Match List-I with List-II in the context of Young's double slit experiment.
| List-I | List-II |
| (A) Fringe width | (I) path difference = \( \displaystyle\frac{(2n+1)\lambda}{2} \) ; n = 0, 1, 2,.. |
| (B) Condition for bright fringe in interference | (II) path difference = nλ; n = 1, 2, 3, … |
| (C) Condition for dark fringe in interference | (III) \( \displaystyle\frac{\lambda D}{d} \) |
| (D) Condition for central maximum | (IV) path difference = 0 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A) - (III), (B) - (II), (C) - (I), (D) - (IV)
In Young’s experiment the path difference \( \Delta x = yd/D \) decides whether the two waves meet in phase or out of phase.
(A) Fringe width — this matches (III) \( \frac{\lambda D}{d} \), the spacing \( y_{n+1} - y_n \) of neighbouring bright fringes.
(B) Condition for bright fringe in interference — this matches (II) path difference = nλ; n = 1, 2, 3, … The waves arrive in phase.
(C) Condition for dark fringe in interference — this matches (I) path difference = \( \frac{(2n+1)\lambda}{2} \); n = 0, 1, 2, … Crest meets trough (destructive interference).
(D) Condition for central maximum — this matches (IV) path difference = 0. The centre is equidistant from both slits.
NCERT (2026-27):
Coherent and Incoherent Addition of Waves – “… To summarise: If we have two coherent sources S1 and S2 vibrating in phase, then for an arbitrary point P whenever the path difference,
\( \displaystyle S_1P \sim S_2P = n\lambda \quad (n = 0, 1, 2, 3, \ldots) \qquad (10.9) \)
we will have constructive interference and the resultant intensity will be 4I0; the sign ~ between S1P and S2P represents the difference between S1P and S2P. On the other hand, if the point P is such that the path difference,
\( \displaystyle S_1P \sim S_2P = \left(n + \frac{1}{2}\right)\lambda \quad (n = 0, 1, 2, 3, \ldots) \qquad (10.10) \)
we will have destructive interference and the resultant intensity will be zero.”
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,
\( \displaystyle x = x_n = \frac{n\lambda D}{d};\quad n = 0, \pm 1, \pm 2, \ldots \qquad (10.13) \)”
Interference of Light Waves and Young’s Experiment – “… Such bands are called fringes. Equations (10.13) and (10.14) show that dark and bright fringes are equally spaced.”