In a YSDE, light of wavelength $\lambda =5000A^0$ is used, which emerges in phase from two slits a distance $d=3\times {10}^{-7}m$ apart. A transparent sheet of thick ness $t=1.5\times {10}^{-7}m$ , refractive index $n=1.17$ , is placed over one of the slits, where does the central maxima of the interference now appear?

                   

 

 

The path difference introduced due to introduction of transparent sheet is given by $\Delta x=\left(m-1\right)t$

If the central occupies position of nth fringes, then

$\left(\mu -1\right)t=n\lambda =d\sin \theta$

$\Rightarrow \sin \theta =\frac{\left(\mu -1\right)t}{d}=\frac{\left(1.17-1\right)\times 1.5\times {10}^{-7}}{3\times {10}^{-7}}$

$=0.085$

Hence, the angular position maxima is

$\theta ={\sin }^{-1}\left(0.085\right)={4.88}^0$

For small angles

$\sin \theta =\theta =\tan \theta$

$\Rightarrow \tan \theta =\frac{y}{D}$

$\therefore \frac{y}{D}=\frac{\left(\mu -1\right)t}{d}$

Shift of central maxima is

$y=$ $\frac{D\left(\mu -1\right)t}{d}$

This formula can be used if D is given