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

\frac{α - 1}{α + 1}, I f I_{1} > I_{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{α + 1}{α - 1}, I f I_{1} > I_{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

{(\frac{1 + \sqrt{α}}{\sqrt{α} - 1})}^{2}, I f I_{1} > I_{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

{(\frac{\sqrt{α} - 1}{\sqrt{α} + 1})}^{2}, I f I_{1} > I_{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C ${(\frac{1 + \sqrt{α}}{\sqrt{α} - 1})}^{2}, I f I_{1} > I_{2}$
The ratio of maximum to minimum intensity due to superposition of two waves

$\frac{I_{m a x}}{I_{m a x}} = \frac{{(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2}}{{(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2}} = α$

$\frac{(\sqrt{I_{1}} + \sqrt{I_{2}})}{(\sqrt{I_{1}} - \sqrt{I_{2}})} = \sqrt{α}$

$\sqrt{I_{1}} + \sqrt{I_{2}} = \sqrt{α} (\sqrt{I_{1}} - \sqrt{I_{2}})$

$\sqrt{\frac{I_{1}}{I_{2}}} = (\frac{\sqrt{α} + 1}{\sqrt{α} - 1}) i f I_{1} > I_{2}$

$\Rightarrow \frac{I_{1}}{I_{2}} = {(\frac{\sqrt{α} + 1}{\sqrt{α} - 1})}^{2} i f I_{1} > I_{2}$

Suggest Corrections

Similar questions

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

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