Question

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\textsuperscript{th} bright fringe. The thickness of the glass plate is ___________ $\mu \text{m}$.

Answer 1

The optical path difference introduced by the glass plate is:
$\Delta x = t(\mu - 1),$
where $t$ is the thickness of the plate and $\mu$ is its refractive index.
The fringe shift is given by:
$\Delta x = n\lambda,$
where $n = 4$ (the shift corresponds to the 4th fringe) and $\lambda = 500 \, \mathrm{nm}$.
Equating: $t(\mu - 1) = n\lambda.$
Substituting $\mu = 1.5$, $n = 4$, and $\lambda = 500 \, \mathrm{nm}$:
$t(1.5 - 1) = 4 \cdot 500.$

Simplify: $t = \frac{2000}{0.5} = 4000 \, \mathrm{nm} = 4 \, \mu \mathrm{m}.$

Answer 2

The optical path difference introduced by the glass plate is:
$\Delta x = t(\mu - 1),$
where $t$ is the thickness of the plate and $\mu$ is its refractive index.
The fringe shift is given by:
$\Delta x = n\lambda,$
where $n = 4$ (the shift corresponds to the 4th fringe) and $\lambda = 500 \, \mathrm{nm}$.
Equating: $t(\mu - 1) = n\lambda.$
Substituting $\mu = 1.5$, $n = 4$, and $\lambda = 500 \, \mathrm{nm}$:
$t(1.5 - 1) = 4 \cdot 500.$

Simplify: $t = \frac{2000}{0.5} = 4000 \, \mathrm{nm} = 4 \, \mu \mathrm{m}.$