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

Answer: (B)

$\frac{9}{4}$

Given,

YDSE experiment, having two slits of width are in the ratio of 1:25

the ratio of intensity,

$\frac{I_1}{I_2}=\frac{W_1}{W_2}=\frac{1}{25}$

$=>\frac{I_2}{I_1}=\frac{25}{1}$

$\frac{I_{max}}{I_{min}}=\frac{(\sqrt{I_2}+\sqrt{I_1}{)}^2}{(\sqrt{I_2}-\sqrt{I_1}{)}^2}=[\frac{\sqrt{\frac{I_2}{I_1}}+1}{\sqrt{\frac{I_2}{I_1}}-1}{]}^2$

$\frac{I_{max}}{I_{min}}=(\frac{5+1}{5-1}{)}^2$

$\frac{I_{max}}{I_{min}}=(\frac{6}{4}{)}^2=\frac{36}{16}=\frac{9}{4}$

The correct option is B.