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In Young's double slit experiment, the slits are 2 mm apart and are illuminated with a mixture of two wavelength lamda_0 = 750 nm and lamda = 900 nm. What is the minimum distance from the common central bright fringe on ascreen 2 m from the slits where a bright fringe from one interference pattern coincides with a bright fringe from the other? |
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Answer» Solution :`n_(1)lambda_(1) = n_(2)lambda_(2)` `(n_(1))/(n_(2))=(lambda_(2))/(lambda_(1))=(900)/(750)` D = 2m d = 2 MM THUS, WEHAVE `(n_(1))/(n_(2)) = (6)/(5)` `5^(th) and 6^(th)` fringes will coincide RESPECTIVELY. The minimum distance is given as `X_(min)=(n_(2)lambda_(2)D)/(d)=(5xx900xx10^(9)xx2)/(2xx10^(-3))` `= 4500 xx 10^(-6) = 4.5 xx 10^(-3) m` `X_(min) = 4.5 mm` |
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