In Young's double slit experiment with a monochromatic light of wavelength $2000A^0$ , the fringe width is found to be$\text{0}\text{.6}\text{mm}$ . When the slits are now illuminated with a light of wavelength $3000A^0$  the fringe width will be

Wavelength ${\lambda }_1=2000A^0$

Wavelength ${\lambda }_2=3000A^0$

Fringe width ${\beta }_1=0.6mm$

Fringe width ${\beta }_2=?$

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

                     $\frac{{\beta }_1}{{\beta }_2}=\frac{{\lambda }_1}{{\lambda }_2}$

                     ${\beta }_2=\frac{{\beta }_1{\lambda }_2}{{\lambda }_1}$

                         $=\frac{0.6\times 3000}{2000}$

               $\therefore {\beta }_2=0.9mm$