The intensity ratio of the maxima and minima | vedclass

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(B) The ratio of maximum intensity to minimum intensity is given by the formula:$\frac{I_{\max }}{I_{\min }}=\frac{(\sqrt{I_1}+\sqrt{I_2})^2}{(\sqrt{I

Vedclass · Accepted Answer

$4$

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(B) The ratio of maximum intensity to minimum intensity is given by the formula:$\frac{I_{\max }}{I_{\min }}=\frac{(\sqrt{I_1}+\sqrt{I_2})^2}{(\sqrt{I_1}-\sqrt{I_2})^2}$Given that $\frac{I_{\max }}{I_{\min }} = \frac{9}{1}$,we have:$\frac{9}{1} = \frac{(\sqrt{I_1}+\sqrt{I_2})^2}{(\sqrt{I_1}-\sqrt{I_2})^2}$Taking the square root on both sides:$\frac{3}{1} = \frac{\sqrt{I_1}+\sqrt{I_2}}{\sqrt{I_1}-\sqrt{I_2}}$Cross-multiplying gives:$3(\sqrt{I_1}-\sqrt{I_2}) = 1(\sqrt{I_1}+\sqrt{I_2})$$3\sqrt{I_1} - 3\sqrt{I_2} = \sqrt{I_1} + \sqrt{I_2}$$2\sqrt{I_1} = 4\sqrt{I_2}$$\sqrt{\frac{I_1}{I_2}} = \frac{4}{2} = 2$Squaring both sides,we get:$\frac{I_1}{I_2} = \frac{4}{1}$Thus,the ratio of the intensities of the light sources is $4: 1$.

The intensity ratio of the maxima and minima in an interference pattern produced by two coherent sources of light is $9: 1$. The intensities of the light sources used are in the ratio (in $: 1$)

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