Question

Two slits in Young's experiment have widths in the ratio $1:25$. The ratio of intensity at the maxima and minima in the interference pattern, i.e. $\frac{I_{max}}{I_{min}}$ is

• A. $\frac{49}{121}$
• B. $\frac{4}{9}$
• C. $\frac{9}{4}$
• D. $\frac{121}{49}$

Answer 1

The correct option is C $\frac{9}{4}$

As,

$I\propto \text{width of the slit}\left(W\right)\propto A^2$

$\therefore \frac{I_1}{I_2}=\frac{W_1}{W_2}=\frac{A_1^2}{A_2^2}$

Given, $\frac{W_1}{W_2}=\frac{1}{25}$

$\therefore \frac{A_1^2}{A_2^2}=\frac{1}{25}or\frac{A_1}{A_2}=\sqrt{\frac{1}{25}}=\frac{1}{5}$

$\therefore \frac{I_{max}}{I_{min}}=\frac{{(A_1+A_2)}^2}{(A_1-A_2{)}^2}=\frac{{(\frac{A_1}{A_2}+1)}^2}{{(\frac{A_1}{A_2}-1)}^2}$

$\Rightarrow \frac{I_{max}}{I_{min}}=\frac{{(\frac{1}{5}+1)}^2}{{(\frac{1}{5}-1)}^2}=\frac{{\left(\frac{6}{5}\right)}^2}{{(-\frac{4}{5})}^2}=\frac{36}{16}=\frac{9}{4}$

Hence, option $\left(C\right)$ is correct.