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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.

1 Answer

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$
answered Apr 19, 2014 by balaji.thirumalai

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