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 $\theta(n)$ with the normal (see the figure). For $n=\sqrt{3}$ the value of $\theta$ is $60^{\circ}$ and $\frac{d \theta}{d n}=m$. The value of $m$ is

JEE Advanced · 2015 · 24 May 2015 · Paper 2 · Geometrical Optics

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 = 3

the value of

θ

6 0^{\circ}

and

\frac{d θ}{d n} = m

. The value of

m

Answer: 2

Applying Snell's law at the first face:

sin 60° = n sin r

At the second face:

sin θ = n sin (60° - r)

Differentiating the second equation with respect to

n

cos θ \frac{d θ}{d n} = - n cos (60° - r) \frac{d r}{d n} + sin (60° - r)

From the first equation:

n cos r \frac{d r}{d n} + sin r = 0 \Rightarrow \frac{d r}{d n} = - \frac{sin r}{n cos r}

Substituting and evaluating at

n = 3

r = 30°

θ = 60°

\frac{d θ}{d n} = \frac{1}{cos 60°} (cos 30° tan 30° + sin 30°) = 2 (\frac{1}{2} + \frac{1}{2}) = 2

Therefore,

m = 2

JEE Advanced · 2015 · 24 May 2015 · Paper 2 · Integer answer