question_answer

If a mica sheet of thickness t and refractive index u 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 \[\Delta {{x}_{1}}={{S}_{2}}P-{{S}_{1}}P=\frac{yd}{D}\] where d is distance between the slits and D the distance between source and screen. Path difference produced by the mica sheet \[\Delta \,{{x}_{2}}=({{\mu }_{2}}-1)\,t\] 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 \[=\frac{(\mu -1)t\,D}{d}\]