April 15, 2010
Optics Of The Atmosphere
Optics Of The Atmosphere
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Physics Geometric Optics question?
Question:
A light ray enters the atmosphere of a large planet and descends vertically to the surface a distance h = 1.00X10^5 km below. The index of refraction where the light enters the atmosphere is, of course, 1.000. As the light descends into the atmosphere the index of refraction n(x) increases linearly with distance x below the top of the atmosphere, reaching the value n(h) = n0 = 1.50 at the surface of the planet (where x = h).
(a) How long after the light enters the atmosphere (t=0) does it reach the surface of the planet?
(b) How does your value for t compare with the time interval required in the absence of an atmosphere?
By definition n = c/cm where cm is the speed of light in the medium and c is speed in vacuum
So c top = 3.00x10^8m/s and c bottom = 3.00x10^8/1.50 = 2.00x10^8m/s
Since the decrease is linear the avg value os cm = (2.00x10^8 + 3.00x10^8)/2 = 2.50x10^8m/s
So the time is x/v = 1.00x10^8/2.50x10^8 = 0.400s
b) without atmosphere t = x/c = 1.00x10^8/3.00x10^8 = 0.333s
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