Two monochromatic light beams have intensities in the ratio 1:9. An interference pattern is obtained by these beams. The ratio of the intensities of maximum to minimum is
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In an interference pattern formed by two sources of intensities \( I_1 \) and \( I_2 \), the maximum intensity is \( (\sqrt{I_1} + \sqrt{I_2})^2 \) and the minimum intensity is \( (\sqrt{I_1} - \sqrt{I_2})^2 \). Use the given ratio of intensities to express one intensity in terms of the other and then calculate the ratio of maximum to minimum intensity.
Let the intensities of the two monochromatic light beams be \( I_1 \) and \( I_2 \). Given the intensity ratio \( \frac{I_1}{I_2} = \frac{1}{9} \), we set \( I_1 = I \) and \( I_2 = 9I \). The maximum intensity \( I_{max} \) is \( (\sqrt{I_1} + \sqrt{I_2})^2 \). Substituting values, \( I_{max} = (\sqrt{I} + \sqrt{9I})^2 = (\sqrt{I} + 3\sqrt{I})^2 = (4\sqrt{I})^2 = 16I \). The minimum intensity \( I_{min} \) is \( (\sqrt{I_1} - \sqrt{I_2})^2 \). Substituting values, \( I_{min} = (\sqrt{I} - \sqrt{9I})^2 = (\sqrt{I} - 3\sqrt{I})^2 = (-2\sqrt{I})^2 = 4I \). The ratio \( \frac{I_{max}}{I_{min}} \) is \( \frac{16I}{4I} = 4 \). Therefore, the ratio of maximum to minimum intensity is 4:1.
