Question

In Young’s double-slit experiment, the ratio of the intensities at the maxima and minima in the interference pattern is 36 : 16. What is the ratio of the widths of the two slits?

Answer 1

Data : \(\cfrac{I_{max}}{I_{min}}\) = \(\cfrac {36}{16}\)

By componendo - dividendo, \(\cfrac{E_{10}}{E_{20}}\) = \(\cfrac{3+2}{3-2}\) = 5

Therefore, the ratio of the slit widths,

\(\cfrac{w_{1}}{w_{2}}\) = \(\cfrac{I_{1}}{I_{2}}\) = \(\left(\cfrac{E_{10}}{E_{20}}\right)^2\) = (5)² = 25