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Soap film of thickness 1 mu m appears bright when seen through reflected light of wavelength 700 nm. What should be the index of refraction of soap solution if it is somewhere between 1.2 and 1.3 ? |
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Answer» Solution :Soap film appears bright by reflected light, hence condition of strong reflection is to be used and the same canbe written as follows : `2 mu t = (2n + 1) lambda//2` We can SEE in the above RELATION that `mu` and `n` are directly RELATED to each other. Hence for MINIMUM `mu` we can calculate minimum `n` and similarly for maximum `mu` we can calculate maximum n. And then in this range we need to find integer. Maximum n : `2nt = (2n+ 1) lambda/2` `implies 2 xx 2 xx 1 xx 10^(-6) = (2n + 1)(700 xx 10^(-9))/(2)` `implies 2n + 1 = (2 xx 1.3 xx 1 xx 10^(-6) xx 2)/(700 xx 10^(-9))` `implies 2n + 1 = 7.43` `implies 2n = 6.43` `implies n = 3.21` Integer between the ranges calculated above is just 3, so we can substitute it back to calculate the exact index of refraction. `2mu t = (2n + 1) lambda//2` `implies 2 xx mu xx 1 xx 10^(-6) = (2 xx 3 + 1) (700 xx 10^(-9))/(2)` `implies mu = 7/(2 xx 10^(-6)) (700 xx 10^(-9))/(2)` `implies mu = (7 xx 0.7)/(4)` `implies mu = 1.225`. |
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