(a) Explanation of diffraction pattern using Huygen's construction (b) Showing the angular width of first diffraction fringe as half of the central fringe (c) Explanation of decrease in intensity with increasing $n$ (a) A plane wavefront from a monochromatic source $S$ is incident on the slit LN. According to Huygens' principle each point of the incident wavefront becomes the source of secondarywavelet. These wavelets emerge with the same phase. These wavelets reach point $C$ with same phase. Due to constructive interference central maximum is formed at $C$. The wavelets which meet at point $P$ (other than point C) have different phasessince they traverse different paths to reach $P$ and accordingly they produce either maxima or minima. Thus a diffraction pattern is generated. (b) Condition for first minimum on the screen $\;a \sin \theta = \lambda$ $\Rightarrow \; \theta = \frac{\lambda}{a}$ $\therefore$ angular width of the central fringe on the screen (from figure) $= 2\theta = \frac{2\lambda}{a}$ Angular width of first diffraction fringe (From fig) $= \frac{\lambda}{a}$ Hence angular width of central fringe is twice the angular width of first fringe. (c) Maxima become weaker and weaker with increasing $n$. This is because the effective part of the wavefront, contributing to the maxima, becomes smaller and smaller, with increasing $n$. OR (a) For small angles $\angle NOM = \tan \angle NOM = \frac{MN}{OM}$ $\angle NCM = \tan \angle NCM = \frac{MN}{MC}$ $\angle NIM = \tan \angle NIM = \frac{MN}{MI}$ In $\triangle NOC$ , $\angle i = \angle NOM + \angle NCM$ $\therefore \; \angle i = \frac{MN}{OM} + \frac{MN}{MC}$ .....(i) Similarly, $\angle r = \angle NCM - \angle NIM$ ......(ii) $= \frac{MN}{MC} - \frac{MN}{MI}$ Using Snell's Law $n_1 \sin i = n_2 \sin r$ For small angles $n_1 i = n_2 r$ Substituting for $i$ and $r$ , we get $\frac{n_1}{OM} + \frac{n_2}{MI} = \frac{n_2 - n_1}{MC}$ Here, $OM = -u, MI = +v, MC = +R$ Substituting these, we get $\Rightarrow \; \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$ Similarly relation for the surface ADC. $\frac{-n_2}{DI_1} + \frac{n_1}{DI} = \frac{n_2 - n_1}{DC_2}$ ......(i) Refraction at the first surface $A B C$ of the lens. $\frac{n_1}{OB} + \frac{n_2}{BI_1} = \frac{n_2 - n_1}{BC_1}$ .....(ii) Adding (i) and (ii) and taking $B I_1 \simeq D I_1$, we get $\frac{n_1}{OB} + \frac{n_1}{DI} = (n_2 - n_1) \left( \frac{1}{BC_1} + \frac{1}{DC_2} \right)$ Here, $OB = -u$ $DI = +v$ $BC_1 = +R_1$ $\Rightarrow DC_2 = -R_2$ $\Rightarrow \; \frac{n_1}{-u} + \frac{n_1}{v} = -R_2$ $\Rightarrow \; n_1 \left( \frac{1}{v} + \frac{1}{u} \right) = (n_2 - n_1)$ $\left( \frac{1}{R_1} + \frac{1}{R_2} \right)$ $\Rightarrow \; \frac{1}{R_1} - \frac{1}{R_2}) = \left( \frac{n_2}{n_1} - 1 \right)$ $\left( \frac{1}{R_1} - \frac{1}{R_2} \right)$