In a Young's double slits experiment, the ratio of amplitude of light coming from slits is $2: 1$. The ratio of the maximum to minimum intensity in the interference pattern is:
Solution
Given the amplitude ratio, $\frac{A_1}{A_2} = \frac{2}{1}$, we can find the maximum and minimum intensities using the formula:
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$\frac{I_{max}}{I_{min}} = \left(\frac{A_1 + A_2}{A_1 - A_2}\right)^2$
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Substituting the given amplitude ratio:
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$$\frac{I_{max}}{I_{min}} = \left(\frac{2 + 1}{2 - 1}\right)^2 = \left(\frac{3}{1}\right)^2 = 9$$
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Thus, the ratio of maximum to minimum intensity in the interference pattern is:
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$\frac{I_{max}}{I_{min}} = 9:1$
About this question
Subject: Physics · Chapter: Optics · Topic: Reflection and Mirrors
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