A) \[\frac{D(\mu -1)t}{2d}\]
B) \[\frac{2D(\mu -1)t}{d}\]
C) \[\frac{D(\mu +1)t}{d}\]
D) \[\frac{D(\mu -1)t}{d}\]
Correct Answer: D
Solution :
[d] The path difference introduced due to introduction of transparent sheet is given by \[\Delta x=(m-1)t.\] If the central maxima occupies position of nth fringe, then \[(\mu -1)t=n\lambda =dsin\theta \] \[\sin \theta =\frac{(\mu -1)t}{d}=\frac{(1.17-1)\times 1.5\times {{10}^{-7}}}{3\times {{10}^{-7}}}=0.085\]Hence the angular position of central maxima is \[\theta ={{\sin }^{-1}}(0.085)=4.88{}^\circ \] For small angles \[\sin \theta \simeq \theta \simeq \tan \theta \] \[\tan \theta =\frac{y}{D}\] \[\frac{y}{D}=\frac{(\mu -1)t}{d}\] Shift of central maxima is \[y=\frac{D(\mu -1)t}{d}\] This formula can be used if D is given.
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