Question

In a Young’s double slit set-up, light from a laser source falls on a pair of very narrow slits separated by $1.0$ micro-meter and bright fringes separated by $1.0$ millimeter are observed on a distant screen. If the frequency of the laser light is doubled, what will be the separation of the bright fringes?
A) $0.25mm$
B) $0.5mm$
C) $1.0mm$
D) $2.0mm$

Answer 1

Hint: Young’s double-slit experiment demonstrates the wave nature of light. Light from two coherent sources is used in the experiment. In this experiment, we will get alternate bright and dark fringes. By using the equation for fringe width in Young’s double-slit experiment, we get the relation between the fringe width and frequency.

Formula used:
$\beta = \dfrac{{\lambda D}}{d}$ (Where $\beta $ stands for the fringe width,$\lambda $the wavelength of the light, $D$ is the separation between the slits and the screen, and $d$ is the separation between the slits)

Complete step by step solution:
The set-up of Young’s double-slit experiment consists of a coherent light source, two slits, and a screen. The slits are separated by a distance and the screen is also placed at a particular distance.

The fringe width by Young’s double-slit method is given by
$\beta = \dfrac{{\lambda D}}{d}$………………………………….(1)
We know that $\lambda = \dfrac{c}{\nu }$ (c is the velocity of light and $\nu $stands for the frequency of the incident light)
Putting the value of $\lambda $in equation (1)
$\beta = \dfrac{{cD}}{{\nu d}}$
$ \Rightarrow \beta \propto \dfrac{1}{\nu }$
From the question we know that,
$\beta = 1mm$ And ${\nu'} = 2\nu $
Let the fringe width that we need to find be ${\beta'}$, corresponding to the new frequency $2\nu $
$\therefore \dfrac{{{\beta'}}}{\beta } = \dfrac{{\dfrac{1}{{2\nu }}}}{{\dfrac{1}{\nu }}} = \dfrac{\nu }{{2\nu }}$
Given that $\beta = 1mm$
Putting the value of $\beta $ as $1$, we get
$\dfrac{{{\beta'}}}{1} = \dfrac{1}{2} $
 $ {\beta'} = 0.5mm $

The answer is Option (B) $0.5mm$.

Note: We can see that as the frequency of the light increases the fringe width decreases. The fringes are formed due to constructive and destructive interference. When the crusts of two waves superimpose and their amplitude adds up to form constructive interference this will lead to the formation of bright fringes. Dark fringes are formed due to the destructive interference where the crust of one wave superimposes with the trough of another wave and they cancel each other leaving the amplitude as zero.