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In a Young's double slits experiment, th...

In a Young's double slits experiment, the slit are 1 mm apart and are illuminated with a mixture of two wavelengths `lamda = 750 nm` and `lamda' = 900 nm` and distance between slit and screen is 2m. At what minimum distance (in nm) form the common central bright fringe on a screen the bright fringe frm one interference pattern coincides with a bright from the other ?

A

6 mm

B

12 mm

C

8 mm

D

9 mm

Text Solution

Verified by Experts

The correct Answer is:
D
The `m^(th)` bright fringe of the `lambda` pattern and the `n^(th)` bright fringe of the `lambda` pattern are located at
`y_(m) = (mDlambda)/(d)` and `y_(m)' = (n'D'lambda')/(d)` ltbr. Equating them `(m)/(n) = (6)/(5)`
Hence the first position at which overlaping occure is
`y_(6) = y_(5)' = (6 xx 2 xx 750 xx 10^(-9))/(1 xx 10^(-3)) m = 9 mm`
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