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Question

In young's double slit experiment the fringe width with light of wave length 6000 A is found to be 4.0mm. What will be the fringe width if light of wavelength 4800 A is used?

A
2.8mm
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B
3.2mm
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C
4.0mm
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D
4.8mm
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Solution

The correct option is B 3.2mm
Given β=4.0mm and λ=6000A.
We know that the fringe width is given by
β=λDd (i)for λ=4800A, the fringe width will be β=λDd (ii)From Eqs. (i) and (ii) we haveβ=βλλ=4.0mm×4800A6000A=3.2mm

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