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White light reflectedat normal incidence from a soap film, has maximum at 6000 Åand the minimum at 4500 Å in the visible region with no minimum in between.If mu = 1.33 for the film, the thickness of the film is : |
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Answer» `0.23 MU m` For MAXIMA, `lambda_(1) = 6000 Å` For MINIMA, `lambda_(2) = 4500 Å` Applying the condition for maxima `2mu T = (2n -1).(lambda_(1))/(2)`....(1) and for minima `2mut = 2n .(lambda_(2))/(2)`...(2) From (1) and (2) `(2n - 1) .(lambda_(1))/(2) = n lambda_(2)` `(2n -1)6000 = 2n XX 4500` `therefore`n = 2 Now from eqn.(2) `2 xx 1.33 xx t = 2 xx 2xx(4500)/(2)` `therefore tapprox 0.34 mu m`. |
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