Question

A cubical block of glass of refractive index $n_1=\frac{3}{2}$ is in contact with the surface of water of refractive index $n_2=\frac{4}{3}$. A beam of light is incident on the vertical face of the block. After refraction, a total internal reflection occurs at the base and refraction at the opposite face takes place. The ray emerges at angle $\theta$ as shown. The value of $\theta$ is given by

• A. $\sin \theta >\frac{\sqrt{17}}{6}$
• B. $\cos \theta >\frac{\sqrt{17}}{6}$
• C. $\sin \theta <\frac{\sqrt{17}}{6}$
• D. $\cos \theta <\frac{\sqrt{17}}{6}$

Answer 1

The correct option is C $\sin \theta <\frac{\sqrt{17}}{6}$


For total internal reflection at $AB$,

$i>{\theta }_c$

$\Rightarrow \sin i>\sin {\theta }_c$
[All angles less than ${90}^{\circ }$]

$\Rightarrow \cos i<\cos {\theta }_c............\left(1\right)$

For refraction at $BC$, from Snell's law,

$n_1\sin r=n_3\sin \theta$

$\Rightarrow \frac{3}{2}\sin ({90}^{\circ }-i)=1\times \sin \theta$

$\Rightarrow \cos i=\frac{2}{3}\sin \theta$

$\Rightarrow \frac{2}{3}\sin \theta <\cos {\theta }_c$ $[$from $\left(1\right)$$]$

$\Rightarrow \frac{2}{3}\sin \theta <\sqrt{1-{\sin }^2{\theta }_c}$

$\Rightarrow \sin \theta <\frac{3}{2}\sqrt{1-{\left(\frac{\frac{4}{3}}{\frac{3}{2}}\right)}^2}[\sin {\theta }_c=\frac{n_2}{n_1}]$

$\therefore \sin \theta <\frac{\sqrt{17}}{6}$

Why this question? Caution - Be careful in inequalities of $\sin \theta$ and $\cos \theta$. For angle less than ${90}^{\circ }$, $\sin \theta$ is an increasing function and $\cos \theta$ is a decreasing function.