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Question

In a Young's double slit experiment, slits are separated by 0.5mm, and the screen is placed 150cm away. A beam of light consisting of two wavelengths, 650nm and 520nm, 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

A
15.6 mm
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B
1.56 mm
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C
7.8 mm
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D
9.75 mm
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Solution

The correct option is C 7.8 mm
Fringe width of pattern due to λ1 wavelength only =λ1Dd=β1
Fringe width of pattern due to λ2 only =λ2Dd=β2

Hence, the distance from central maximum to point where the bright fringes first meet must be the least integral multiple common to both.
n1β1=n2β2
n1λ1Dd=n2λ2Dd
n1(650)=n2(520)
n1=4,n2=5
Hence the required distance =n1λ1Dd
=4×(150×102)×(650×109)(0.5×103)
=7.8 mm

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