QJEEM23Ph7945: JEE Main 2023 Physics - Class 12 JEEM23Ph7945 Solution: The width of fringe is 2   mm on the screen in a double…

In the double-slit experiment, the fringe width ( $\beta$ ) is given by the formula :

$\beta = \frac{\lambda D}{d}$

where:

• $\lambda$ is the wavelength of the light,
• $D$ is the distance between the screen and the double-slit,
• $d$ is the separation between the two slits.

If we are considering a change in wavelength but the fringe width is changing and the setup of the experiment (the values of $D$ and $d$ ) stays the same, we can see that the fringe width is directly proportional to the wavelength. This is because $D$ and $d$ are constants in this case, so we can write :

$\frac{\beta_1}{\beta_2} = \frac{\lambda_1}{\lambda_2}$

Here, the given wavelengths are $\lambda_1 = 400 \, \text{nm}$ and $\lambda_2 = 600 \, \text{nm}$ , and the given fringe width for the light of wavelength $400 \, \text{nm}$ is $\beta_1 = 2 \, \text{mm}$ . We are asked to find the fringe width $\beta_2$ for the light of wavelength $600 \, \text{nm}$ .

Substituting the given values into the proportionality equation, we get :

$\frac{2 \, \text{mm}}{\beta_2} = \frac{400 \, \text{nm}}{600 \, \text{nm}}$

Solving this equation for $\beta_2$ gives:

$\beta_2 = 2 \, \text{mm} \times \frac{600 \, \text{nm}}{400 \, \text{nm}} = 3 \, \text{mm}$