A monochromatic beam of light is incident at $60^{\circ}$ 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 $60^{\circ}$ and $\frac{d θ}{d n} = m$ . The value of m is .

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Solution

By snells law

$1 s i n 60 = n s i n r_{1} \dots (1)$

d.w.r n

$0 = s i n r_{1} + n c o s r_{1} \frac{d r_{1}}{d n} (r 1 = 30^{\circ}, n = \sqrt{3})$

$\frac{d r_{1}}{d n} = \frac{1}{3} \dots (2)$

By apply snells law

$n s i n r_{2} = 1 s i n θ$

$n s i n (60 - r_{1}) = 1 s i n θ$

$s i n (160 - r_{1}) + n c o s (60 - r_{1}) \frac{d r_{1}}{d n} = c o s θ \frac{d θ}{d n}$

By solving above equation $\frac{d θ}{d n} = 2$
$\Rightarrow \sqrt{3} \times s i n 30^{\circ} = s i n θ \Rightarrow θ = 60^{\circ}$

Also, $s i n θ = n s i n 30^{\circ} = \frac{n}{2}$

$∴ c o s θ \frac{d θ}{d n} = \frac{1}{2} \Rightarrow \frac{d θ}{d n} = \frac{1}{2 c o s 60^{\circ}}$

$\Rightarrow m = 1$ .

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