Wave Optics

JEE Main level, numerical answer, medium

10 of 191

Question 10 of 191

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 xx is ______.

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Answer: 24

Imax⁡=(I1+I2)2=(4I+9I)2=25IImin⁡=(I1−I2)2=(4I−9I)2=IImax⁡−Imin⁡=24Ix=24\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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