i) A parallel beam of light of 600 nm falls on a narrow slit and the resulting diffraction pattern is observed on a screen 1.2 m away. It is observed that the first minimum is at a distance of 3 mm from the centre of the screen. Calculate the width of the slit. 

ii)For a single slit of width “a”, the first minimum of the interference pattern of monochromatic light of wavelength $\mathrm{\lambda }$ occurs at an angle of $\frac{\mathrm{\lambda }}{\mathrm{a}}$ .At the same angle of $\frac{\mathrm{\lambda }}{\mathrm{a}}$ , we get a maximum for two narrow slits separated by a distance “a”. Explain.

i) 

$$\begin{gathered} \mathrm{\lambda }=600 \mathrm{nm}=600\times {10}^{-9}\mathrm{m}, \mathrm{D}=1.2 \mathrm{m} \\ \mathrm{First} \mathrm{minima} \mathrm{at} {\mathrm{x}}_1=3 \mathrm{mm}=3\times {10}^{-3} \mathrm{m} \\ \mathrm{Diffraction} \mathrm{angle} \mathrm{for} \mathrm{first} \mathrm{minima} : \\ {\mathrm{\theta }}_1=\frac{{\mathrm{x}}_1}{\mathrm{D}} \\ {\mathrm{\theta }}_1=\frac{3\times {10}^{-3}\times 10}{12}=2.5\times {10}^{-3} \mathrm{rad} \\ \mathrm{We} \mathrm{know}, \mathrm{a} \sin {\mathrm{\theta }}_1=\mathrm{n\lambda }, \mathrm{n}=1 \\ \mathrm{a} \sin {\mathrm{\theta }}_1=\mathrm{n\lambda } \\ \mathrm{Since} \mathrm{angle} \mathrm{is} \mathrm{very} \mathrm{small} \mathrm{so} \sin {\mathrm{\theta }}_1~\mathrm{\theta } \\ \mathrm{a}=\frac{\mathrm{\lambda }}{{\mathrm{\theta }}_1}=\frac{600\times {10}^{-9}}{2.5\times {10}^{-3}} \\ \mathrm{a}=\frac{6}{2.5}\times {10}^{-4}\mathrm{m}=2.4\times {10}^{-4}\mathrm{m}=0.24 \mathrm{mm} \\ \therefore \boxed{\mathrm{a}=0.24 \mathrm{mm}} \end{gathered}$$ii) $\mathrm{For} \mathrm{a} \mathrm{single} \mathrm{slit} \mathrm{of} \mathrm{width} \text{'}\mathrm{a}\text{'},$

$\mathrm{the} {\mathrm{n}}^{\mathrm{th}} \mathrm{minimum}, {\mathrm{sin\theta }}_{\mathrm{n}}=\frac{\mathrm{n\gamma }}{\mathrm{a}}$

$$\begin{gathered} \mathrm{or} {\mathrm{\theta }}_{\mathrm{n}}=\frac{\mathrm{n\lambda }}{\mathrm{a}}(\mathrm{when} \mathrm{\theta } \mathrm{is} \mathrm{small}) \\ \mathrm{\theta }=\frac{\mathrm{\lambda }}{\mathrm{a}}(\mathrm{when} \mathrm{n}=1 \mathrm{for} \mathrm{the} \mathrm{first} \mathrm{minimum}) \\ \mathrm{Now}, \mathrm{the} \mathrm{maximum} \mathrm{of} \mathrm{two} \mathrm{narrow} \mathrm{slits} \mathrm{separated} \\ \mathrm{by} \mathrm{a} \mathrm{distance} \text{'}\mathrm{a}\text{'} \\ \mathrm{Path} \mathrm{difference}, \left(\mathrm{x}\right)=\frac{\mathrm{\lambda D}}{\mathrm{d}} \\ \mathrm{angle} \left(\mathrm{\theta }\right)=\frac{\mathrm{x}}{\mathrm{D}}=\frac{\mathrm{\lambda }}{\mathrm{d}}\mathrm{or} \frac{\mathrm{\lambda }}{\mathrm{a}} (\therefore \mathrm{d}=\mathrm{n}) \\ \mathrm{This} \mathrm{is} \mathrm{why}, \mathrm{at} \mathrm{the} \mathrm{same} \mathrm{angle} \frac{\mathrm{\lambda }}{\mathrm{a}}, \mathrm{we} \mathrm{get} \mathrm{a} \\ \mathrm{maximum} \mathrm{for} \mathrm{two} \mathrm{narrow} \mathrm{slits} \mathrm{separated} \mathrm{by} \mathrm{a} \\ \mathrm{distance} \text{'}\mathrm{a}\text{'}. \end{gathered}$$