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 $\Delta ADC,\cos \alpha =\frac{DC}{AC}=\frac{t^{{}^{\prime }}}{\sqrt{2}t}$ ___(2)
From (1) and (2)
$\frac{\sqrt{2{\mu }^2-1}}{\sqrt{2}\mu }=\frac{{\mathrm{t}}^{\prime }}{\sqrt{2}\mathrm{t}}\Rightarrow t^{\prime }=\frac{\left(\sqrt{2{\mu }^2-1}\right)\mathrm{t}}{\mu }$