Question

A glass prism has a right-triangular cross-section ABC, with $\angle$A$={90}^o$. 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. Find the range of 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

The correct option is A $\sqrt{\frac{3}{2}}<\mu <\sqrt{2}$

Since of is parallel to $BC$.

at $AB$

(1) $\sin (90-\theta )=\mu \sin r$

$\frac{\cos \theta }{\mu }=\sin r$

at $AC$

$\begin{array}{c}\mu \sin (90-r)=\left(1\right)\sin 90\\ \mu \cos r=1\\ {\sin }^{2}r+{\cos }^{2}r=1\\ \begin{array}{c}\frac{{\cos }^2\theta }{{\mu }^2}+\frac{1}{{\mu }^2}=1\\ {\cos }^2\theta +1={\mu }^2-\left(A\right)\end{array}\\ 2-{\sin }^2\theta ={\mu }^2\Rightarrow {\sin }^2\theta =2-{\mu }^2-\left(A\right)\end{array}$

at $AC$

$\left(1\right)\sin \theta =\mu \sin r-0$

$\mu \sin (90-\gamma )>1$ ( at $AB$ condition for $TIR)$

$\cos x>\frac{1}{\mu }⟶3$

By (1)

$\sin r=\frac{\sin \theta }{\mu }\Rightarrow \cos r=\frac{\sqrt{M^2-{\sin }^2\theta }}{\mu }$

$\frac{\sqrt{{\mu }^2-{\sin }^2\theta }}{\mu }>\frac{1}{\mu }$

${\mu }^2>{\sin }^2\theta +1$

By equation $A$

$\begin{array}{c}{\mu }^2>2{\mu }^2+1\\ 2{\mu }^2>3\\ \mu >\sqrt{\frac{3}{2}}\end{array}$

for refraction to be happened At AC

$\sin \theta =\mu \sin r$

$\mu \sin r<1$

${\sin }^{2}x<\frac{1}{{\mu }^2}$

By equation (3) $\cos \gamma >\frac{1}{\mu }\Rightarrow 1-{\sin }^2\gamma >\frac{1}{{\mu }^2}$

SO

$1-\frac{1}{{\mu }^2}<\frac{1}{{\mu }^2}$

$1-\frac{2}{{\mu }^2}<O\Rightarrow \mu <\sqrt{2}$

$\text{so}\sqrt{\frac{3}{2}}<\mu <\sqrt{2}$

So option (A) is correct