, applying Snell's Law:

$\therefore \: \mu1 \times \sin(i) = \mu2 \times \sin(r)$

$\implies \: 1 \times \sin( {45}^{ \circ} ) = \mu\times \sin( {30}^{ \circ} )$

$\implies \: \dfrac{1}{ \sqrt{2} } = \mu\times \dfrac{1}{2}$

$\implies \: \mu = \sqrt{2}$

So, speed of light in that medium :

$\therefore \: v = \dfrac{3 \times {10}^{8} }{ \mu}$

$\implies \: v = \dfrac{3 \times {10}^{8} }{ \sqrt{2} }$

$\implies \: v = \dfrac{3 \times {10}^{8} }{ 1.41 }$

$\implies \: v = 2.13 \times {10}^{8} \: m {s}^{ - 1}$

So, Velocity of light in the prism is 2.13 × 10⁸ m/s.