Question

In YDSE, light of wavelength $\lambda$ is used, which emerges in phase from two slits a distance $d$ apart. A transparent sheet of thickness $t$, refractive index $\mu =1.17$, 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

The correct option is D $\frac{D(\mu -1)t}{d}$

Due to the introduction of sheet in front of any one slit, the fringes get shifted. This lateral shift is given as

$\overline{X}=\frac{\beta }{\lambda }(\mu -1)t=\frac{D}{d}(\mu -1)t$