Ray Optics and Optical Instruments - Result Question 18
18. Light enters at an angle of incidence in a transparent rod of refractive index $n$. For what value of the refractive index of the material of the rod the light once entered into it will not leave it through its lateral face whatsoever be the value of angle of incidence?
[1998]
(a) $n>\sqrt{2}$
(b) $n=1$
(c) $n=1.1$
(d) $n=1.3$
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Answer:
Correct Answer: 18. (a)
Solution:
- (a) Let a ray of light enter at $A$ and the refracted beam is $A B$. This is incident at an angle $\theta$. For no refraction at the lateral face, $\theta>C$
or, $\sin \theta>\sin C$ But $\theta+r=90^{\circ} \Rightarrow \theta=(90^{\circ}-r)$
$\therefore \sin (90^{\circ}-r)>\sin C$
or $\cos r>\sin C$
From Snell’s law, $n=\frac{\sin i}{\sin r} \Rightarrow \sin r=\frac{\sin i}{n}$
$\therefore \cos r=\sqrt{1-\sin ^{2} r}=\sqrt{(1-\frac{\sin ^{2} i}{n^{2}})}$
$\therefore$ equation (1) gives
$\sqrt{1-\frac{\sin ^{2} i}{n^{2}}}>\sin C \Rightarrow 1-\frac{\sin ^{2} i}{n^{2}}>\sin ^{2} C$
Also, $\sin C=\frac{1}{n}$
$\therefore 1-\frac{\sin ^{2} i}{n^{2}}>\frac{1}{n^{2}}$ or $1>\frac{\sin ^{2} i}{n^{2}}+\frac{1}{n^{2}}$
or $\frac{1}{n^{2}}(\sin ^{2} i+1)<1$ or $n^{2}>(\sin ^{2} i+1)$
Maximum value of $\sin i=1$
$\therefore n^{2}>2 \Rightarrow n>\sqrt{2}$
