A beam of light consisting of two wavelengths, 650 nm and 520 nm is used to obtain interference in YDSE. If bright fringes due to both the wavelengths coincide at any point P then least distance of P from central maxima is (Slit separation is 2 mm and distance between slits and screen is 1.2 m)
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a
1.56 mm
b
2.36 mm
c
1.8 mm
d
3 mm
answer is A.
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Detailed Solution
condition for two maxima overlap nλ1=(n+1)λ2 n 650 =n+1 520 solving n=4 4th maxima of 650 nm distance from cenral maxima 4β= 4λ Dd =4x 650x10-9x 1.2 2x 10-3= 1.56 mm