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  • In the Young's double slit experiment, the distance between the slits varies in time as where <img alt="d_{0}, \omega$

In the Young's double slit experiment, the distance between the slits varies in time as d(t)=d_{0}+a_{0} \sin \omega t; 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 :
 
Option: 1 \frac{2 \lambda \mathrm{D}\left(\mathrm{d}_{0}\right)}{\left(\mathrm{d}_{0}^{2}-\mathrm{a}_{0}^{2}\right)}
Option: 2 \frac{2 \lambda \mathrm{Da}_{0}}{\left(\mathrm{~d}_{0}^{2}-\mathrm{a}_{0}^{2}\right)}
Option: 3 \frac{\lambda \mathrm{D}}{\mathrm{d}_{0}^{2}} \mathrm{a}_{0}
Option: 4 \frac{\lambda \mathrm{D}}{\mathrm{d}_{0}+\mathrm{a}_{0}}

Answers (1)

best_answer

d\left ( t \right )= d_{0}+a_{0}\sin\omega t
Fring\, width= \beta = \frac{\lambda D}{d}
\beta _{max}= \frac{\lambda D}{d_{min}}
\beta _{min}= \frac{\lambda D}{d_{max}}
d _{min}= d_{0}+a_{0}\left (- 1 \right )=d_{0}-a_{0}
d _{max}= d_{0}+a_{0}(1)=d_{0}+a_{0}
\Delta \beta = \beta _{max}-\beta _{min}=\frac{\lambda D}{d_{min}}-\frac{\lambda D}{d_{max}}
= \lambda D\left [ \frac{1}{d_{0}-a_{0}}-\frac{1}{d_{0}+a_{0}} \right ]
\Delta \beta =\frac{\lambda D\left ( 2a_{0} \right )}{d_{0}^{2}-a_{0}^{2}}
The correct option is (2)


 

Posted by

vishal kumar

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