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Q.

A man is standing on a straight road in a summer noon. The refractive index of the air changes with height as $μ = μ_{o} (1 + \frac{y}{3 (m e t e r)})$ due to variation of the vertical temperature where $μ_{o}$ is the refractive index of the air at road surface and y height. Eyes of the man are at height of 1.5 m above the road. What is the apparently visible length of the road?

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a

$\frac{2}{\sqrt{5}}$

b

$\frac{3}{\sqrt{5}}$

c

$\frac{4}{\sqrt{5}}$

d

$\frac{6}{\sqrt{5}}$

answer is A.

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Detailed Solution

$μ_{o} \sin 90^{0} = μ_{0} (1 + \frac{1.5}{3}) \sin θ$

Question Image

$\sin θ = \frac{2}{3}, \cot θ = \frac{\sqrt{5}}{2}, \cot θ = \frac{h}{l}$

$l = \frac{h}{\cot θ} = 1.5 \times \frac{2}{\sqrt{5}} = \frac{3}{\sqrt{5}} m$

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A man is standing on a straight road in a summer noon. The refractive index of the air changes with height as μ=μo1+y3meter due to variation of the vertical temperature where μo is the refractive index of the air at road surface and y height. Eyes of the man are at height of 1.5 m above the road. What is the apparently visible length of the road?