In single-slit diffraction, how does the intensity of the secondary maxima vary as the order of the maximum increases?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → It decreases.
In single-slit diffraction there is a central maximum at \( \theta=0 \), with secondary maxima on both sides at \( \theta\approx(n+1/2)\lambda/a \) and minima at \( \theta\approx n\lambda/a \).
Option (1) It increases. — Incorrect. The brightest part of the pattern is the central maximum. The secondary maxima away from it are fainter.
Option (2) It decreases. — Correct. The secondary maxima go on becoming weaker and weaker as the order \( n \) increases, so each maximum is fainter than the one before it.
Option (3) It remains the same. — Incorrect. Maxima of equal intensity belong to the interference pattern of two coherent point sources. In single-slit diffraction different maxima have different intensities.
Option (4) It first decreases and then increases. — Incorrect. The intensity never rises again. Going outward from the centre, every secondary maximum is weaker than the previous one.
NCERT (2026-27):
The Single Slit – “It is observed that the intensity has a central maximum at \( \theta=0 \) and other secondary maxima at \( \theta\approx(n+1/2)\,\lambda/a \), which go on becoming weaker and weaker with increasing n. The minima (zero intensity) are at \( \theta\approx n\lambda/a \), n = ±1, ±2, ±3, ….”