Wave Optics

JEE Main level, single correct, medium

86 of 191

Question 86 of 191

In the Young's double slit experiment, the distance between the slits varies in time as d(t)=d0+a0sin⁡ωtd\left(t\right)={d}_{0}+{a}_{0}\sin \omega t; where d0,ω{d}_{0},\omega and a0{a}_{0} are constants. The difference between the largest fringe width and the smallest fringe width obtained over time is given as:

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Answer: (B) 2λDa0(d02−a02)\dfrac{2\lambda D{a}_{0}}{\left({d}_{0}^{2}-{a}_{0}^{2}\right)}

Fringe width,
β=λDd\beta =\dfrac{\lambda D}{d}
βmax⁡⇒dmin⁡{\beta }_{\max }\Rightarrow {d}_{\min } and βmin⁡⇒dmax⁡{\beta }_{\min }\Rightarrow {d}_{\max }
d=d0+a0sin⁡ωtd={d}_{0}+{a}_{0}\sin \omega t
dmax⁡=d0+a0{d}_{\max }={d}_{0}+{a}_{0} and dmin⁡=d0−a0{d}_{\min }={d}_{0}-{a}_{0}
∴βmin⁡=λDd0+a0∴{\beta }_{\min }=\dfrac{\lambda D}{{d}_{0}+{a}_{0}} and ∴βmax⁡=λDd0−a0∴{\beta }_{\max }=\dfrac{\lambda D}{{d}_{0}-{a}_{0}}
βmax⁡−βmin⁡=λDd0−a0−λDd0+a0=2λDa0d02−a02{\beta }_{\max }-{\beta }_{\min }=\dfrac{\lambda D}{{d}_{0}-{a}_{0}}-\dfrac{\lambda D}{{d}_{0}+{a}_{0}}=\dfrac{2\lambda D{a}_{0}}{{d}_{0}^{2}-{a}_{0}^{2}}

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