InterviewSolution
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InterviewSolution
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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? |
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Answer» Solution :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` 
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