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=4.0mmandλ=6000A.
We know that the fringe width is given by β=λDd(i)forλ′=4800A,thefringewidthwillbeβ′=λ′Dd(ii)FromEqs.(i)and(ii)wehaveβ′=βλ′λ=4.0mm×4800A6000A=3.2mm