Question

A monochromatic beam of width $t$ is incident at $45^\circ $ on an air-water surface. The refractive index of water is $\mu $ . The width of the beam in water is,

• A. $\left(\mu - 1\right)t$
• B. $\mu t$
• C. $\frac{\sqrt{\mu^{2} - 1}}{\mu}$
• D. $\frac{\left(\sqrt{2 \mu^2-1}\right)}{\mu} t$

Answer 1

Answer: (D)

Snell's law,
$\sin 45^{o}=\mu \sin ⁡ \alpha $
$\sin \alpha =\frac{1}{\sqrt{2} \, \mu }$
$\cos \alpha = \frac{\sqrt{2 \mu ^{2} - 1}}{\sqrt{2} \mu }$ ___(1)
From $\Delta ABC, \, AC=\sqrt{2} \, t$
From $\Delta ADC,\cos \alpha =\frac{D C}{A C}=\frac{t^{'}}{\sqrt{2} \, t}$ ___(2)
From (1) and (2)
$\frac{\sqrt{2 \mu^2-1}}{\sqrt{2} \mu}=\frac{\mathrm{t}^{\prime}}{\sqrt{2} \mathrm{t}} \Rightarrow t^{\prime}=\frac{\left(\sqrt{2 \mu^2-1}\right) \mathrm{t}}{\mu}$