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Separation between the slits of YDSE set-up is 0.1 cm. Films of same thickness 0.5 mm is pasted on both the slits. Refractive index of one of the films 1.52 and for the other it is 1.48 Wavelength of the light used is 600 nm. Sereen is placed at a distance 1 m from the slits. Calculate fringe width. What will be the distance of first available maxima from the centre? |
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Answer» Solution :FRINGE width is not affected by the films and is WRITTEN as follows : `omega = (lambda)/d` `omega = (600 xx 10^(-9) xx 1)/(0.1 xx 10^(-2)) = 6.0 xx 10^(-4) m` Optical PATH difference introduced at the centre due to thin films can be written as follows: `Delta x = (mu_1 - mu_2)t` `= (1.52 - 1.48) xx 0.5 xx 10^(-3)` `= 0.04 xx 0.5 xx 10^(-3) = 2 xx 10^(-5) m` Now equating `Deltax = n lambda`, we get the following: `n = (Delta x)/(lambda) = (2 xx 10^(-5))/(600 xx 10^(-9)) = 33.33` From value of n obtained, we can understand that on one side maxima is at distance `y_1 = 0.33 W` and on the other side it is at a distance `y_2 = 0.67 w`, where `w` is fringe width of the pattern. `y_1 = 0.33 xx 6 xx 10^(-4) = 1.98 xx 10^(-4) m` `y_2 = 0.67 xx 6 xx 10^(-4) = 4.02 xx 10^(-4) m`. |
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