Question

A monochromatic beam of light falls on $YDSE$ apparatus at some angle $\left(\text{Assume}\theta \right)$ as shown in figure. A thin sheet of glass of thickness $t$ is inserted in front of the lower slit $S_2$. The central bright fringe $(\text{path difference}=0)$ will be obtained

• A. $\text{at O}$
• B. $\text{above O}$
• C. $\text{below O}$
• D. $\text{anywhere depending on angle}\theta ,\text{thickness of plate t, and refractive}\text{index of glass}\mu$

Answer 1

The correct option is D $\text{anywhere depending on angle}\theta ,\text{thickness of plate t, and refractive}\text{index of glass}\mu$

The path difference in reaching a point on the screen (without glass slab),
$\Delta x=d\sin \theta$

The path difference due to introduction of glass slab is,
$\Delta x=(\mu -1)t$

So,
If $d\sin \theta =(\mu -1)t$, central fringe is obtained at $O$

If $d\sin \theta >(\mu -1)t$, central fringe is obtained above $O$

If $d\sin \theta <(\mu -1)t$, central fringe is obtained above $O$

Hence, The central bright fringe $(\text{path difference}=0)$ will be obtained anywhere depending on angle $\theta$, thickness of plate $t$ and refractive index of slab $\mu$.

Hence, $\left(D\right)$ is the correct answer.

Key Note: Because of beam inclination means light reaches $S_1$ before it reaches $S_2$, hence an induced time lag before the slits. It means that, without any glass sheet, the central fringe would be above point $O$ because of this inclination. But due to presence of glass sheet, it can be anywhere depending on angle $\theta$, thickness of plate $t$, and refractive index of glass $mu$.