Two waves of intensity $l_{2} and I_{2}$ interfere. The maximum intensity produced to the minimum intensity is in the ratio a . Then, $I_{1} / I_{2}$ is equal to

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Solution

The correct option is C

$\frac{I_{m a x}}{I_{m i n}} = \frac{(\sqrt{I_{1}} + \sqrt{I_{2}})^{2}}{(\sqrt{I_{1}} - \sqrt{I_{2}})^{2}} = a \frac{I_{1}}{I_{2}} = {[\frac{\sqrt{a} + 1}{\sqrt{a} + 1}]}^{2} i f I_{1} > I_{2}$

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Two waves of intensity l2 and I2 interfere. The maximum intensity produced to the minimum intensity is in the ratio a . Then, I1/I2is equal to

The correct option is C ImaxImin=(√I1+√I2)2(√I1−√I2)2=aI1I2=[√a+1√a+1]2 if I1>I2

Two waves of intensity $l_{2} and I_{2}$ interfere. The maximum intensity produced to the minimum intensity is in the ratio a . Then, $I_{1} / I_{2}$ is equal to

The correct option is C

$\frac{I_{m a x}}{I_{m i n}} = \frac{(\sqrt{I_{1}} + \sqrt{I_{2}})^{2}}{(\sqrt{I_{1}} - \sqrt{I_{2}})^{2}} = a \frac{I_{1}}{I_{2}} = {[\frac{\sqrt{a} + 1}{\sqrt{a} + 1}]}^{2} i f I_{1} > I_{2}$