Question

In Young’s double slit experiment, the fringe width is $\beta $. If the entire arrangement is placed in the refractive index $n$, the fringe width becomes,
A) $n\beta$
B) $\dfrac{\beta }{{n + 1}}$
C) $\dfrac{\beta }{{n - 1}} $
D) $\dfrac{\beta }{n}$

Answer 1

Hint: Whenever the light source is placed in a different medium of relative refractive index of n, there is a change in wavelength, given by:
$\lambda ^{'} = \dfrac{\lambda }{n}$

Complete step by step solution:
In Young’s experiment, a coherent light source, such as a laser beam, illuminates a plate pierced by two parallel slits, and the light passing through the slits is observed on a screen behind the plate. The wave nature of light causes the light waves passing through the two slits to interfere, producing bright and dark bands on the screen. The width of each fringe produced has a definite value across the entire band and it is given the below formula.
The formula for fringe width in a double’s slit experiment is given by:
$\beta = \dfrac{{\lambda D}}{d} $
Where,
$\lambda $= wavelength of light
$D$ is the distance of the light source from the screen
$d$ is the distance between the slits.
Rearranging the equation, we can obtain
$\lambda = \dfrac{{\beta d}}{D} \to (1)$
When the whole experiment is conducted in water, there is a shift in the wavelength because whenever the light source is placed in a different medium of the relative refractive index of $n$, there is a change in wavelength, given by:
${\lambda ^{'}} = \dfrac{\lambda }{n}$
Substitute the value of $\lambda $ in equation (1). We get:
${\lambda ^{'} } = \dfrac{\lambda }{n} = \dfrac{{\beta d}}{{nD}}$
Since there is a change in the wavelength, there will be a shift in the fringe width, too.
The new fringe width, ${\beta ^{'} } = \dfrac{{{\lambda ^{'} }D}}{d}$
Substituting the equation for new wavelength, we have:
$ {\beta ^{'} } = \dfrac{{{\lambda ^{'} }D}}{d} $
$ {\beta ^{'} } = \dfrac{{\beta \not d}}{{n\not D}} \times \dfrac{{\not D}}{{\not d}} $
$ \Rightarrow {\beta ^{'} } = \dfrac{\beta }{n} $

$\therefore$ The fringe width is $\dfrac{\beta }{n}$. Hence, the correct option is Option (D).

Note:
The wavelength of light in the denser medium will always be lesser than that of air since refractive index of any denser medium like water, glass etc. is greater than one:
${\lambda ^{'} } = \dfrac{\lambda }{n}$
Since $n > 1\therefore {\lambda ^{'}} < \lambda $
So, if you are doing actual calculations with numbers, you should make sure that the wavelength in the denser medium should always be lesser, to ensure that you are moving on the right path.