Optics Question 475
Question: In a YDSE light of wavelength $ \lambda =5000 $
$ \overset{\text{o}}{\mathop{\text{A}}} $ is used which emerges in phase from two slits a distance $ d=3\times {{10}^{-7}}m $ apart. A transparent sheet of thickness $ t=1.5\times {{10}^{-7}}m $ , is refractive index n = 1.17, is placed over one of the slits. Where does the central maxima of the interference now appear?
Options:
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} $
Show Answer
Answer:
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.
