Question

In the figure shown below, find the minimum value of $i$ for which total internal reflection never takes place at $P$.
[Medium surrounding the slab has refractive index ${\mu }_2$]

• A. ${\cos }^{-1}\left(\frac{\sqrt{18}}{17}\right)$
• B. ${\sin }^{-1}\left(\frac{\sqrt{8}}{17}\right)$
• C. ${\cos }^{-1}\left(\frac{\sqrt{17}}{8}\right)$
• D. ${\sin }^{-1}\left(\frac{\sqrt{17}}{8}\right)$

Answer 1

The correct option is D ${\sin }^{-1}\left(\frac{\sqrt{17}}{8}\right)$

For no total internal reflection at $P$:
$\beta \le {\theta }_c$
In critical case, $\beta ={\theta }_c$


So, $r={90}^{\circ }-\beta ={90}^{\circ }-{\theta }_c....\left(1\right)$

We know ,
$\sin {\theta }_c=\frac{{\mu }_2}{{\mu }_1}$

$\Rightarrow \sin {\theta }_c=\frac{\frac{4}{3}}{\frac{3}{2}}=\frac{8}{9}............\left(2\right)$

Applying snell's law at point $Q$

${\mu }_2\sin i={\mu }_1\sin r$

Substituting the given data we get,

$\frac{4}{3}\sin i=\frac{3}{2}\sin ({90}^{\circ }-{\theta }_c)$ $[$ From $\left(1\right)]$

$\Rightarrow \frac{8}{9}\sin i=\cos {\theta }_c$

$\Rightarrow \frac{8}{9}\sin i=\sqrt{1-{\sin }^2{\theta }_c}$

$\Rightarrow \frac{8}{9}\sin i=\sqrt{1-\frac{64}{81}}$ $[$ From $\left(2\right)]$
$\Rightarrow \frac{8}{9}\sin i=\frac{\sqrt{17}}{9}$
$\Rightarrow \sin i=\frac{\sqrt{17}}{8}$
$\Rightarrow i={\sin }^{-1}\left(\frac{\sqrt{17}}{8}\right)$

Hence, option (d) is the correct answer.

Why this question? This question check your ability to apply geometry and trigonometry in the Snell's law to get the required data. Such questions are generally asked in JEE Main.