InterviewSolution
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InterviewSolution
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A monochromatic light of `lambda = 500 Å` is incident on two indentical slits separated by a distance of `5 xx 10^(-4)` m. The interference pattern is seen on a screen placed at a distance of 1 m from the plane of slits. A thin glass plate of thickness `1.5 xx 10^(-6)` m and refractive index `mu = 1.5` is placed between one of the slits and the screen . Find the intensity at the center of the screen. |
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Answer» In case of interference `I_(R) = I_(1) + I_(2) + 2 (sqrt (I_(1) I_(2))) cos phi` Now, as for identical slits `I_(1) = I_(2) = I,` so `I_(R) = 2 (1 + cos phi) = 4 I cos^(2) (phi // 2)` But for central maxima, `phi = 0^(@),` and here `I_(R) = I_(0)` (given) `I_(0) = 4 I cos (0^(@)) = 4 I` Hence, `I_(R) = I_(0) cos^(2) ((phi)/(2))` Now, when the glass plate is introduced, path difference between the waves at the position of central maxima will become `Delta x = ( mu - 1) t = (1.5 - 1) 1.5 xx 10^(-6) = 7.5 xx 10^(-7) m` `:. phi = (2 pi)/(lambda) (Delta x) = (2 pi)/(5 xx 10^(-7)) xx 7.5 xx 10^(-7) = 3 pi` So, intensity at central maxima will now be `I_(R) = I _(0) cos^(2) ((3 pi)/(2)) = I_(0) xx 0 = 0` Also, from theory of intergerence, fringe shift `y_(0) = (D)/(d) (mu - 1) t` Which in the light of Eq. (ii) and given data becomes `y_(0) = 7.5 xx 10^(-7) = 1.5 xx 10^(-3) m = 1.5 mm`. |
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