In a Young's double slit experiment, slits are separated by 0.5 mm, and the screen is placed 150 cm away. A beam of light consisting of two wavelengths, 650 n m and 520 n m, is used to obtain interference fringes on the screen. The least distance from the common central maximum to the point where the bright fringes due to both the wavelengths coincide is:
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a
9.75 m m
b
15.6 m m
c
1.56 m m
d
7.8 mm
answer is D.
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Detailed Solution
n1λ1=n2λ2n1n2=λ2λ1=520650=454th bright of 650 nm coincide with 5th bright of 520 n mx=nλDd=4×650×10−9×150×10−20.5×10−3=7.8 m m