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,
$\frac{I_{max}}{I_{min} } = \frac{\frac{36}{25}}{\frac{16}{25}} = \frac{9}{4} = 9 : 4$