Question

(i) Plot a graph to show variation of the angle of deviation as a function of angle of incidence for light passing through a prism. derive an expression for refractive index of the prism in terms of angle of minimum deviation and angle of prism.

(ii) What is dispersion of light? What is its cause?

(iii) A ray of light incident normally on one face of right isosceles prism is totally reflected as shown in fig. What must be the minimum value of refractive index of glass? Give relevant calculations.

Answer 1

(i)

 

From figure δ = D_m, i = e which implies r₁ = r₂

2r = A, or r = A/2

Using δ = i + e − A

D_m = 2i − A

i = \(\frac{A+D_m}{2}\)

μ = \(\frac{sin\,i}{sin\,r}=\frac{sin(\frac{A+D_m}{2})}{sin(A/2)}\)

(ii) The phenomenon of splitting of white light into its constituent colours.

Cause: Refractive index of the material is different for different colours According to the equation, δ ≇ (μ – 1) A, where A is the angle of prism, different colours will deviate through different amount.

For total internal reflection, ∠i ≥ ∠i_c (critical angle)

⇒ 45º ≥ ∠i_c, i.e., ∠i_c ≤ 45º

sin i_c ≤ sin45º

≤ \(\frac{1}{√2}\)

\(\frac{1}{sin\,i_c}\) ≥ √2

⇒ μ ≥ √2

Hence, the minimum value of refractive index must be √2.