• question_answer In a single slit diffraction experiment, a slit of width d is illuminated by red light of wavelength 650 nm. For what value of d will (i) the first minimum fall is at an angle of diffraction of $60{}^\circ$ and (ii) the first maximum fall is at an angle of diffraction of $60{}^\circ ?$

(i) $7.5\times {{10}^{-7}}m$    (ii) $1.13\times {{10}^{-6}}m$ Give, wavelength of red light, $\lambda =650\,\text{nmz}\,\,\text{650}\times {{10}^{-9}}m$             (i) for first minimum of the differection pattern,                         $d\,\,\sin \theta =\lambda ;\,\theta =60{}^\circ$ (given)             $\therefore \,\,d=\frac{\lambda }{\sin }\theta =\frac{650\times {{10}^{-9}}}{\sin 60{}^\circ }$ (Putting volues)             $=\frac{650\times 2\times {{10}^{-9}}}{\sqrt{3}}$                   $\left( \because \sin 60{}^\circ =\frac{\sqrt{3}}{3} \right)$             $=750.55\times {{10}^{-9}}m=7.5\times {{10}^{-7}}m$             (ii) for first maximum of the differention pattern                         $d\,\,\sin \theta =\frac{3\lambda }{2}$    $\therefore \,\,d=\frac{3\lambda }{2\sin \theta }$                         $\theta =60{}^\circ$               (given)                         $d=\frac{3\times 650\times {{10}^{-9}}}{2\times \sin 60{}^\circ }=\frac{3\times 650\times {{10}^{-9}}}{2\times \frac{\sqrt{3}}{2}}$ $=\sqrt{3}\times 650\times {{10}^{-9}}=1125.8\times {{10}^{-9}}=1.115\times {{10}^{-6}}m$