1. Path difference introduced by the glass plate

When a glass plate of refractive index $\mu$ and thickness $t$ is introduced in one path, the additional optical path difference is

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

Here,

• $\mu = 1.5$
• $t = x\lambda$

So,

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

2. Condition for intensity at old central maximum to remain unchanged

Originally, at the central maximum, path difference was zero, so intensity was maximum.

After inserting the plate, the phase difference changes due to the extra path difference $\Delta$.

For the intensity at that same point to remain unchanged (i.e. still maximum), the new path difference must be an integral multiple of wavelength:

$\Delta = n\lambda, \quad n = 0,1,2,\dots$

Substitute $\Delta = 0.5x\lambda$:

$0.5x\lambda = n\lambda$

Cancelling $\lambda$,

$0.5x = n$

$x = 2n$

So possible values are $x = 0, 2, 4, \dots$

Among the given options, the valid one is

$x = 2$

3. Checking options

• A: $x=3 \Rightarrow \Delta = 1.5\lambda$ → not same maximum intensity
• B: $x=2 \Rightarrow \Delta = 1\lambda$ → same maximum intensity
• C: $x=1.5 \Rightarrow \Delta = 0.75\lambda$ → intensity changes
• D: $x=0.5 \Rightarrow \Delta = 0.25\lambda$ → intensity changes

Hence, the correct option is B.