Question

If the frequency of the source is doubled in Young’s double slit experiment, then fringe width $(\beta )$ will be:
A. unchanged
B. ${\beta \mathord{\left/ {\vphantom {\beta 2}} \right.} 2}$
C. $2\beta$
D. $3\beta$

Answer 1

Hint
The given question is based on the concept of the relation between the fringe width and the frequency. To find the fringe width in the Young’s double slit experiment, we use the formula, $\beta = \dfrac{{\lambda D}}{d}$ where we can substitute the given values to find the fringe width.
$\Rightarrow \beta = \dfrac{{\lambda D}}{d}$
where $\beta$ is the fringe width
$\lambda$ is the wavelength of the light
$D$ is the distance between the slits and the screen and
$d$ is the distance between the slits.

Complete step by step answer
The formula that relates the fringe width with the wavelength of the source is given as,
$\Rightarrow \beta = \dfrac{{\lambda D}}{d}$
So, we can notice that keeping the distance between the screen and the slits, that is, D and the distance between the slits, that is, d constant, the fringe width becomes directly proportional to the wavelength of the light.
Therefore, we get,
$\Rightarrow \beta \propto \lambda$…… (1)
The formula that relates the frequency with the wavelength of the source is given by,
$\Rightarrow f = \dfrac{c}{\lambda }$
Where $\lambda$ is the wavelength of the light, $c$ is the speed of the light in air and $f$ is the frequency of the light/source.
So, here, the speed of the light in air is constant, so, the frequency becomes inversely proportional to the wavelength of the light.
Hence, we get,
$\Rightarrow f \propto \dfrac{1}{\lambda }$
Hence we can write,
$\Rightarrow \lambda \propto \dfrac{1}{f}$…… (2)
Let us compare the equations (1) and (2), thus, we get,
$\Rightarrow \beta \propto \dfrac{1}{f}$
Finally, the relation between the fringe width and the frequency of the source is obtained. So, we can say that the fringe width and the frequency of the source are inversely proportional to each other.
Now, we need to find the change in the fringe width when the frequency doubles.
$\Rightarrow f' = 2f$
So in place of $f'$ we can write $\dfrac{1}{{\beta '}}$
Hence we get,
$\Rightarrow \dfrac{1}{{\beta '}} = 2\dfrac{1}{\beta }$
On taking inverse, we get
$\Rightarrow \beta ' = \dfrac{\beta }{2}$
$\therefore$ The fringe width gets halved, when the frequency is doubled.
Thus, the option (B) is correct.

Note
The fringe width in the Young’s double slit experiment is the distance between 2 adjacent light or dark bands. When monochromatic light is used, the fringe width is constant. But if white light is used, then it breaks up into the components and we get fringes of the red colour more than that of the blue.