A monochromatic beam of light is incident at 60 o 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 = 3 the value of θ is 60 o and d θ d n = m . The value of m is

Step-by-step solution

By Snell's law 1 sin ⁡ 60 = n sin ⁡ r 1 ⇒ sin ⁡ r 1 = 1 2 r 1 = 30 o ...(i) By differentiating 'w.r.t' n O = sin ⁡ r 1 + n cos ⁡ r 1 d r 1 d n = 1 2 + 3 3 2 d r 1 d n = d r 1 d n = 1 3 ...(ii) By applying Snell's law n sin ⁡ r 2 = 1 sin ⁡ θ n sin ⁡ 60 - r 1 = 1 sin ⁡ θ ∴ A + r 1 + r 2 By differentiating 'w.r.t' n sin ⁡ ( 60 - r 1 ) - n cos ⁡ 60 - r 1 d r 1 d n = c o s θ d θ d n By substituting value of ' r 1 ' and d r 1 d n from (i) and (ii) d θ d n = 2