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Waves

JEE Main level, numerical answer, medium

51 of 142

Question 51 of 142

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 _____ $I$ .

Show answer and solution

Answer: 24

The resultant intensity is given by

{I}_{net}={I}_{1}+{I}_{2}+2\sqrt{{I}_{1}}\sqrt{{I}_{2}}\cos \phi

Now, for

{I}_{\max }

, phase angle is

\phi =0

and for

{I}_{\min }

, phase angle is

\phi =\pi

.
Thus,

{I}_{\max }={\left(\sqrt{{I}_{1}}+\sqrt{{I}_{2}}\right)}^{2}={\left(\sqrt{9I}+\sqrt{4I}\right)}^{2}=25I

And

{I}_{\min }={\left(\sqrt{{I}_{1}}-\sqrt{{I}_{2}}\right)}^{2}={\left(\sqrt{9I}-\sqrt{4I}\right)}^{2}=I

The difference of resultant intensities is

{I}_{\max }-{I}_{\min }=25I-I=24I

JEE Main question paper PDF

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JEE Main 2025 Apr 2 Shift 1 question paper

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JEE Main Physics practice question on Waves (Interference and Superposition of Waves), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.

Two light beams of intensities 4I4I4I and 9I9I9I interfere on a screen. The phase difference between these beams on the screen at point AAA is zero and at point BBB is π\piπ. The difference of resultant intensities, at the point AAA and BBB, will be _____ III.

JEE Main question paper PDF

More Physics practice

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 _____ $I$ .