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