Optics Question 137
Find the condition on q for constructive interference at P between the ray BP and reflected ray AP
[IIT-JEE (Screening) 2003]
Options:
A) cosq = 3l/2d
B) cosq = l/4d
C) secq - cosq = l/d
D) secq - cosq = 4l/d
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Answer:
Correct Answer: B
Solution:
$ \because $ PR = r
Therefore PO = d secθ and CO = PO cos 2θ $ =d\sec \theta \cos 2\theta $ is Path difference between the two rays D = CO + PO = (d secθ + d secθ cos 2θ)
Phase difference between the two rays is Δφ = 2π(p/λ) (One is reflected, while another is direct)
Therefore condition for constructive interference should be $ \Delta =\lambda ,2\lambda ,3\lambda …… $ or $ d\sec \theta (1+\cos 2\theta )=\lambda $ or $ \frac{d}{\cos \theta }(2{{\cos }^{2}}\theta )=\lambda $
Therefore $ \cos \theta =\frac{\lambda }{2d} $
BETA
