The 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 $\frac{d θ}{d n} = m .$ The value of $m$ is _______ [2015]

Question

Smart Achievers · Accepted Answer

(2)

Applying Snell's law at $A$ ,

$\sin 60 ° = n \sin r$ $\dots (i)$

Differentiating w.r.t. $' n'$ , we get

$0 = \sin r + n \cos r \frac{d r}{d n}$ $\dots (i i)$

Again, applying Snell's law at $B$ ,

$\sin θ = n \sin (60 ° - r)$ $\dots (i i i)$

Differentiating w.r.t. $' n'$ , we get

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

$∴$ $\cos θ \frac{d θ}{d n} = \sin (60 ° - r) - n \cos (60 ° - r) (- \frac{\tan r}{n})$ $[from (ii)]$

$∴$ $\frac{d θ}{d n} = \frac{1}{\cos θ} [\sin (60 ° - r) + \cos (60 ° - r) \tan r]$ $\dots (i v)$

From eq. (i), for $n = \sqrt{3}$ , we get $r = 30 °$

From eq. (iii), for $n = \sqrt{3}$ and $r = 30 °$ , we get $θ = 60 °$

Substituting the values of $r$ and $θ$ in eq. (iv), we get

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

Solution

Want Us to Email you About Special Offers & Updates?

Solution

Q. The 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 is 60° and dθdn=m. The value of m is _______ [2015]

Want Us to Email you About Special Offers & Updates?