CBSE 2026 · Region 5 · Set 2 · Q18 · 2 marks
In a Young's double-slit experiment, two waves each of intensity I superpose each other and produce an interference pattern. Prove that the resultant intensities at maxima and minima are $\displaystyle 4$ I and zero respectively.
Marking-scheme solution
$\displaystyle I_{1}=I_{2}=I$
$\displaystyle I_{\max }=\left(\sqrt{I_{1}}+\sqrt{I_{2}}\right)^{2}$
$\displaystyle I_{\max }=(2 \sqrt{I})^{2}=4 I$
$\displaystyle I_{\min }=\left(\sqrt{I_{1}}-\sqrt{I_{2}}\right)^{2}$
$\displaystyle I_{\min }=0$
Wave OpticsInterference of Light Waves and Young’s ExperimentApplyvery_short_answermedium
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CBSE Class 12 Physics past-paper question from the 2026board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.