First slide

Total internal reflection - Ray optics and optical instruments

Question

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 what so ever be the value of angle of incidence?

Moderate

A

$n > \sqrt{2}$
B

n = 1
C

n = 1.1
D

n = 1.3

Solution

Let a ray of light enter at A and the refracted beam is AB. This is incident at angle $θ$ . For no refraction at the lateral face, $θ > C$ or, $\sin θ > \sin C$ But $θ + r = 90^{0}$
$\Rightarrow θ = (90^{0} - r)$
$∴ \sin (90^{0} - r) > \sin C$
Or $\cos r > \sin C .................................. (i)$
From Snell’s law, $n = \frac{\sin i}{\sin r} \Rightarrow \sin r = \frac{\sin i}{n}$
$∴ \cos r = \sqrt{1 - \sin^{2} r} = \sqrt{(1 - \frac{\sin^{2} i}{n^{2}})}$
$∴$ equation (i) we get $\sqrt{1 - \frac{\sin^{2} i}{n^{2}}} > \sin C \Rightarrow 1 - \frac{\sin^{2} i}{n^{2}} > \sin^{2} C$
Also, $\sin C = \frac{1}{n}$
$∴$ $1 - \frac{\sin^{2} i}{n^{2}} > \frac{1}{n^{2}} o r 1 > \frac{\sin^{2} i}{n^{2}} + \frac{1}{n^{2}}$
Or $\frac{1}{n^{2}} (\sin^{2} i + 1) < 1$ or $n^{2} > (\sin^{2} i + 1)$
Maximum value of $\sin i = 1$
$H e n c e n^{2} > 2 \Rightarrow n > \sqrt{2}$

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