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In a Young's double slit experiment the ...

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

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The correct Answer is:
9
The `m^(th)` bright fringe of the `lamda` pattern and the `n^(th)` bright fringe of the `'lamda'` pattern are located at
`y_(m)=(mDlamda)/d` and `y_(m)=(n^(')Dlamda^('))/d`
Equating them `m/n=6/5`
Hence the first positon at which overlaping occurs is
`y_(6)=y_(5)^(')=(6xx2xx750xx10^(-9))/(1xx10^(-3))m=9mm`
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