Question

A transparent cube of side \(d\), made of a material of refractive index \(\mu_2\), is immersed in a liquid of refractive index \(\mu_1(\mu_1< \mu_2)\). A ray is incident on the face \(AB\), at an angle \(\theta\) (shown in the figure). Total internal reflection takes place at point \(E\) on the face \(BC\). Then \(\theta\) must satisfy,


My

• A. $\theta <{\sin }^{-1}\left(\frac{{\mu }_1}{{\mu }_2}\right)$
• B. $\theta >{\sin }^{-1}\left(\sqrt{\frac{{\mu }_2^2}{{\mu }_1^2}-1}\right)$
• C. $\theta <{\sin }^{-1}\left(\sqrt{\frac{{\mu }_2^2}{{\mu }_1^2}-1}\right)$
• D. $\theta >{\sin }^{-1}\left(\frac{{\mu }_1}{{\mu }_2}\right)$

Answer 1

The correct option is C $\theta <{\sin }^{-1}\left(\sqrt{\frac{{\mu }_2^2}{{\mu }_1^2}-1}\right)$


The critical angle, for the given pair of media, is,
${}_r{\mu }_d=\frac{1}{\sin C}$

$\Rightarrow \frac{{\mu }_2}{{\mu }_1}=\frac{1}{\sin C}$

$\therefore {\mu }_1={\mu }_2\sin C......\left(i\right)$

For the first refraction, according to Snell's law,
$\frac{{\mu }_2}{{\mu }_1}=\frac{\sin \theta }{\sin ({90}^{\circ }-C)}$

$\Rightarrow {\mu }_1\sin \theta ={\mu }_2\cos C$

${\mu }_2\sin C\sin \theta ={\mu }_2\cos C$

$\sin C\sin \theta =\sqrt{1-{\sin }^{2}C}$

$\sin \theta =\frac{1}{\sin C}\sqrt{1-{\sin }^{2}C}$

$\sin \theta =\frac{{\mu }_2}{{\mu }_1}\sqrt{1-{\left(\frac{{\mu }_1}{{\mu }_2}\right)}^2}$

$\therefore \sin \theta =\sqrt{{\left(\frac{{\mu }_2}{{\mu }_1}\right)}^2-1}$

$\therefore \theta ={\sin }^{-1}\left(\sqrt{{\left(\frac{{\mu }_2}{{\mu }_1}\right)}^2-1}\right)$

This is the maximum possible value of $\theta$ for total internal reflection.

So, for $TIR$ to occur,

$\theta <{\sin }^{-1}\left(\sqrt{{\left(\frac{{\mu }_2}{{\mu }_1}\right)}^2-1}\right)$

Hence, $\left(C\right)$ is the correct answer.