Question

In Young’s double slit interference experiment, the slit widths are in the ratio 1:25. Then the ratio of intensity at the maxima and minima in the interference pattern is

• A. 3 : 2
• B. 1 : 25
• C. 9 : 4
• D. 1 : 5

Answer 1

Answer: (C)

We know that,
$\frac{I_{max}}{I_{min}}=\frac{{\left(\sqrt{\frac{{\omega }_1}{{\omega }_2}+1}\right)}^2}{{\left(\sqrt{\frac{{\omega }_1}{{\omega }_2}-1}\right)}^2}$
$I_{max}$ and $I_{min}$ are maximum and min ium intensity
${\omega }_1$ and ${\omega }_2$ are widths of two slits
$\therefore \frac{I_{max}}{I_{min}}=\frac{{\left(\sqrt{\frac{1}{25}}+1\right)}^2}{{\left(\sqrt{\frac{1}{25}}-1\right)}^2}\left(\frac{{\omega }_1}{{\omega }_2}=\frac{1}{25}given\right)$
On solving we get,
$\frac{I_{max}}{I_{min}}=\frac{\frac{36}{25}}{\frac{16}{25}}=\frac{9}{4}=9:4$