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

Answer 1

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

Intensity is proportional to square of the width of the slit. Thus, $\frac{I_1}{I_2}=\frac{1}{25}$
or $\frac{a_1}{a_2}=\sqrt{\frac{I_1}{I_2}}=\frac{1}{5}$ or $a_2=5a_1$
now, $\frac{I_{max}}{I_{min}}=\frac{(a_1+a_2)}{(a_1-a_2{)}^2}=\frac{(a_1+5a_1{)}^2}{(a_1-5a_2{)}^2}=36/16=9/4$