JEE Main 2023 — Wave Optics Question with Solution

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} $