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

\frac{α - 1}{α + 1}, i f l_{1} > l_{2}

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\frac{α + 1}{α - 1}, i f l_{1} > l_{2}

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{[\frac{1 + \sqrt{α}}{\sqrt{α - 1}}]}^{2}, i f l_{1} > l_{2}

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{[\frac{\sqrt{α - 1}}{1 + \sqrt{α}}]}^{2}, i f l_{1} > l_{2}

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Solution

The correct option is C ${[\frac{1 + \sqrt{α}}{\sqrt{α - 1}}]}^{2}, i f l_{1} > l_{2}$
$\frac{l_{m a x}}{l_{m i n}} = \frac{(\sqrt{l_{1}} + \sqrt{l_{2}})^{2}}{(\sqrt{l_{1}} - \sqrt{l_{2}})^{2}} = α \frac{l_{1}}{l_{2}} = {[\frac{\sqrt{a} + 1}{\sqrt{a} - 1}]}^{2} i f l_{1} > l_{2}$

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Two waves of intensity I2 and I2 interfere. The maximum intensity produced to the minimum intensity is in the ratio α. Then,I1I2 is equal to

The correct option is C [1+√α√α−1]2,if l1>l2lmaxlmin=(√l1+√l2)2(√l1−√l2)2=αl1l2=[√a+1√a−1]2 if l1>l2

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

The correct option is C ${[\frac{1 + \sqrt{α}}{\sqrt{α - 1}}]}^{2}, i f l_{1} > l_{2}$
$\frac{l_{m a x}}{l_{m i n}} = \frac{(\sqrt{l_{1}} + \sqrt{l_{2}})^{2}}{(\sqrt{l_{1}} - \sqrt{l_{2}})^{2}} = α \frac{l_{1}}{l_{2}} = {[\frac{\sqrt{a} + 1}{\sqrt{a} - 1}]}^{2} i f l_{1} > l_{2}$