Question

In the ideal double slit experiment, when a glass plate $(\mu =1.5$) of thickness $t$ is introduced in the path of one of the interfering beams (wavelength $\lambda$), the intensity at the position where central maxima occurred previously remains unchanged. The minimum thickness of glass plate is $k\lambda$, then the value of $k$ is (Integer only)

Answer 1

In order to obtain a maxima at the previous central maxima position, the path differences between interfering waves,

$\Delta x=n\lambda$

The path difference induced by glass slab is,

$\Delta x_{slab}=(\mu -1)t$

According to question,
$(\mu -1)t=n\lambda$

For minimum thickness of slab, the permissible value of $n$ is 1.

$\Rightarrow t=\frac{\lambda }{\mu -1}=\frac{\lambda }{1.5-1}$

$\therefore t=2\lambda$

Given, $t=k\lambda$ $\Rightarrow k=2$

Correct answer: $2$

Why this question? Caution: $n=0$ can't be taken for minimum thickness as it yields $t=0$.