A parallel beam of monochromatic light falls normally on a narrow slit and the diffraction pattern is seen on a distant screen. If the light is replaced by light of a shorter wavelength, which of the following is correct?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → The angular width of the central maximum decreases.
The minima of a single-slit diffraction pattern lie at \( \theta\approx n\lambda/a \), \( n=\pm1,\pm2,\ldots \). The central maximum spreads between the first minima at \( \theta\approx\pm\lambda/a \). So its angular width is \( 2\lambda/a \). For a given slit, this width is proportional to the wavelength.
Option (1) No diffraction pattern is observed. — Incorrect. Light of a shorter wavelength still has its minima at \( \theta\approx n\lambda/a \), so the pattern is still formed. It is only narrower.
Option (2) The angular width of the central maximum increases. — Incorrect. A width that grows with \( \lambda \) cannot increase when \( \lambda \) is made smaller.
Option (3) The angular width of the central maximum decreases. — Correct. With the same slit width \( a \), a shorter wavelength brings the first minima at \( \pm\lambda/a \) closer to the centre. So the central maximum becomes narrower.
Option (4) The angular width of the central maximum remains the same. — Incorrect. The width depends on the wavelength as well as on the slit width, so it changes when the wavelength changes.
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, ….”