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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 = 2.0 xx 10^(-6)m^(-1). A person of height h = 2.0m 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 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_0and 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))` Onsubstituting`y=2.0m anda=2 xx 10^(-6) m^(-1)` we have `x_("max") = 2 sqrt((2)/(2 xx 10^(-6)) =2000 m` |
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