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 assume a number of thin 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. From figure, `theta = 90-i`.
The slope of tangent at point P is
`theta = dy//dx = coti`,
From Snell.s law, n sin i = constant
At the surface, `n = n_0 and i = 90^@ .`

`n_0 sin 90^@ =n sin i= (n_0sqrt(1+ ay))sin i`
`sin i = (1)/( sqrt(1+ay))implies cot i = (dy)/(dx) = sqrt(ay)`
`int_(0)^(y) (dy)/(sqrt((ay)) = int_(0)^(x) dx implies x=2 sqrt((y)/(a))`
On substituting `y=2.0 m and a=2 xx 10^(-6) m^(-1)`
we have `x_("max") = 2 sqrt((2)/(2 xx 10^(-6)) =2000 m`