In a Young’s double slit experiment, the fringe width
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → increases if the screen is moved away from the slits
In Young’s experiment the fringe width is
\( \displaystyle \beta = \frac{\lambda D}{d} \)
where \( \lambda \) is the wavelength, D the distance of the screen from the slits and d the separation of the slits. So \( \beta \) is proportional to D: moving the screen away from the slits widens the fringes.
Why the other options are incorrect
Option (1) increases, if the separation between two slits is increased — Incorrect. \( \beta \) is inversely proportional to d, so increasing the slit separation makes the fringes narrower.
Option (2) decreases if the wavelength of light is increased — Incorrect. \( \beta \) is proportional to \( \lambda \), so a longer wavelength gives wider fringes, not narrower ones.
Option (4) increases if the wavelength of light is decreased — Incorrect. For the same reason, a shorter wavelength gives narrower fringes.
NCERT (2026-27):
Interference of Light Waves and Young’s Experiment – “… We will have constructive interference resulting in a bright region when \( xd/D = n\lambda \). That is,
\( \displaystyle x = x_n = \frac{n\lambda D}{d}; \quad n = 0, \pm 1, \pm 2, \ldots \qquad (10.13) \)
… Thus dark and bright bands appear on the screen, as shown in Fig. 10.13. Such bands are called fringes. Equations (10.13) and (10.14) show that dark and bright fringes are equally spaced.”