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Q.
In Young's double slit experiment, interference pattern is found to have an intensity ratio between bright and dark fringes as $9$. Then amplitude ratio will be:
Wave Optics
Solution:
As we know maximum and minimum amplitudes at locations of bright and dark fringes are given as
$ a_{\max }=a_{1}+a_{2} $ and $ a_{\min }=a_{1}-a_{2} $
$\Rightarrow \frac{I_{\max }}{I_{\min }}=\frac{\left(a_{1}+a_{2}\right)^{2}}{\left(a_{1}-a_{2}\right)^{2}}=\frac{9}{1} $
$\Rightarrow a_{1}=2 $ and $ a_{2}=1 $
$\Rightarrow \frac{a_{1}}{a_{2}}=\frac{2}{1}=2$