Question

The Refractive index μ atmosphere around a planet varies with altitude h according to equation $\mu ={\mu }_0-kh$. Here ${\mu }_0$ and $k$ are positive constants. Denoting radius of planet by $R$, altitude at which a point flash of light spreads around the planet in a spherical layer is $\frac{n({\mu }_0-kR)}{k}$, then value of $n$ is

• A. 0.5
• B. 1
• C. 1.5
• D. 2

Answer 1

The correct option is A 0.5

The point flash of light needs to flash at critical ic so that it spreads around the planet in spherical layers.
Thus, $\sin i_c=\frac{{\mu }_r}{{\mu }_i}=\frac{{\mu }_0-k(h+dh)}{{\mu }_0-kh}$ ………….(i)
By geometry we can find
$\sin i_c=\frac{(R+h)}{R+h+dh}$ ……………(ii)
On comparing (i) and (ii) we get, $\frac{R+h}{(R+h)[1+\frac{dh}{(R+h)}]}$= $\frac{{\mu }_0-kh}{{\mu }_0-kh}-\frac{kdh}{{\mu }_0-kh}$
${(1+\frac{dh}{R+h})}^{-1}=1-\frac{kdh}{{\mu }_0-kh}$
$1-\frac{\left(dh\right)}{R+h}=1-\frac{k\left(dh\right)}{{\mu }_0-kh}$
On solving
h = $\frac{({\mu }_0-kR)}{2k}$