(a) (i) A beam of light of wavelength 600 nm from a distance source falls on a single slit 1 mm wide and a resulting diffraction pattern is observed on a screen 2m away.
What is the distance between the first dark fringes on either side of central bright fringe?
(ii) The width of a slit is 0.012 mm. Monochromatic light is incident on it. The angular position of first bright line is 5.2º. The wavelength of incident light is [sin 5.2º = 0.0906]. (b) In Young’s double slit experiment intensity at a point is (1/4) of the maximum intensity. What is the angular position of this point?
OR
(i) What is the effect on the interference fringes to a Young’s double slit experiment, when (a) the width of the source slit is increased and (b) the monochromatic source is replaced by a source of white light? Justify your answer in each case. (ii) The intensity at the central maxima in Young’s double slit experiment set-up is $I_o$
Show that the intensity at a point, where the path difference is $\lambda /3$ is $I_o/ 4$.

(a) (i)

$$\begin{gathered} \mathrm{The} \mathrm{distance} \mathrm{between} \mathrm{the} \mathrm{first} \mathrm{dark} \mathrm{fringe} \mathrm{on} \mathrm{either} \\ \mathrm{side} \mathrm{of} \mathrm{the} \mathrm{central} \mathrm{bright} \mathrm{fringe} \\ =\mathrm{width} \mathrm{of} \mathrm{central} \mathrm{maximum} \\ =\frac{2\mathrm{D\lambda }}{\mathrm{a}}=\frac{2\times 2\times 600\times {10}^{-9}}{{10}^{-3}} \\ =2.4\times {10}^{-3}\mathrm{m}=2.4 \mathrm{mm} \end{gathered}$$

(a)(ii)

$$\begin{gathered} \mathrm{It} \mathrm{is} \mathrm{a} \mathrm{one} \mathrm{of} \mathrm{Fraunhoffer} \mathrm{diffraction} \mathrm{from} \mathrm{single} \mathrm{slit}. \mathrm{so} \\ \mathrm{for} \mathrm{bright} \mathrm{fringe} \mathrm{where} \mathrm{a} \mathrm{is} \mathrm{the} \mathrm{width} \mathrm{of} \mathrm{slit}. \\ \mathrm{a} \sin \mathrm{\theta }=(2\mathrm{n}+1)\frac{\mathrm{\lambda }}{2} \\ \mathrm{\lambda }=\frac{2\mathrm{a} \sin \mathrm{\lambda }}{2\mathrm{n}+1}=\frac{2\times 1.2\times {10}^{-5}\times 0.0906}{2\times 1+1} \\ =724\times {10}^{-10} \mathrm{m}=7248\stackrel{°}{\mathrm{A}.} \end{gathered}$$

(b)

$$\begin{gathered} \mathrm{Let} \mathrm{P} \mathrm{be} \mathrm{the} \mathrm{point} \mathrm{on} \mathrm{the} \mathrm{central} \mathrm{maxima} \mathrm{whose} \\ \mathrm{intensity} \mathrm{is} \mathrm{one} \mathrm{fourth} \mathrm{of} \mathrm{the} \mathrm{maximum} \mathrm{intensity}. \\ \mathrm{For} \mathrm{interference} \mathrm{we} \mathrm{know} \mathrm{that} \\ \mathrm{I}={\mathrm{I}}_1+{\mathrm{I}}_2+2\sqrt{{\mathrm{I}}_1{\mathrm{I}}_2}\mathrm{cos\varphi } \\ \mathrm{Where} \mathrm{I} \mathrm{is} \mathrm{the} \mathrm{intensity} \mathrm{at} \mathrm{P} \mathrm{and} {\mathrm{I}}_1, {\mathrm{I}}_2 \mathrm{are} \mathrm{the} \mathrm{intensity} \\ \mathrm{of} \mathrm{light} \mathrm{originating} \mathrm{from} \mathrm{A} \mathrm{and} \mathrm{B} \mathrm{respectively} \mathrm{and} \mathrm{\varphi } \mathrm{is} \\ \mathrm{the} \mathrm{phase} \mathrm{difference} \mathrm{at} \mathrm{P}. \end{gathered}$$

$$\begin{gathered} \mathrm{In} \mathrm{YDSE}, {\mathrm{I}}_1={\mathrm{I}}_2=\mathrm{I} \mathrm{and} {\mathrm{I}}_{\max }=4\mathrm{I} \\ \mathrm{We} \mathrm{are} \mathrm{concentrating} \mathrm{at} \mathrm{a} \mathrm{point} \mathrm{where} \mathrm{the} \mathrm{intensity} \\ \mathrm{is} \mathrm{one} \mathrm{fourth} \mathrm{of} \mathrm{the} \max \mathrm{intensity}. \\ \therefore \mathrm{I}=\mathrm{I}+\mathrm{I}+2\mathrm{I} \cos \mathrm{\varphi } \\ \Rightarrow -\frac{1}{2} \cos \mathrm{\varphi } \Rightarrow \mathrm{\varphi }=\frac{2\mathrm{\pi }}{3} \\ [\mathrm{P}1 \mathrm{note} \mathrm{that} \mathrm{we} \mathrm{take} \mathrm{the} \mathrm{least} \mathrm{value} \mathrm{of} \mathrm{the} \mathrm{angle} \mathrm{as} \mathrm{the} \\ \mathrm{point} \mathrm{is} \mathrm{in} \mathrm{central} \mathrm{maxima}] \\ \mathrm{For} \mathrm{a} \mathrm{phase} \mathrm{difference} \mathrm{of} 2\mathrm{\pi }, \mathrm{the} \mathrm{path} \mathrm{difference} \mathrm{is} \mathrm{\lambda }. \\ \mathrm{For} \mathrm{phase} \mathrm{difference} \mathrm{of} \frac{2\mathrm{\pi }}{3}, \mathrm{the} \mathrm{path} \mathrm{difference} \mathrm{is} \\ \frac{\mathrm{\lambda }}{2\mathrm{\pi }}\times \frac{2\mathrm{\pi }}{3}=\frac{\mathrm{\lambda }}{3} \\ \mathrm{But} \mathrm{the} \mathrm{path} \mathrm{difference} (\mathrm{in} \mathrm{terms} \mathrm{of} \mathrm{P} \mathrm{and} \mathrm{Q}) \mathrm{is} \mathrm{d} \sin \mathrm{\theta } \\ \mathrm{as} \mathrm{shown} \mathrm{in} \mathrm{fig}. \\ \therefore \mathrm{d} \sin \mathrm{\theta }=\frac{\mathrm{\lambda }}{3} \Rightarrow \sin \mathrm{\theta }=\frac{\mathrm{\lambda }}{3\mathrm{d}} \\ \Rightarrow \mathrm{\theta }={\sin }^{-1}\left(\frac{\mathrm{\lambda }}{3\mathrm{d}}\right) \end{gathered}$$

i) (a) For interference fringes to be seen, s / S $\le \mathrm{\lambda }$/ d condition should be satisfied, 

where s = size of the source and d = distance of the source from the plane of two slits.
As, the source slit width increases, fringe pattern gets less and less sharp. When the source slit is too wide, the above condition does not get satisfied and the interference pattern disappears.
(b) The interference pattern due to the different colour components of white light overlap. The central bright fringes for different colours are at the same position. Therefore, central fringe is white. And on the either side of the central fringe (i.e. central maxima),coloured bands will appear. The fringe close to either side of central white fringe is red and the farthest will be blue.

(ii) Given, $\mathrm{OP}={\mathrm{y}}_{\mathrm{n}}$

$$\begin{gathered} \mathrm{The} \mathrm{distance} \mathrm{OP} \mathrm{equals} \mathrm{one}-\mathrm{third} \mathrm{of} \mathrm{fringe} \mathrm{width} \mathrm{of} \mathrm{the} \\ \mathrm{pattern}. \\ \mathrm{i}.\mathrm{e}. {\mathrm{y}}_{\mathrm{n}}=\frac{\mathrm{\beta }}{3}=\frac{1}{3}\left(\frac{\mathrm{D\lambda }}{\mathrm{d}}\right)=\frac{\mathrm{D\lambda }}{3\mathrm{d}} \\ \Rightarrow \frac{{\mathrm{dy}}_{\mathrm{n}}}{\mathrm{D}}=\frac{\mathrm{\lambda }}{3} \\ \mathrm{Path} \mathrm{difference} ={\mathrm{S}}_2\mathrm{P}-{\mathrm{S}}_1\mathrm{P}=\frac{{\mathrm{dy}}_{\mathrm{n}}}{\mathrm{D}}=\frac{\mathrm{\lambda }}{3} \\ \therefore \mathrm{Phase} \mathrm{difference}, \mathrm{\varphi }=\frac{2\mathrm{\pi }}{\mathrm{\lambda }}\times \mathrm{path} \mathrm{difference} \\ =\frac{2\mathrm{\pi }}{\mathrm{\lambda }}\times \frac{\mathrm{\lambda }}{3}=\frac{2\mathrm{\pi }}{3} \\ \mathrm{If} \mathrm{intensity} \mathrm{at} \mathrm{central} \mathrm{fringe} \mathrm{is} {\mathrm{I}}_0, \mathrm{then} \mathrm{intensity} \mathrm{at} \mathrm{a} \mathrm{point} \\ \mathrm{P}, \mathrm{where} \mathrm{phase} \mathrm{difference} \mathrm{\varphi }, \mathrm{is} \mathrm{given} \mathrm{by} \\ \mathrm{I}={\mathrm{I}}_0 {\cos }^2 \mathrm{\varphi } \\ \Rightarrow \mathrm{I}={\mathrm{I}}_0{\left(\cos \frac{2\mathrm{\pi }}{3}\right)}^2 \\ ={\mathrm{I}}_0{\left(-\cos \frac{\mathrm{\pi }}{3}\right)}^2 \\ ={\mathrm{I}}_0{\left(-\frac{1}{2}\right)}^2=\frac{{\mathrm{I}}_0}{4} \\ \mathrm{Hence}, \mathrm{the} \mathrm{intensity} \mathrm{at} \mathrm{point} \mathrm{P} \mathrm{would} \mathrm{be} \frac{{\mathrm{I}}_0}{4}. \end{gathered}$$