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In the Young's double slit experiment,

Question:

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

  1. $\frac{2 \lambda D\left(d_{0}\right)}{\left(d_{0}^{2}-a_{0}^{2}\right)}$

  2. $\frac{2 \lambda \mathrm{Da}_{0}}{\left(\mathrm{~d}_{0}^{2}-\mathrm{a}_{0}^{2}\right)}$

  3. $\frac{\lambda \mathrm{D}}{\mathrm{d}_{0}^{2}} \mathrm{a}_{0}$

  4. $\frac{\lambda \mathrm{D}}{\mathrm{d}_{0}+\mathrm{a}_{0}}$


Correct Option: , 2

Solution:

Fringe Width, $\beta=\frac{\lambda D}{d}$

$\beta_{\max } \Rightarrow d_{\min }$ and $\beta_{\min } \Rightarrow d_{\max }$

$\mathrm{d}=\mathrm{d}_{0}+\mathrm{a}_{0} \sin \omega \mathrm{t}$

$\mathrm{d}_{\max }=\mathrm{d}_{0}+\mathrm{a}_{0}$ and $\mathrm{d}_{\min }=\mathrm{d}_{0}-\mathrm{a}_{0}$

$\therefore \beta_{\min }=\frac{\lambda \mathrm{D}}{\mathrm{d}_{0}+\mathrm{a}_{0}}$ and $\therefore \beta_{\max }=\frac{\lambda \mathrm{D}}{\mathrm{d}_{0}-\mathrm{a}_{0}}$

$\beta_{\max }-\beta_{\min }=\frac{\lambda \mathrm{D}}{\mathrm{d}_{0}-\mathrm{a}_{0}}-\frac{\lambda \mathrm{D}}{\mathrm{d}_{0}+\mathrm{a}_{0}}=\frac{2 \lambda \mathrm{Da}_{0}}{\mathrm{~d}_{0}^{2}-\mathrm{a}_{0}^{2}}$

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