Question

In a YDSE, the light of wavelength $I=5000\mathring{A}$ 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 center 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)

The path difference when transparent sheet is introduced
$\Delta x=(\mu -1)t$
If the central maxima occupies position of nth fringe ,then
$(\mu -1)t=n\lambda =d\sin \theta$
$\Rightarrow \sin \theta =\frac{(\mu -1)t}{d}$
$=\frac{(1.17-1)\times 1.5\times 10^{-7}}{3\times 10^{-7}}=0.085$
Therefore, angular position of central maxima
$\theta ={\sin }^{-1}(0.085)=4.88^{\circ }\approx 4.9$
For For small angles, $\sin \theta \approx \theta \approx \tan \theta$
$\Rightarrow \tan \theta =\frac{y}{D}$
$\therefore \frac{y}{D}=\frac{(\mu -1)t}{d}$
$\Rightarrow y=\frac{D(\mu -1)t}{D}$