Question

A transparent solid cylindrical rod has a refractive
index u and is surrounded by air. A light ray is
incident at mid point of one end of rod as shown.
If 0 is angle of incident then u for grazing a light
along the walls is(sin theta =1/√3
1) 2/√3
(2) 1
3) 1\√3
(4) √3​

Answer 1

Answer:

Refractive index of the material is given as

$\mu = \frac{2}{\sqrt3}$

Explanation:

If light is grazing along the walls then we have

$\mu sin\theta_c = 1 sin 90$

now we have

$sin\theta_c = \frac{1}{\mu}$

now at the entering face of the cylinder

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

$sin\theta = \mu cos\theta_c$

$sin\theta = \mu\sqrt{1 - (\frac{1}{\mu})^2}$

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

as we know that

$sin\theta = \frac{1}{\sqrt3}$

so we have

$\frac{1}{\sqrt3} = \sqrt{\mu^2 - 1}$

$\frac{1}{3} + 1 = \mu^2$

$\mu = \frac{2}{\sqrt3}$

#Learn

Topic : Total internal reflection

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