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Question

# In Young's double slit experiment the slits are 0.5 mm apart and interference is observed on a screen placed at a distance of 100 cm from the slits. It is found that the 9th bright fringe and 3rd dark fringe from the centre are 9mm apart. What is the wavelength of light used?

A

2000A

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B

4000A

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C

6000A

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D

8000A

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Solution

## The correct option is C 6000A The distance of the mth bright fringe from the central fringe is ym=mλDd=mβwhere β=λD/d is the fringe width.∴y9=9β (i) The distance of the mth dark fringe from the central fringe is y′m=(m−12)λDd=(m−12)β∴y′2=32β (ii) From Eqs. (i) and (ii), we get y9−y′2=9β−32β =152β It is given that y9−y′2=9.0mm. Henceβ=9.0×215=1.2mmNow λ=βd/D. Substituting for β, d and D, we getλ=6×10−7m=6000A

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