Question

(i) Draw the ray diagram showing refraction of light through a glass prism and hence obtain the relation between the refractive index μ of the prism, angle of prism and angle of minimum deviation.

(ii) Determine the value of the angle of incidence for a ray of light travelling from a medium of refractive index μ₁ = √2 into the medium of refractive index μ₂ = 1, so that it just grazes along the surface of separation.

Answer 1

(i)

From fig ∠A + ∠QNR = 180º  ......(1)

From triangle ∠QNR r₁ + r₂ + ∠QNR = 180º  .....(2)

Hence from eq (1) and (2)

∴ ∠A = r₁ + r₂

The angle of Deviation

δ = (i − r₁) + (e − r₂) = i + ρ – A

At minimum deviation i = e and r₁ and r₂

∴ r = \(\frac{A}{2}\)

And i = \(\frac{A+δ_m}{2}\)

Hence refractive index

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

(ii) From Snell’s law μ₁ sin i = μ₂ sin r

Given μ₁ = √2, μ₂ = 1 and = 90º (just grazing)

∴ √2 sin i = 1 × sin 90º ⇒ sin i = \(\frac{1}{√2}\) or i = 45º