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, $\frac{I_{\max }}{I_{\min }}$ is

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

Answer 1

Answer: (C)

As, intensity $I\propto$ width of slit W
Also, intensity $I\propto$ square of amplitude $A$
$\therefore \frac{I_1}{I_2}=\frac{W_1}{W_2}=\frac{A_1{}^2}{A_2{}^2}$
But $\frac{W_1}{W_2}=\frac{1}{25}$ (given)
$\therefore \frac{A_1{}^2}{A_2{}^2}=\frac{1}{25}\text{ or }\frac{A_1}{A_2}=\sqrt{\frac{1}{25}}=\frac{1}{5}$
$\therefore \frac{I_{\max }}{I_{\min }}=\frac{{\left(A_1+A_2\right)}^2}{{\left(A_1-A_2\right)}^2}=\frac{{\left(\frac{A_1}{A_2}+1\right)}^2}{{\left(\frac{A_1}{A_2}-1\right)}^2}$
$=\frac{{\left(\frac{1}{5}+1\right)}^2}{{\left(\frac{1}{5}-1\right)}^2}=\frac{{\left(\frac{6}{5}\right)}^2}{{\left(-\frac{4}{5}\right)}^2}=\frac{36}{16}=\frac{9}{4}$