A) \[\frac{D}{d}\mu t\]
B) \[\frac{D}{d}(\mu -1)t\]
C) \[\frac{D}{d}(\mu +1)t\]
D) \[\frac{D}{d}({{\mu }^{2}}-1)t\]
Correct Answer: B
Solution :
We can realize the situation as shown. Geometric path difference between \[{{S}_{2}}P\]and \[{{S}_{1}}P\]is \[\Delta {{x}_{1}}={{S}_{21}}P-{{S}_{1}}P=\frac{yd}{D}\] where d is distance between the slits and D the distance between source and screen. Path difference produced by the mica sheet \[\Delta {{x}_{2}}=(\mu -1)t\] Therefore, net path difference between the two rays is \[\Delta x=\Delta {{x}_{1}}-\Delta {{x}_{2}}\] or \[\Delta x=\frac{yd}{D}-(\mu -1)t\] For nth maxima on upper side, \[\Delta x=n\lambda \] \[\therefore \] \[\frac{yd}{D}-(\mu -1)t=n\lambda \] \[\therefore \] \[y=\frac{n\lambda D}{d}+\frac{(\mu -1)t\,D}{d}\] Earlier it was\[\frac{n\lambda D}{d}.\] \[\therefore \]Displacement of fringes\[=\frac{(\mu -1)tD}{d}\]You need to login to perform this action.
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