A double-slit arrangement that gives interference fringes with sodium light is now illuminated with white light, the slits and the screen being kept in the same positions. The pattern seen on the screen will have
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → a white central fringe with coloured fringes on either side of it
White light is a mixture of wavelengths, and each wavelength forms its own set of fringes. The bright fringes of a wavelength \( \lambda \) lie at \( x_n=n\lambda D/d \). At the centre (\( n=0 \)) the path difference is zero for every wavelength, so all colours are bright there together. For \( n\neq0 \) the position depends on the wavelength. So the bright fringes of violet and blue lie closer to the centre than those of red.
Option (1) all the bright fringes white — Incorrect. The maxima of all the colours fall together only at the centre. Away from it, each colour has its maxima at a different place, so the fringes cannot stay white.
Option (2) a white central fringe with coloured fringes on either side of it — Correct. At the centre every wavelength interferes constructively, and the colours mix to give white. On either side the fringes of different colours separate, so these fringes look coloured.
Option (3) a coloured central fringe with white fringes on either side of it — Incorrect. This reverses the pattern. The centre is the one point where all the colours reach a maximum together, and the colours separate more and more as \( n \) increases.
Option (4) coloured fringes with a dark central fringe — Incorrect. A dark fringe needs a path difference of \( (n+\frac{1}{2})\lambda \). At the centre the path difference is zero, so the central fringe is bright for every colour.
NCERT (2026-27):
Interference of Light Waves and Young’s Experiment – “We will have constructive interference resulting in a bright region when \( \displaystyle\frac{xd}{D}=n\lambda \). That is,
\( \displaystyle x=x_n=\frac{n\lambda D}{d} \)\( ;\ n=0,\pm1,\pm2,\ldots \qquad (10.13) \)”
Seeing the Single Slit Diffraction Pattern – “… With slight adjustment of the width of the slit and the parallelism of the edges, the pattern should be seen with its bright and dark bands. Since the position of all the bands (except the central one) depends on wavelength, they will show some colours.”