JEE PHYSICS MCQ: WAVE OPTICS

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The wavelength of the light used in Young’s double slit experiment is

\lambda

. The intensity at a point on the screen is

I

, where the path difference is

\frac{\lambda }{6}

. If

{{I_0}}

denotes the maximum intensity, then the ratio of

{I\;and\;{I_0}}

View answer and explanation

Correct option: D

0.75

Option D (0.75) is correct because the ratio of intensities is calculated using the phase difference and the intensity formula in Young's double-slit experiment.

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