Question

A glass prism has a right-triangular cross-section $ABC$, with $\angle A = 90$°. A ray of light parallel to the hypotenuse $BC$ and incident on the side $AB$ emerges grazing the side $AC$. Another ray, again parallel to the hypotenuse $BC$, incident on the side $AC$ suffers total internal reflection at the side $AB$. Which one of the following must be true about the refractive index $\mu$ of the material of the prism?

• A. $\sqrt{\frac{3}{2}<}\mu<\sqrt{2}$
• B. $\mu>\sqrt{3}$
• C. $\mu<\sqrt{\frac{3}{2}}$
• D. $\sqrt{2}<\mu<\sqrt{3}$

Answer 1

Answer: (D)

When light is made incident over face $AB$, refraction occurs as shown below.

From geometry of figure,
$r_{1}=90-\theta_{c}$
$i=90-\alpha$
Refractive index $\mu$ of prism is
$\mu=\frac{1}{\sin\theta_{c}}=\frac{\sin i}{\sin r_{1}}$
$\Rightarrow \mu =\frac{1}{\sin \theta_{c}}=\frac{\sin\left(90-\alpha\right)}{\sin\left(90-\theta_{c}\right)}$
$\Rightarrow \mu=\frac{1}{\sin \theta_{c}}=\frac{\cos \alpha}{\cos\theta_{c}}$
Also, when incidence is made over face $AC$, refraction occurs as shown below.

Again, from geometry of figure, At surface $AB$, $TIR$ occurs
$\Rightarrow i>\theta_{c}$
$\Rightarrow 90-r>\theta_{c}$
$\Rightarrow 90-\theta_{c}>r_{2}$
$\Rightarrow \sin\left(90-\theta_{c}\right)>\sin r_{2}$
$\Rightarrow \cos\left(\theta_{c}\right)>\frac{\sin\alpha}{\mu}$
$\left[\therefore \mu=\frac{\sin i}{\sin r_{2}}=\frac{\sin \alpha}{\sin r_{2}}\right]$
$\Rightarrow \cos\theta_{c}>\sqrt{1-\frac{\cos^{2}\alpha}{\mu}}$
$\Rightarrow \cos \theta_{c}>\frac{\sqrt{1-\mu^{2} \cos ^{2} \theta_{c}}}{\mu}$
$\Rightarrow 2 \mu^{2} \cos ^{2} \theta_{c}>1$
$ \Rightarrow 2 \mu^{2}\left(1-\sin ^{2} \theta_{c}\right)>1$
$\Rightarrow 2 \mu^{2}\left(1-\frac{1}{\mu^{2}}\right)>1$
$\Rightarrow 2 \mu^{2}-2>1$
$\Rightarrow 2 \mu^{2}>3$
$\Rightarrow \mu>\sqrt{\left(\frac{3}{2}\right)}$