Q 12-10-136JEE MainJEE Main 2019 (10 Apr, Shift 2)Easy
In a Young's double-slit experiment, the ratio of the slits' widths is $4:1$. The ratio of the intensity of maxima to minima, close to the central fringe on the screen, will be
Answer: (B) $9:1$
Intensity from each slit is proportional to its width, so $I_1 : I_2 = 4 : 1$ and the amplitudes are $2 : 1$.
$$\frac{I_{max}}{I_{min}} = \left(\frac{2+1}{2-1}\right)^2 = 9:1$$
Solution by Sreeraj P, M.Sc Physics