Question

A ray of light enters a spherical drop of water of refractive index $n=\sqrt{3}$ as shown in figure. If $\varphi ={60}^{\circ }$, then

• A. Angle of deviation $\delta =\pi -2\varphi$
• B. Angle of deviation $\delta =\pi -4\theta +2\varphi$
• C. Angle of deviation $\delta =\pi$
• D. For minimum angle of deviation $\varphi <{45}^{\circ }$

Answer 1

Answer: (B), (C), (D)

The correct options are
B Angle of deviation $\delta =\pi -4\theta +2\varphi$
C Angle of deviation $\delta =\pi$
D For minimum angle of deviation $\varphi <{45}^{\circ }$

$\delta ={\delta }_{refraction1}+{\delta }_{reflection}+{\delta }_{refraction2}=(\varphi -\theta )+(\pi -2\theta )+(\varphi -\theta )=\pi +2\varphi -4\theta$

By snells law,
$\frac{n}{1}=\frac{sin\varphi }{sin\theta }\Rightarrow \sqrt{3}=\frac{sin60}{sin\theta }\Rightarrow \theta ={30}^{\circ }$

$\delta =\pi +2\times {60}^{\circ }-4\times {30}^{\circ }=\pi$

$\delta =\pi +2\varphi -4\theta \dots \dots \dots \left(1\right)$ and $n=\frac{sin\varphi }{sin\theta }$

$\frac{d\varphi }{d\theta }=n\frac{cos\theta }{cos\varphi }$
$\Rightarrow \frac{d\varphi }{d\theta }=n\frac{\sqrt{1-si{n}^2\theta }}{cos\varphi }$
$\Rightarrow \frac{d\varphi }{d\theta }=\frac{n}{cos\varphi }\sqrt{1-\frac{si{n}^2\varphi }{n^2}}$

By (1) $\frac{d\delta }{d\varphi }=0\Rightarrow 2d\varphi =4d\theta$

$\frac{d\varphi }{d\theta }=2\Rightarrow \frac{n}{cos\varphi }\sqrt{1-\frac{si{n}^2\varphi }{n^2}}=2$

$1-\frac{si{n}^2\varphi }{n^2}=\frac{4co{s}^2\varphi }{n^2}\Rightarrow cos\varphi =\sqrt{\frac{2}{3}}$

$\varphi <{45}^{\circ }$ $[\varphi \approx {35}^{\circ }]$