In an interference pattern produced by two identical slits, the intensity of the central maximum is I. If one slit is closed, the intensity of the central maximum changes to \( I_0 \). Then I and \( I_0 \) are related by
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( I=4I_0 \)
Each identical slit alone sends a wave of the same amplitude \( a \). The intensity is proportional to the square of the amplitude, so one slit gives \( I_0\propto a^{2} \). At the central maximum the path difference is zero. The two waves arrive in phase, and their amplitudes add to \( 2a \).
\( \begin{aligned} \frac{I}{I_0} &= \frac{(2a)^{2}}{a^{2}} \\ &= 4 \end{aligned} \)
Therefore, \( I=4I_0 \).
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. In fact at any point on the perpendicular bisector of S1S2, the intensity will be 4I0.”