Match List-I with List-II in the context of Young's double slit exper…

The correct answer is Option (3) → (A) - (II), (B) - (IV), (C) - (I), (D) - (III)

In Young’s double slit experiment the path difference at a point y on the screen is \( yd/D \); bright fringes need \( n\lambda \) and dark fringes need \( (n + \frac{1}{2})\lambda \).

(A) Dλ/d — this matches (II) linear fringe width, the distance between two consecutive bright fringes, \( \beta = D\lambda/d \).

(B) (n + ½) Dλ/d — this matches (IV) position of dark fringe, from \( yd/D = (n + \frac{1}{2})\lambda \).

(C) λ/d — this matches (I) Angular fringe width, \( \theta = \beta/D \).

(D) 2nπ — this matches (III) phase difference for constructive interference: a path difference \( n\lambda \) gives \( \frac{2\pi}{\lambda} \times n\lambda = 2n\pi \).

Note: Items (B) and (D) of this question are from the latest textbook (2026-27). Items (A), the linear fringe width, and (C), the angular fringe width, are printed only in the pre-rationalised textbook (2022-23). They were removed from the latest book, but Young’s double slit experiment and the expression for fringe width are still in the CUET 2026 syllabus, so do not skip them.

NCERT (2026-27):
Coherent and Incoherent Addition of Waves – “… If \( \phi = 0, \pm 2\pi, \pm 4\pi, \ldots \) which corresponds to the condition given by Eq. (10.9) we will have constructive interference leading to maximum intensity.”

Interference of Light Waves and Young’s Experiment – “… On the other hand, we will have destructive interference resulting in a dark region when \( \frac{xd}{D} = (n + \frac{1}{2})\lambda \) that is

    \( \displaystyle x = x_n = \left(n + \frac{1}{2}\right)\frac{D\lambda}{d}; \quad n = 0, \pm 1, \pm 2 \qquad (10.14) \)

Thus dark and bright bands appear on the screen, …. Such bands are called fringes. Equations (10.13) and (10.14) show that dark and bright fringes are equally spaced.”

NCERT (Pre-rationalised edition, 2022-23):
Interference of Light Waves and Young’s Experiment – “… Equations (10.18) and (10.19) show that dark and bright fringes are equally spaced and the distance between two consecutive bright and dark fringes is given by

    \( \displaystyle \beta = x_{n+1} - x_n \quad \text{or} \quad \beta = \frac{D\lambda}{d} \qquad (10.20) \)

which is the expression for the fringe width.”

Summary – “… 4. Young’s double slit of separation d gives equally spaced fringes of angular separation \( \lambda/d \).”