Question

In a YDSE, light of wavelength $\lambda =5000\stackrel{\circ }{A}$ 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

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


By using the slit having thickness $t$ and having refractive index of $\mu$.
then the total optical path difference,
$(\mu t-t)-\frac{yd}{D}=0$ (for central maxima)

so, $(\mu t-t)=\frac{yd}{D}(\mu -1)t=\frac{yd}{D}\Rightarrow y=\frac{(\mu -1)tD}{d}$

Hence, optioin D is correct.