Two light beams of intensities 4I and 9I… - JEE Main 2022 Physics PYQ

JEE Main 2022 — Physics Waves & Oscillations

2022integereasy

Two light beams of intensities $4I$ and $9I$ interfere on a screen. The phase difference between these beams on the screen at point $A$ is zero and at point $B$ is $\pi$ . The difference of resultant intensities, at the point $A$ and $B$ , will be _____ $I$ .

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Held on 29 Jul 2022 · Verified 6 Jul 2026.

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Solution

The resultant intensity is given by ${I}_{\mathrm{net}}={I}_{1}+{I}_{2}+2\sqrt{{I}_{1}}\sqrt{{I}_{2}}\mathrm{cos}\phi$

Now, for ${I}_{\mathrm{max}}$ , phase angle is $\phi =0$ and for ${I}_{\mathrm{min}}$ , phase angle is $\phi =\pi$ .

Thus, ${I}_{\mathrm{max}}={(\sqrt{{I}_{1}}+\sqrt{{I}_{2}})}^{2}={(\sqrt{9I}+\sqrt{4I})}^{2}=25I$

And ${I}_{\mathrm{min}}={(\sqrt{{I}_{1}}-\sqrt{{I}_{2}})}^{2}={(\sqrt{9I}-\sqrt{4I})}^{2}=I$

The difference of resultant intensities is ${I}_{\mathrm{max}}-{I}_{\mathrm{min}}=25I-I=24I$

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