In Young’s double-slit experiment, the ratio of the intensities of the bright and dark fringes in the interference pattern is 9. The ratio of the amplitudes of the two interfering waves will be
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 2
By the superposition principle, the resultant displacement is \( y=y_1+y_2 \), and the intensity is proportional to the square of the amplitude.
At a bright fringe the two waves arrive in phase, so their amplitudes add: \( a_1+a_2 \).
At a dark fringe they arrive out of phase, so the displacements oppose: \( a_1-a_2 \) (taking \( a_1\gt a_2 \)).
The ratio of the intensities then gives:
\( \displaystyle \frac{I_{\mathrm{bright}}}{I_{\mathrm{dark}}}=\frac{(a_1+a_2)^2}{(a_1-a_2)^2} \)
\( \displaystyle \frac{(a_1+a_2)^2}{(a_1-a_2)^2}=9 \)
\( \displaystyle \Rightarrow\; \frac{a_1+a_2}{a_1-a_2}=3 \)
\( a_1+a_2=3a_1-3a_2 \)
\( 2a_1=4a_2 \)
\( \Rightarrow\; a_1/a_2=2 \)
NCERT (2026-27):
Coherent and Incoherent Addition of Waves – “… Since the intensity is proportional to the square of the amplitude, the resultant intensity will be given by
\( I=4I_0 \)
where I0 represents the intensity produced by each one of the individual sources; I0 is proportional to a2.”