Question

A man is standing on a straight road in a summer noon. The refractive index of the air changes with height as  \mu=\mu_{0}\left(1+\frac{y}{3(\text { metre })}\right)μ=μ0​(1+3( metre )y​) due to variation of the vertical temperature where \mu_{0}μ0​ is the refractive index of the air at road surface and yy height. Eyes of the man are at height of 1.5 \mathrm{m}m above the road. What is the apparently visible length of the road.

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Answer 1

y = 1.5m ; x = 100 m  

Explanation:

Given: On a hot summer day in a desert, the refractive index of the atmosphere changes with height (y) above the surface of the earth as μ= m0 (1+by)^1/2 where m0 is the refractive index at the surface and b = 6* x10^−4 m^−1. A man of height 1.5 m is standing on a straight level road.

Find: The distance beyond which he cannot see a point on the road.

Solution:

i = angle of incidence at height y

μo = μ sin i

μo = μo (1 + by)^1/2 sin i

sin i = 1/ (1 + by)^1/2

cot i = √by

dy/ dx = tan (90 - i) = cot i

dy/ dx = √by

dy / √y [0 to y] = √b * dx [0 to y]

2√y  [0 to y]  = √bx

Therefore x = 2√y  / b

y = 1.5m ; x = 100 m