In a double slit experiment, instead of taking slits of equal widths, one slit is made twice as wide as the other. Then, in the interference pattern:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → the intensity of both the maxima & the minima increase.
The intensity of light from a slit is proportional to its width. In an interference pattern the amplitudes add at a maximum and subtract at a minimum.
\( \displaystyle I_{\max} = \left(\sqrt{I_1} + \sqrt{I_2}\right)^{2} \qquad I_{\min} = \left(\sqrt{I_1} - \sqrt{I_2}\right)^{2} \)
With two equal narrow slits, each giving \( I_0 \):
\( I_{\max} = 4I_0 \)
\( I_{\min} = 0 \)
With one slit twice as wide, \( I_1 = I_0 \) and \( I_2 = 2I_0 \):
\( \displaystyle I_{\max} = \left(1 + \sqrt{2}\right)^{2} I_0 \approx 5.8\,I_0 \)
\( \displaystyle I_{\min} = \left(\sqrt{2} - 1\right)^{2} I_0 \approx 0.17\,I_0 \)
The maxima become brighter (\( 5.8\,I_0 \gt 4I_0 \)), and the minima are no longer completely dark, because the two waves have unequal amplitudes and cannot cancel fully. So both intensities increase.
Why the other options are incorrect
Option (2) the intensity of the maxima increases & the minima have zero intensity — Incorrect. The maxima do become brighter, but unequal amplitudes cannot cancel, so the minima are not zero.
Option (3) the intensity of the maxima decreases and that of the minima increases — Incorrect. The wider slit adds light, so the maxima become brighter, not dimmer.
Option (4) the intensity of the maxima decreases and the minima have zero intensity — Incorrect. Both parts are wrong: the maxima become brighter and the minima are not zero.