A) \[\frac{\omega M}{M+m}\]
B) \[\frac{\omega (M-2M)}{M+2m}\]
C) \[\frac{\omega (M+2M)}{M}\]
D) \[\frac{\omega M}{M+2m}\]
Correct Answer: A
Solution :
In Youngs double slit experiment, path difference is \[\frac{bx}{d}\] and for minima, this path difference should be \[\left( n+\frac{1}{2} \right)\lambda \] \[\therefore \] \[\frac{bx}{d}=\left( n+\frac{1}{2} \right)\lambda \] here \[x=\frac{b}{2}\] and \[x=0\] \[\frac{b}{d}.\frac{b}{1}=\frac{\lambda }{2}\] \[\frac{{{b}^{2}}}{d}=\lambda \]You need to login to perform this action.
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