Question

In a 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 $t=1.5\times 10^{-7}m$, is refractive index $n=1.17$, is placed over one of the slits. Where does the central maxima of the interference now appear?

• 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 $\Delta x=(\mu -1)t$.
If the central maxima occupies position of nth fringe, then $(\mu -1)t=n\lambda =d\sin \theta$
$\sin \theta =\frac{(\mu -1)t}{d}=\frac{(1.17-1)\times 1.5\times 10^{-7}}{3\times 10^{-7}}=0.085$
Hence the angular position of central maxima is
$\theta ={\sin }^{-1}(0.085)=4.88^{\circ }$
For small angles $\sin \theta \simeq \theta \simeq \tan \theta$
$\tan \theta =\frac{y}{D}$
$\frac{y}{D}=\frac{(\mu -1)t}{d}$
Shift of central maxima is $y=\frac{D(\mu -1)t}{d}$