Due to a vertical temperature gradient in the atmosphere, the index of refraction varies. Suppose index of refraction varies as `n=n_(0)``sqrt(1+ay)`, where `n_(0)` is the index of refraction at the surface and `a=2.0xx10^(-6)m^(-1)` . A person of height `h=2.0 m` stands on a level surface. Beyond what distance will he not see the runway?

Explanation

As refractive index is changing along y-direction, we can assum a number of thing layers of air placed parallel to x-axis. Let O be the distant object just visible to the man. Consider a layer of air at a distance y from the ground. Let P be a point on the trajectory of the ray. Form Fig. , `theta =90^(@)-i` .
The slope of trangent at point P is `tan theta=dy//dx =cot i` .
From Snell's law, n sin i= constant
At the surface, `n =n_(0)` and `i= 90^(@)` .
`n_(0)sin90^(@)=nsini=(n_(0)sqrt(1+ay))sini`
`sin i=(1)/(sqrt(1+ay))rArr cot i=(dy)/(dx)=sqrt(ay)`
`int_(0)^(y)(dy)/(sqrt(ay))=int_(0)^(x)dxrArr x=2sqrt((y)/(a))`
On substituting `y=2.0 m` and `a=2xx10^(-1)` , we have
`x_(max)=2sqrt((2)/(2xx10^(-6)))=2000m`