In young's double slit experiment performed using a monochromatic light of wavelength $$\lambda$$, when a glass plate ($$\mu = 1.5$$) of thickness $$x\lambda$$ is introduced in the path of the one of the interfering beams, the intensity at the position where the central maximum occurred previously remains unchanged. The value of $$x$$ will be

In Young's double slit experiment, a glass plate of refractive index $$\mu = 1.5$$ and thickness $$x\lambda$$ is introduced in one of the beams. The intensity at the original central maximum position remains unchanged.

Calculate the extra path difference introduced by the glass plate.

$$\Delta = (\mu - 1)t = (1.5 - 1)(x\lambda) = 0.5 x\lambda$$

Apply the condition for unchanged intensity.

For the intensity at the central maximum to remain unchanged, the phase difference introduced must be an integer multiple of $$2\pi$$. This means the extra path difference must be an integer multiple of $$\lambda$$:

$$\Delta = n\lambda \quad (n = 1, 2, 3, ...)$$

$$0.5 x\lambda = n\lambda$$

$$x = 2n$$

Find the minimum value of $$x$$.

For $$n = 1$$: $$x = 2$$

The correct answer is Option B.