When monochromatic light of wavelength 600   nm is used in Young's double slit experiment, the…

In Young's double slit experiment, fringe width is given by $\beta =\frac{\lambda D}{d}$. Therefore, fringe width is proportional to the wavelength of the light used.

Now fringe width in the first case is ${\beta }_1=\frac{6}{5}=1.2 \mathrm{mm}$.

Therefore,

$$\begin{gathered} \frac{{\beta }_2}{{\beta }_1}=\frac{{\lambda }_2}{{\lambda }_1} \\ \Rightarrow {\beta }_2=\frac{400}{600}\times 1.2=0.8 \mathrm{mm} \end{gathered}$$

Therefore, the location of the third bright fringe will be $3\times 0.8=2.4 \mathrm{mm}$.