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In Young's double slit experiment, the fringe width is 1×10−4m if the distance between the slit and screen is doubled and the distance between the two slit is reduced to half and wavelength is changed from 6.4×107m to 4.0×10−7m, the value of new fringe width will be

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a

0.15×10−4m

b

2.0×10−4m

c

1.25×10−4m

d

2.5×10−4m

answer is D.

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Detailed Solution

β=λDd⇒β2β1=λ2D2d1λ1D1d2⇒β2=2.5×10−4m

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