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) \[d=\frac{3\times 650\times {{10}^{-9}}}{2\times \sin 60{}^\circ }=\frac{3\times 650\times {{10}^{-9}}}{2\times \frac{\sqrt{3}}{2}}\] \[=\sqrt{3}\times 650\times {{10}^{-9}}=1125.8\times {{10}^{-9}}=1.115\times {{10}^{-6}}m\]
You need to login to perform this action.
You will be redirected in
3 sec