JEE Mains · Physics · STD 12 - Wave optics
- JEE Mains 2025
- Medium
- Numerical
Two coherent monochromatic light beams of intensities 4I and 9I are superimposed. The difference between the maximum and minimum intensities in the resulting interference pattern is xI. The value of \(x\) is ______.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) 24
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \mathrm{I}_{\max }=\left(\sqrt{\mathrm{I}_1}+\sqrt{\mathrm{I}_2}\right)^2 \\ & =(\sqrt{4 \mathrm{I}}+\sqrt{9 \mathrm{I}})^2=25 \mathrm{I} \\ & \mathrm{I}_{\min }=\left(\sqrt{\mathrm{I}_1}-\sqrt{\mathrm{I}_2}\right)^2 \\ & =(\sqrt{4 \mathrm{I}}-\sqrt{9 \mathrm{I}})^2=\mathrm{I} \\ & \mathrm{I}_{\max }-\mathrm{I}_{\min }=24 \mathrm{I} \\ & \mathrm{x}=24\end{aligned}\)
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