The width of fringe is $2 \mathrm{~mm}$ on the screen in a double slits experiment for the light of wavelength of $400 \mathrm{~nm}$. The width of the fringe for the light of wavelength 600 $\mathrm{nm}$ will be:

<p>In the double-slit experiment, the fringe width ($\beta$) is given by the formula :</p> <p>$ \beta = \frac{\lambda D}{d} $</p> <p>where:</p> <ul> <li>$\lambda$ is the wavelength of the light,</li> <li>$D$ is the distance between the screen and the double-slit,</li> <li>$d$ is the separation between the two slits.</li> </ul> <p>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 :</p> <p>$ \frac{\beta_1}{\beta_2} = \frac{\lambda_1}{\lambda_2} $</p> <p>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 $600 \, \text{nm}$.</p> <p>Substituting the given values into the proportionality equation, we get :</p> <p>$ \frac{2 \, \text{mm}}{\beta_2} = \frac{400 \, \text{nm}}{600 \, \text{nm}} $</p> <p>Solving this equation for $\beta_2$ gives:</p> <p>$ \beta_2 = 2 \, \text{mm} \times \frac{600 \, \text{nm}}{400 \, \text{nm}} = 3 \, \text{mm} $</p>