Question

In YDSE, light of wavelength $\lambda=5000\,\mathring{A}$ 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 $n=1.17$, is placed over one of the slits. Where does the central maxima of the interference now appear from the center of the screen? (Find the value of $y$ ?)

• A. $\frac{D(\mu-1) t}{2 d}$
• B. $\frac{2 D(\mu-1) t}{d}$
• C. $\frac{D(\mu+1) t}{d}$
• D. $\frac{D(\mu-1) t}{d}$

Answer 1

Answer: (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 $n$th 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}} $
$=0.085$
Hence, the angular position of central maxima is
$\theta=\sin ^{-1}(0.085)=4.88^{\circ}$
For small angles,
$\sin \theta=\theta=\tan \theta$
$\Rightarrow \tan \theta=\frac{y}{D} $
$\therefore \frac{y}{D}=\frac{(\dot{\mu}-1) t}{d}$
Shift of central maxima is
$y=\frac{D(\mu-1) t}{d}$
This formula can be used if $D$ is given.