In Young's Double Slit experiment, the intensities of two coherent monochromatic waves are I and 4I. The ratio of intensity at maxima to minima will be:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 9 : 1
In interference, the amplitudes add at a maximum and subtract at a minimum, and the intensity is proportional to the square of the amplitude.
\( \displaystyle \frac{a_1}{a_2} = \sqrt{\frac{I_1}{I_2}} = \sqrt{\frac{I}{4I}} = \frac{1}{2} \)
Taking \( a_1 = 1 \) and \( a_2 = 2 \),
\( \displaystyle \frac{I_{\max}}{I_{\min}} = \frac{(a_1 + a_2)^{2}}{(a_2 - a_1)^{2}} = \frac{(1 + 2)^{2}}{(2 - 1)^{2}} = \frac{9}{1} \)