A man of height 1.47 m stands on a straight road on a hot day. The variation of temperature in the air results in a variation of refractive index with height y as μ=μ0(1+ay) where μ0 is the refractive index of air near the road a=1.5×10−6/m What is the apparent length of the road ( in m ) man is able to see ?

O→P:μ0sin⁡π2=μsin⁡i at O,y=0⇒μ=μ0(1+0)=μ0 μ0=μ01+aysin⁡i⇒sin⁡i=11+ay Slope dydx=cot⁡i=ay ∫dyy=a∫dx⇒2y=ax⇒x=2ya ymax=1.47m⇒xmax =21.471.5×10−6≈2000m

A man of height 1.47 m stands on a straight road on a hot day. The variation of temperature in the air results in a variation of refractive index with height y as μ=μ0(1+ay) where μ0 is the refractive index of air near the road a=1.5×10−6/m What is the apparent length of the road ( in m ) man is able to see ?

O→P:μ0sin⁡π2=μsin⁡i at O,y=0⇒μ=μ0(1+0)=μ0 μ0=μ01+aysin⁡i⇒sin⁡i=11+ay Slope dydx=cot⁡i=ay ∫dyy=a∫dx⇒2y=ax⇒x=2ya ymax=1.47m⇒xmax =21.471.5×10−6≈2000m

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A man of height 1.47 m stands on a straight road on a hot day. The variation of temperature in the air results in a variation of refractive index with height y as $μ = μ_{0} \sqrt{(1 + ay)} where μ_{0}$ is the refractive index of air near the $road a = 1.5 \times 10^{- 6} / m$ What is the apparent length of the road ( in m ) man is able to see ?

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$700 \sqrt{2}$

700 m

infinite distance

2000 m

answer is B.

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Detailed Solution

$O \to P : μ_{0} \sin \frac{π}{2} = μ \sin i [at O, y = 0 \Rightarrow μ = μ_{0} (1 + 0) = μ_{0}]$

$μ_{0} = μ_{0} \sqrt{1 + ay} \sin i \Rightarrow \sin i = \frac{1}{\sqrt{1 + ay}} Slope \frac{dy}{dx} = \cot i = \sqrt{ay} \int \frac{dy}{\sqrt{y}} = \sqrt{a} \int dx \Rightarrow 2 \sqrt{y} = \sqrt{a} x \Rightarrow x = 2 \sqrt{\frac{y}{a}} y_{max} = 1 .47 m \Rightarrow x_{max}$
$\begin{array}{l} = 2 \sqrt{\frac{1.47}{1.5 \times 10^{- 6}}} \\ \approx 2000 m \end{array}$