1.

In Young's experiment for the interference of light, the separation between the slits is d and the distance of the screen from the slits is D. if D is increased by 0.5% and d is decreased by 0.3% then for the light of a given wavelength, which one of the following is true ? "The fringe width. . . "

Answer»

INCREASES by 0.8%
decreases by 0.8%
increases by 0.2%
decreases by 0.2%

Solution :We KNOW that, fringe WIDTH, `BETA=(lamdaD)/(d)`
On differentiating, we get `(Deltabeta)/(beta)=(DeltaD)/(D)-(Deltad)/(d)`
`implies (Deltabeta)/(beta)xx100%=(DeltaD)/(D)xx100-(Deltad)/(d)xx100`
`=(0.5)-(-0.3)=0.8%`
THEREFORE, fringe width increases by 0.8%.


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