Monochromatic light of wavelength $$500 \text{ nm}$$ is used in Young's double slit experiment. An interference pattern is obtained on a screen. When one of the slits is covered with a very thin glass plate (refractive index $$= 1.5$$), the central maximum is shifted to a position previously occupied by the $$4^{th}$$ bright fringe. The thickness of the glass-plate is ______ $$\mu m$$.

We need to find the thickness of a glass plate that shifts the central maximum to the position of the 4th bright fringe in Young's double slit experiment.

Wavelength: $$\lambda = 500 \text{ nm}$$

Refractive index of glass plate: $$\mu = 1.5$$

Central maximum shifts to the position of the 4th bright fringe.

When a glass plate of thickness $$t$$ and refractive index $$\mu$$ is placed in front of one slit, it introduces an additional optical path. Inside the glass, the wavelength becomes $$\lambda/\mu$$, so the optical path through the glass is $$\mu t$$ instead of $$t$$. The extra path difference introduced is:

$$\Delta = \mu t - t = (\mu - 1)t$$

The central maximum originally occurs where the path difference from both slits is zero. With the glass plate, the central maximum shifts to the point where the extra path difference due to the glass exactly compensates. For the central maximum to shift to the position of the $$n^{th}$$ bright fringe, the extra path difference must equal $$n\lambda$$:

$$(\mu - 1)t = n\lambda$$

Here, $$n = 4$$ (shift to 4th bright fringe position):

$$(1.5 - 1) \times t = 4 \times 500 \text{ nm}$$

$$0.5 \times t = 2000 \text{ nm}$$

$$t = \frac{2000}{0.5} = 4000 \text{ nm} = 4 \text{ } \mu m$$

The answer is 4 $$\mu$$m.