In Young's double slits experiment, if the ratio of the width of the two slits is 25 : 1, what is the ratio of intensity at the maxima and minima in the interference pattern?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 9 : 4
The light passing through a slit is proportional to its width, so the intensities from the two slits are in the ratio of the widths. The amplitude varies as the square root of the intensity.
\( \displaystyle \frac{I_1}{I_2} = \frac{25}{1} \;\Rightarrow\; \frac{a_1}{a_2} = \sqrt{\frac{25}{1}} = \frac{5}{1} \)
At a maximum the amplitudes add and at a minimum they subtract:
\( \displaystyle \frac{I_{\max}}{I_{\min}} = \frac{(a_1 + a_2)^{2}}{(a_1 - a_2)^{2}} = \frac{(5 + 1)^{2}}{(5 - 1)^{2}} = \frac{36}{16} = \frac{9}{4} \)