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A double slit apparatus is immersed in a liquid of refractive index 1.33. It has slit separation of 1mm and distance between the plane of slit and screen 1.33m. The slits are illuminated by a parallel beam of light whose wavelength in air is 6300 Å. (a) Calculate the fringe width (b) One of the slits of the apparatus is covered by a thin glass sheet of refractive index 1.53. Find the smallest thickness of the sheet to bring the adjacent minima on the axis. |
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Answer» Solution :(a) As FRINGE WIDTH `beta = (D lamda // d)`and by presence of medium the wavelength becomes `(lamda//mu)` , so the fringe-width in LIQUID will be `beta. = (D lamda)/(mu d) = (1.33 xx 6300 xx 10^(-10))/(1.33 xx 1 xx 10^(-3))` ` =0.63 mm` (b) Now as the distance of a minima from ADJACENT maxima is `beta. //2 `so ACCORDING to given problem, shift `y_0 = D/d (mu_r - 1)t = (beta.)/(2)` with `mu_r = (mu_P)/(mu_M)` `D/d [ (mu_P)/(mu_M) - 1] t = (D lamda)/(2mu_M d) i.e., t = (lamda)/(2(mu_P - mu_M))` so `t = (6300 xx 10^(-10))/(2(1.53 - 1.33)) = 1.575 mu m` |
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