In Young's double slit experiment, the distance d between the slits S1 and S2 is 1.0 mm. What should the width of each slit be so as to obtain 10 maxima of the two slit interference pattern within the central maximum of the single slit diffraction pattern? 

(b) Let the width of each slit be a. The linear separation between m bright fringes in the double slit experiment is 

${\mathrm{y}}_{\mathrm{m}}=\frac{\mathrm{n\lambda D}}{\mathrm{d}}$

Since $\mathrm{y}<<\mathrm{D},$ the angular separation between m bright fringes will be

${\mathrm{\theta }}_{\mathrm{m}}=\frac{{\mathrm{y}}_{\mathrm{n}}}{\mathrm{D}}=\frac{\mathrm{n\lambda }}{\mathrm{d}}$

For 10 bright fringes we have

${\mathrm{\theta }}_{10}=\frac{10\mathrm{\lambda }}{\mathrm{d}} \dots \dots \left(\mathrm{i}\right)$

Now the angular width of the principal maximum in the diffraction pattern due to a slit of width a is $2{\mathrm{\theta }}_1=\frac{2\mathrm{\lambda }}{\mathrm{a}} \dots \dots \left(\mathrm{ii}\right)$

                                  $2{\mathrm{\theta }}_1=\frac{2\mathrm{\lambda }}{\mathrm{a}}$

Equating (i) and (ii), we get

$$\begin{gathered} \frac{10\mathrm{\lambda }}{\mathrm{d}}=\frac{2\mathrm{\lambda }}{\mathrm{a}} \\ \mathrm{or} \mathrm{a}=\frac{\mathrm{d}}{5}=\frac{1.0\mathrm{mm}}{5}=0.2 \mathrm{mm} \end{gathered}$$