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Oil film of thickness 1 mu m is deposited on a glass plate. Refractive index of oil is 1.25 and that of the glass is 1.5. Which wavelength in the visible region (400nm to 700 nm) will be strongly reflected by this film? |
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Answer» Solution :There is denser medium (glass) on one side of the FILM, hence condition for strong reflection in this case will be written as follows: `2 mu t = n lambda` `implies n lambda = 2 xx 1.25 xx 1 xx 10^(-6)` `implies n lambda = 2.25 xx 10^(-6)` Now we can substitute MAXIMUM and minimum wavelengths of given range to get corresponding minimum and maximum values of n. `n_("min") xx 700 xx 10^(-9) = 2.5 xx 10^(-6)` `implies n_("min") = 3.57` `n_("max") xx 400 xx 10^(-9) = 2.5 xx 10^(-6)` `n_("max") = 6.25` Integers in the above-calculated range are 4,5 and 6. Corresponding wavelengths can be calculated as follows: `n lambda = 2.5 xx 10^(-6) implies 4 lambda_1 = 2.5 xx 10^(-6) implies lambda_1 = 625 nm` `n lambda = 2.5 xx 10^(-6) implies 5 lambda_2 = 2.5 xx 10^(-6) implies lambda_2 = 500 nm` `n lambda = 2.5 xx 10^(-6) implies 6 lambda_3 = 2.5 xx 10^(-6) implies lambda_3 = 417 nm`. |
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