Question

An isosceles triangular glass prism stands with its base in water as shown. The angles that its two equal sides make with the base are $\theta$ each. An incident ray of light parallel to the water surface internally reflects at the glass-water interface and subsequently reemerges into the air. Take the refractive indices of glass and water to be $3/2$ and $4/3$ respectively. The least value of $\theta$ for this to happen must be

• A. ${\tan }^{-1}\frac{2}{\sqrt{17}}$
• B. ${\tan }^{-1}\frac{8}{\sqrt{17}}$
• C. ${\tan }^{-1}\frac{2}{\sqrt{3}}$
• D. ${\tan }^{-1}\frac{8}{\sqrt{3}}$

Answer 1

The correct option is A ${\tan }^{-1}\frac{2}{\sqrt{17}}$


Using Snell's law for initial refraction,

$\sin (90-\theta )=\frac{3}{2}\sin r$

$\cos \theta =\frac{3}{2}\sin r$

$\sin r=\frac{2}{3}\cos \theta$

For total internal reflection,

$90-x>c$, where $c$ is critical angle.

$\Rightarrow r+\theta >c$

($90-x=r+\theta$ from the geometry of the triangle)

$\Rightarrow \sin (r+\theta )>\sin c$

$\Rightarrow \sin r\cos \theta +\cos r\sin \theta >\frac{8}{9}$

(${\mu }_{gw}=\frac{3}{2}\times \frac{3}{4}=\frac{8}{9}$
$sinc=\frac{1}{{\mu }_{gw}}=\frac{8}{9}$)

$\Rightarrow \frac{2}{3}{\cos }^2\theta +\frac{\sqrt{9-4{\cos }^2\theta }}{3}\cdot \sin \theta >\frac{8}{9}$

$\Rightarrow \sqrt{9-4{\cos }^2\theta }\cdot \sin \theta >(\frac{8}{3}-2{\cos }^2\theta )$

$\Rightarrow (9-4{\cos }^2\theta )(1-{\cos }^2\theta )>\frac{64}{9}+4{\cos }^4\theta -\frac{32}{3}{\cos }^2\theta$

$\Rightarrow {\cos }^2\theta <\frac{17}{21}\Rightarrow {\sec }^2\theta <\frac{21}{17}\Rightarrow \tan \theta >\frac{2}{\sqrt{17}}$

($\because ta{n}^2\theta =se{c}^2\theta -1$)

As we have got $tan\theta >\frac{2}{\sqrt{17}}$

Hence, least value of $\theta$ will be ${tan}^{-1}\frac{2}{\sqrt{17}}$