question_answer

If a mica sheet of thickness t and refractive index \[\mu \] is placed in the path of one of interfering beams in a double slit experiment, then displacement of fringes will be:

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\[=\frac{(\mu -1)tD}{d}\]