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
$\Rightarrow i>{\theta }_c$
$\Rightarrow 90-r>{\theta }_c$
$\Rightarrow 90-{\theta }_c>r_2$
$\Rightarrow \sin \left(90-{\theta }_c\right)>\sin r_2$
$\Rightarrow \cos \left({\theta }_c\right)>\frac{\sin \alpha }{\mu }$
$\left[\therefore \mu =\frac{\sin i}{\sin r_2}=\frac{\sin \alpha }{\sin r_2}\right]$
$\Rightarrow \cos {\theta }_c>\sqrt{1-\frac{{\cos }^2\alpha }{\mu }}$
$\Rightarrow \cos {\theta }_c>\frac{\sqrt{1-{\mu }^2{\cos }^2{\theta }_c}}{\mu }$
$\Rightarrow 2{\mu }^2{\cos }^2{\theta }_c>1$
$\Rightarrow 2{\mu }^2\left(1-{\sin }^2{\theta }_c\right)>1$
$\Rightarrow 2{\mu }^2\left(1-\frac{1}{{\mu }^2}\right)>1$
$\Rightarrow 2{\mu }^2-2>1$
$\Rightarrow 2{\mu }^2>3$
$\Rightarrow \mu >\sqrt{\left(\frac{3}{2}\right)}$