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Wave Optics

JEE Main level, single correct, medium

117 of 191

Question 117 of 191

In a Young's double slit experiment the intensity at a point where the path difference is $\frac{\lambda}{6}$ ( $\lambda$ being the wavelength of the light used) is $\mathrm{I}$ . If $\mathrm{I}_0$ denotes the maximum intensity, $\frac{\mathrm{I}}{\mathrm{I}_0}$ is equal to

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Answer: (D) $3 / 4$

\frac{\mathrm{I}}{\mathrm{I}_{\max }}=\cos ^2\left(\frac{\phi}{2}\right)

JEE Main question paper PDF

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JEE Main 2025 Apr 2 Shift 1 question paper

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More Physics practice

JEE Main Physics practice question on Wave Optics (Youngs Double Slit Experiment), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.

In a Young's double slit experiment the intensity at a point where the path difference is λ6\frac{\lambda}{6}6λ​ (λ\lambdaλ being the wavelength of the light used) is I\mathrm{I}I. If I0\mathrm{I}_0I0​ denotes the maximum intensity, II0\frac{\mathrm{I}}{\mathrm{I}_0}I0​I​ is equal to

JEE Main question paper PDF

More Physics practice

In a Young's double slit experiment the intensity at a point where the path difference is $\frac{\lambda}{6}$ ( $\lambda$ being the wavelength of the light used) is $\mathrm{I}$ . If $\mathrm{I}_0$ denotes the maximum intensity, $\frac{\mathrm{I}}{\mathrm{I}_0}$ is equal to