Question

In Young's double slit experiment the ratio of maximum and minimum intensity in the interference experiment is $9$. It means that the

• A. ratio of their amplitudes is $4$
• B. ratio of their amplitudes is $2$
• C. intensities due to two slits are $4$ units and $1$ unit respectively
• D. intensities due to two slits are $5$ units and $4$ units respectively

Answer 1

The correct option is C intensities due to two slits are $4$ units and $1$ unit respectively

let the intensities be $x$ and $y$
ratio is $9:1$ let it be $9k$ and $k$
maximum intensity is $(\sqrt{x}+\sqrt{y}{)}^2=9k$
minimum intensity is $(\sqrt{x}-\sqrt{y}{)}^2=k$
$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}=\frac{9}{1}$
$\frac{x}{y}=\frac{4}{1}$
option $C$ is correct