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# In a Young's Double Slit experiment, the width of one of the two slits is double the other. If the amplitude of light coming from a slit is proportional to the width, find the ratio of the maximum to minimum intensity in the interference pattern.

Let the amplitude of light wave coming from the narrower slit be $a \rightarrow$ the amplitude of light wave from the wider slit $= 2a$
The maximum intensity occurs where the constructive interference takes place and the minimum intensity where the destructive interference takes place.
$\Rightarrow a_{max} = 2a + a = 3a$
$\Rightarrow a_{min} = 2a - a = a$
$\Rightarrow$ Ratio of maximum to minimum intensity $=\large\frac{I_{max}}{I_{min}}$$=\large\frac{a_{max}^2}{a_{min}^2}$$=\large\frac{3^2a^2}{a^2}$$=9$