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Q.

In YDSE a light containing two wavelengths 500 nm and 700 nm are used. The minimum distance from central maxima (in mm) where maxima of two wavelengths coincide is “P”. Then 2P is:
Given D/d=103 , where D is the distance between the slits and the screen and d is the distance between the slits 

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answer is 7.

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Detailed Solution

At the place where maxima for both the wavelengths coincide, y will be same for both the maxima, e.e,
 n1λ1Dd=n2λ2Ddn1n2=λ2λ1=700500=75
Minimum distance of maxima of the two wavelengths from central fringe is=5×700×109×103=3.5mm
`  2*P = 2*3.5 =7mm

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