A monochromatic beam of light is incident at 60° on one face of an equilateral prism of refractive index n and emerges from the opposite face making an angle θ(n) with the normal (see the figure). For n the value of θ is 60° and . The value of m is

Engineering

Physics

Prism

Question

A monochromatic beam of light is incident at 60° on one face of an equilateral prism of refractive index n and emerges from the opposite face making an angle θ(n) with the normal (see the figure). For n $= \sqrt{3}$ the value of θ is 60° and $\frac{dθ}{dn} = m$ . The value of m is

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sin 60° = n sin r₁

n sin r₂ = sin Q

n sin (60° – r₁) = sin Q

n (sin 60° cos r₁ – cos 60° sin r₁) = sin Q

$\frac{\sqrt{3} n}{2}$ cos r₁ – $\frac{n}{2}$ sin r₁ = sin Q

$\frac{\sqrt{3} n}{2} \sqrt{1 - \frac{3}{4 n^{2}}} - \frac{1}{2} \frac{\sqrt{3}}{2} = \sin Q$

$\frac{\sqrt{3}}{2} \frac{n}{2 n} \sqrt{4 n^{2} - 3} - \sqrt{4 n^{2} - 3} - \frac{\sqrt{3}}{4} = \sin Q$

$\frac{4}{\sqrt{3}} sin Q = \sqrt{4 n^{2} - 3} - 1$

On differentiatig

$\frac{4}{\sqrt{3}} \cos Q \frac{dQ}{dn} = \frac{1}{2 \sqrt{4 n^{2} - 3}} 8 n$

Putting the values

$\frac{4}{\sqrt{3}} \frac{1}{2} \frac{dθ}{dn} = \frac{1}{2 \sqrt{12 - 3}} 8 \sqrt{3} = \frac{4 \sqrt{3}}{3} = \frac{4}{3}$

$\frac{dθ}{dn} = 2$