A) cosq = 3l/2d
B) cosq = l/4d
C) secq ? cosq = l/d
D) secq ? cosq = 4l/d
Correct Answer: B
Solution :
\[\because \]PR = d Þ PO = d secq and CO = PO cos 2q \[=d\sec \theta \cos 2\theta \] is Path difference between the two rays D = CO + PO = (d secq + d secq cos 2q) Phase difference between the two rays is f = p (One is reflected, while another is direct) Therefore condition for constructive interference should be \[\Delta =\frac{\lambda }{2},\frac{3\lambda }{2}......\] or \[d\sec \theta (1+\cos 2\theta )=\frac{\lambda }{2}\] or \[\frac{d}{\cos \theta }(2{{\cos }^{2}}\theta )=\frac{\lambda }{2}\] Þ \[\cos \theta =\frac{\lambda }{4d}\]You need to login to perform this action.
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