Question

In a $YDSE$, the light of wavelength $\lambda =5000\stackrel{\circ }{A}$ is used, which emerges in phase from two slits a distance $d=3\times {10}^{-7}m$ apart. A transparent sheet of thickness $t=1.5\times {10}^{-7}m$ refractive index $\mu =1.17$ is placed over one of the slits. What is the new angular position of the central maxima of the interference pattern, from the centre of the screen? Find the value of $y$.

• A. ${4.9}^{\circ }$ and $\frac{D(\mu -1)t}{2d}$
• B. ${4.9}^{\circ }$ and $\frac{D(\mu -1)t}{d}$
• C. ${3.9}^{\circ }$ and $\frac{D(\mu +1)t}{d}$
• D. ${2.9}^{\circ }$ and $\frac{D(\mu +1)t}{d}$

Answer 1

Answer: (B)

${4.9}^{\circ }$ and $\frac{D(\mu -1)t}{d}$

The path difference when transparent sheet is introduced $△x=(\mu -1)t$
If the central maxima occupies position of $nth$ fringe, then $(\mu -1)t=n\lambda =d\sin \theta$
$\Rightarrow \sin \theta =\frac{(\mu -1)t}{d}=\frac{(1.71-1)\times 1.5\times {10}^{-7}}{3\times {10}^{-7}}=0.085$
Therefore, angular position of central maxima
$\theta ={\sin }^{-1}\left(0.085\right)={4.88}^{\circ }\approx 4.9$
For small angles, $\sin \theta \approx \theta \approx \tan \theta$
$\Rightarrow \tan \theta =\frac{y}{D}$
$\therefore \frac{y}{D}=\frac{(\mu -1)t}{d}\Rightarrow y=\frac{D(\mu -1)t}{d}$.