A parallel beam of light of wavelength 6000 $\stackrel{0}{A}$ is incident normally on a slit of width 0.2 mm. The diffraction  pattern is obtained on a screen which is placed at the focal plane of a convex lens of focal length 50 cm. If the lens is placed close to the slit, the distance between the minima on both sides of central maximum will be

The angular separation of minima on both side of central maximum of single slit is 2 $\theta$ where $\theta$ is given by
$$\begin{gathered} \mathrm{dsin}\mathrm{\theta }=\mathrm{\lambda }\text{ or }\sin \mathrm{\theta }=\mathrm{\lambda }/\mathrm{d} \\ \sin \mathrm{\theta }=\frac{6000\times {10}^{-10}}{0.2\times {10}^{-3}}=3\times {10}^{-3} \\ \mathrm{or} \mathrm{\theta }\approx 3\times {10}^{-3}\mathrm{rad}(\because \sin \mathrm{\theta }\approx \mathrm{\theta }) \\ \mathrm{when} \mathrm{the} \mathrm{lens} \mathrm{is} \mathrm{placed} \mathrm{close} \mathrm{to} \mathrm{the} \mathrm{slit}, \mathrm{then} \\ \mathrm{x}=\mathrm{ftan}\mathrm{\theta }\approx \mathrm{f\theta } \\ \mathrm{wherc} \mathrm{x} \mathrm{is} \mathrm{the} \mathrm{distarce} \mathrm{of} \mathrm{first} \mathrm{minlmum} \mathrm{from} \mathrm{the} \mathrm{central} \\ \mathrm{maximum}.\mathrm{The} \mathrm{distance} \mathrm{between} \mathrm{two} \mathrm{minima} \mathrm{on} \mathrm{both} \mathrm{sides} \mathrm{of} \\ \mathrm{central} \mathrm{maximum} \mathrm{is} \\ 2\mathrm{x}=2\mathrm{f\theta }=2\times 0\cdot 5\times \left(3\times {10}^{-3}\right)=3\times {10}^{-3}\mathrm{m}=3 \mathrm{mm} \end{gathered}$$