Question

A ray of light passes through an equilateral prism (μ = 1.5). The angle of minimum deviation is(a) 45°(b) 37°12'(c) 20°(d) 30°

Answer 1

The angle of deviation is equal to difference between the angle of incidence and the angle of refraction of a ray of light passing through prism .

There is an angle of incidence at which the deflection is minimal . This is the angle of incidence at which minimum deviation of light occurs .

This is called the angle of minimum deviation .

The angle of minimum deviation is given by formula -

μ  =  [sin(A + D)/2]/sin(A/2)

where, μ = the refractive index of prism = 1.5

            A  =  internal angle of prism = 60°

             D = angle of minimum deviation

=>  1.5  = [sin(60 + D)/2]/sin(60/2)

=>   1.5 =  [sin(60 + D)/2]/sin(30)

=>  1.5 =  [sin(60 + D)/2]/0.5

=>   0.75 = [sin(60 + D)/2]

=>   (60 + D)/2  =  48.6°

=>  D  =  37°12'

Thus,  The angle of minimum deviation is 37°12'

Thus, option (b) 37°12' is correct .

Answer 2

answer : option (b) 37°12'

a/c to question,

A ray of light passes through an equilateral prism.

angle of prism, A = 60°

refractive index , $\mu$ = 1.5

use formula, $\mu=\frac{sin\frac{(A+\delta)}{2}}{sin\frac{A}{2}}$

where $\delta$ is minimum deviation.

so, $1.5=\frac{sin\frac{(60^{\circ}+\delta)}{2}}{sin\frac{60^{\circ}}{2}}$

or, $1.5=\frac{sin(30^{\circ}+\frac{\delta}{2})}{sin30^{\circ}}$

or, 1.5 = sin(30° + $\delta$/2)/(1/2)

or, 3/4 = sin(30° +$\delta$/2)

or, 30° + $\delta$/2 = sin^-1(3/4)

or, 30° + $\delta$/2 = 48.59°

or, $\delta$/2 = 18.59°

or, $\delta$ = 37.18°

hence, it is approximately similar to option (b) 37°12'