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

Solution

Angle of prism, A = 60^o
Refractive index of prism = 1.5
Given, angle of incidence , i₁=i₂ = $\frac{3}{4} A$ = $\frac{3}{4} 45^{o}$ = 45^oAs, A + $δ$ = i₁ + i₂
$\Rightarrow$ 60^o+ $δ$ = 45^o + 45^oThen, angle of minimum deviation, $δ$ =  90^o - 60^o = 30^oWe know that, velocity of light in vacuum = 3 $\times$ 10⁸ m/s
Now, using the formula,

   $μ = \frac{\sin (A + δ_{m}) / 2}{\sin A / 2} = \frac{c}{v}$
That is,
   $v = c \frac{\sin A / 2}{\sin (A + δ_{m}) / 2} = 3 \times 10^{8} \frac{\sin 30^{o}}{\sin (60^{o} + 30^{o}) / 2} = 3 \times 10^{8} \frac{\sin 30^{o}}{\sin 45^{o}} = \frac{3 \times 10^{8} \times 0.5000}{0.7071} = 2.12 \times 10^{8} m / s$

is the required speed of light in the prism.

For Daily Free Study Material Join wiredfaculty

Wave Optics

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

Some More Questions From Wave Optics Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Phone Number

Email Address

Loaction