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 $\frac{D(\mu-1) t}{2 d}$
• 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, $\sin \theta \approx \theta \approx \tan \theta$
$\Rightarrow \tan \theta=\frac{y}{D}$
$\therefore \frac{y}{D} =\frac{(\mu-1) t}{d}$
$\Rightarrow y=\frac{D(\mu-1) t}{D}$