A) lf d is \[\frac{3\lambda }{2}\] at O, minima will be observed.
B) If d is \[\frac{11\lambda }{6},\] then intensity at O will be \[\frac{3}{4}\] of maximum intensity.
C) If d is \[3\lambda ,\] \[O\] will be a maxima.
D) If d is \[\frac{7\lambda }{6},\] the intensity at O will be equal to maximum intensity.
Correct Answer: D
Solution :
[d] \[I={{I}_{\max }}{{\cos }^{2}}\left( \frac{\phi }{2} \right)={{I}_{\max }}{{\cos }^{2}}\left( \frac{\pi }{\lambda }d \right)\] In option (1), \[d=\frac{3\lambda }{2}I={{I}_{\max }}{{\cos }^{2}}\left( \frac{\pi }{\lambda }\times \frac{3\lambda }{2} \right)=0,\] at O minima will be observed In option (2). \[d=\frac{11\lambda }{6}\] \[I={{I}_{\max }}{{\cos }^{2}}\left( \frac{\pi }{\lambda }\times \frac{11\lambda }{6} \right)=\frac{3}{4}{{I}_{\max }},\] Intensity at O will be \[\frac{3}{4}\] of maximum intensity In option (3). \[d=3\lambda :I={{I}_{\max }}{{\cos }^{2}}\left( \frac{\pi }{\lambda }\times 3\lambda \right)={{I}_{\max }},\] at O maximum will be observed In option (4). \[d=\frac{7\lambda }{6}:\] \[I={{I}_{\max }}{{\cos }^{2}}\left( \frac{\pi }{\lambda }\times \frac{7\lambda }{6} \right)=\frac{3}{4}{{I}_{\max }},\] Intensity at O will be \[\frac{3}{4}\] of maximum intensity. Hence option (4) is incorrect.
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