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JEE Mains · Physics · STD 12 - Wave optics
  • JEE Mains 2020
  • Medium
  • MCQ

In a double slit experiment, at a certain point on the screen the path difference between the two interfering waves is \(\frac{1}{8} ^{th}\) of a wavelength. The ratio of the intensity of light at that point to that at the centre of a bright fringe is

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Answer & Solution

Correct Answer

(D) \(0.583\)

Step-by-step Solution

Detailed explanation

\(I=I_{0} \cos ^{2}\left(\frac{\Delta \phi}{2}\right)\)

\(\frac{I}{I_{0}}=\cos ^{2}\left(\frac{\Delta \phi}{2}\right)\)

\(\frac{I}{I_{0}}=\cos ^{2}\left(\frac{\frac{2 \pi}{\lambda} \times \frac{\lambda}{8}}{2}\right)\)

\(\frac{\mathrm{I}}{\mathrm{I}_{0}}=\cos ^{2}\left(\frac{\pi}{8}\right)\)

\(\frac{\mathrm{I}}{\mathrm{I}_{0}}=0.853\)

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