The ratio of slit widths in a Young’s double slit experiment is \( 1 : 25 \). The ratio of intensities at the maxima and minima in the interference pattern is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 9/4 \)
The intensity of light from each slit is proportional to the slit width, and the amplitude is proportional to the square root of the intensity.
\( I_{1} : I_{2} = 1 : 25 \)
\( \displaystyle a_{1} : a_{2} = \sqrt{1} : \sqrt{25} = 1 : 5 \)
At a maximum the amplitudes add, and at a minimum they subtract:
\( \displaystyle \frac{I_{\mathrm{max}}}{I_{\mathrm{min}}} = \frac{(a_{1} + a_{2})^{2}}{(a_{2} - a_{1})^{2}} = \frac{(1 + 5)^{2}}{(5 - 1)^{2}} = \frac{36}{16} = \frac{9}{4} \)