If frac{I_1}{I_2} = frac{16}{1}, then frac{I_ | vedclass

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(B) Let the intensities of the two waves be $I_1$ and $I_2$.Given the ratio of intensities: $\frac{I_1}{I_2} = \frac{16}{1}$.The ratio of maximum to m

Vedclass · Accepted Answer

$\frac{25}{9}$

Vedclass · Answer

(B) Let the intensities of the two waves be $I_1$ and $I_2$.Given the ratio of intensities: $\frac{I_1}{I_2} = \frac{16}{1}$.The ratio of maximum to minimum intensity in interference is given by the formula:$\frac{I_{\max}}{I_{\min}} = \left( \frac{\sqrt{I_1} + \sqrt{I_2}}{\sqrt{I_1} - \sqrt{I_2}} \right)^2$.Dividing the numerator and denominator by $\sqrt{I_2}$:$\frac{I_{\max}}{I_{\min}} = \left( \frac{\sqrt{\frac{I_1}{I_2}} + 1}{\sqrt{\frac{I_1}{I_2}} - 1} \right)^2$.Substituting the given value $\frac{I_1}{I_2} = 16$:$\frac{I_{\max}}{I_{\min}} = \left( \frac{\sqrt{16} + 1}{\sqrt{16} - 1} \right)^2 = \left( \frac{4 + 1}{4 - 1} \right)^2 = \left( \frac{5}{3} \right)^2 = \frac{25}{9}$.

If $\frac{I_1}{I_2} = \frac{16}{1},$ then $\frac{I_{\max}}{I_{\min}} = ?$

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