Wave Optics

JEE Main level, single correct, medium

96 of 191

Question 96 of 191

In a double – slit experiment, at a certain point on the screen the path difference between the two interfering waves is 18th\dfrac{1}{8}th of a wavelength. The ratio of the intensity of light at that point to that at the center of a bright fringe is:

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Answer: (A) 0.8530.853

I=I0cos2(∆ϕ2)I={I}_{0}{cos}^{2}\left(\dfrac{∆\phi }{2}\right)
II0=cos2[2πλ×∆X2]=cos2(π8);II0=0.853\dfrac{I}{{I}_{0}}={cos}^{2}\left[\dfrac{\dfrac{2\pi }{\lambda }\times ∆X}{2}\right]={cos}^{2}\left(\dfrac{\pi }{8}\right);\dfrac{I}{{I}_{0}}=0.853

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