In the Young's double slit experiment, the distance between the slits varies in time as{d}({t})={d}_{0}+{a}_{0} \sin\\omegat; where {d}_{0}, \\omega and {a}_{0} are constants. The difference between the largest fringe width and the smallest fringe width obtained over time is given as : [25 Jul 2021 Shift 1]

Correct Answer is : 2λDa0(d02−a02)

{2 \\lambdaD(d_{0})}/{(d_{0}^{2}-a_{0}^{2})}

{2 \\lambda{Da}_{0}}/{({d}_{0}^{2}-{a}_{0}^{2})}

{\\lambda{D}}/{{d}_{0}^{2}} {a}_{0}

JEE Main 25 July 2021 Shift 1 Solved Paper

Section: Physics

Question : 14 of 90

Marks: +1, -0

In the Young's double slit experiment, the distance between the slits varies in time asd(t)=d0+a0‌sin‌ω‌t; where d0,ω and a0 are constants. The difference between the largest fringe width and the smallest fringe width obtained over time is given as :

[25 Jul 2021 Shift 1]

2λD(d0)
(d02−a02)
2λDa0
(d02−a02)
λD
d02
a0
λD
d0+a0

Validate

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