JEE Main 2024 — Wave Optics Question with Solution

To solve this problem, we need to understand how the interference pattern shifts due to the introduction of a thin glass plate in Young's double slit experiment. This shift occurs because the light passing through the glass experiences a different optical path length compared to the light passing through the other slit without glass.

### Step-by-Step Analysis:

1. Calculate the Path Difference Due to the Glass Plate:

• The path difference is influenced by the optical thickness of the glass, which is the product of the physical thickness $t$ of the glass and the refractive index $n$ minus the path it would have in air (i.e., $n\times t-t$).


• The effective additional path difference in the medium of the glass plate is $(n-1)\times t$.

1. Determine the Shift in Fringes:

• The shift of the central maximum to the position previously occupied by the fourth bright fringe indicates that the optical path difference created by the glass plate corresponds to four fringe spacings.


• Each fringe width corresponds to a change in path difference of one wavelength ($\lambda$).

1. Calculating the Thickness $t$ of the Glass Plate:

• Since the central maximum shifts by four fringes, the path difference $\Delta$ caused by the glass plate must be equal to $4\lambda$.


• Therefore, $(n-1)\times t=4\lambda$.

1. Insert Values and Solve for $t$:

• Given:


• $n=1.5$ (refractive index of the glass)


• $\lambda =500\text{nm}=500\times 10^{-9}\text{m}$ (wavelength of the light)


• $(1.5-1)\times t=4\times 500\times 10^{-9}\text{m}$


• $0.5\times t=2000\times 10^{-9}\text{m}$


• $t=\frac{2000\times 10^{-9}\text{m}}{0.5}$


• $t=4000\times 10^{-9}\text{m}$

Conversion to Micrometers:

• $4000\times 10^{-9}\text{m}$ equals $4000\text{nm}$


• Since $1\mu \text{m}=1000\text{nm}$,


• $t=4\mu \text{m}$.

Conclusion:

The thickness of the glass plate required to shift the central maximum to the position previously occupied by the fourth bright fringe is $4\mu \text{m}$.