Red, violet and green light are used one after another in Young’s double-slit experiment. Which one gives the maximum fringe width?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → red
In Young’s experiment the bright fringes lie at \( x_n=n\lambda D/d \), so the gap between neighbouring fringes, the fringe width, is
\( \begin{aligned} \beta &= x_{n+1}-x_n \\ &= \lambda D/d \end{aligned} \)
For the same slits and screen it is proportional to the wavelength.
Option (1) red — Correct. Red light has the longest wavelength of the three, so it gives the widest fringes.
Option (2) violet — Incorrect. Violet has the shortest wavelength of the three, so it gives the narrowest fringes.
Option (3) green — Incorrect. Green lies between violet and red in wavelength, so its fringe width is also in between.
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,
\( \displaystyle x=x_n=\frac{n\lambda D}{d} \)\( \displaystyle ;\ n=0,\pm1,\pm2,\ldots \qquad (10.13) \)
… Thus dark and bright bands appear on the screen, … Such bands are called fringes. … dark and bright fringes are equally spaced.”