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 \[60{}^\circ ?\]

Answer:

(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)                         \[d=\frac{3\times 650\times {{10}^{-9}}}{2\times \sin 60{}^\circ }=\frac{3\times 650\times {{10}^{-9}}}{2\times \frac{\sqrt{3}}{2}}\] \[=\sqrt{3}\times 650\times {{10}^{-9}}=1125.8\times {{10}^{-9}}=1.115\times {{10}^{-6}}m\]