In Young’s double-slit experiment, d is the slit separation, D is the distance between the slits and the screen, λ is the wavelength and n = 0, 1, 2, ... (small angles). Match List-I with List-II.
| List-I | List-II |
| 1. \( y_n \), position of the nth bright fringe | A. \( \frac{n\lambda}{d} \) |
| 2. \( \theta_n \), angular position of the nth bright fringe | B. \( \frac{(2n+1)\lambda}{2d} \) |
| 3. \( \beta \), fringe width | C. \( \frac{nD\lambda}{d} \) |
| 4. \( \theta'_n \), angular position of the dark fringe of order n | D. \( \frac{D\lambda}{d} \) |
| 5. \( y'_n \), position of the dark fringe of order n | E. \( \frac{(2n+1)D\lambda}{2d} \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 1-C, 2-A, 3-D, 4-B, 5-E
At a point a distance \( y \) from the centre of the screen the path difference is \( yd/D \). A bright fringe needs a path difference \( n\lambda \) and a dark fringe \( (n+1/2)\lambda \). For small angles the angular position is \( \theta=y/D \).
1. \( y_n \), position of the nth bright fringe — This matches C. \( nD\lambda/d \). From \( y_nd/D=n\lambda \), the position is \( y_n=nD\lambda/d \).
2. \( \theta_n \), angular position of the nth bright fringe — This matches A. \( n\lambda/d \). Since \( \theta_n=y_n/D \), the angle is \( \theta_n=n\lambda/d \).
3. \( \beta \), fringe width — This matches D. \( D\lambda/d \). The gap between neighbouring bright fringes is \( y_{n+1}-y_n \), which equals \( D\lambda/d \) for every \( n \).
4. \( \theta'_n \), angular position of the dark fringe of order n — This matches B. \( (2n+1)\lambda/2d \). A dark fringe needs a path difference \( (n+1/2)\lambda \), so \( \theta'_n \) is \( (2n+1)\lambda/2d \).
5. \( y'_n \), position of the dark fringe of order n — This matches E. \( (2n+1)D\lambda/2d \). Since \( y'_n=D\theta'_n \), the position is \( y'_n=(2n+1)D\lambda/2d \).
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,
\( \displaystyle x=x_n=\frac{n\lambda D}{d} \)\( ;\ n=0,\pm1,\pm2,\ldots \qquad (10.13) \)
On the other hand, we will have destructive interference resulting in a dark region when \( \displaystyle\frac{xd}{D}=\left(n+\frac{1}{2}\right)\lambda \) that is
\( \displaystyle x=x_n=\left(n+\frac{1}{2}\right)\frac{\lambda D}{d} \)\( ;\ n=0,\pm1,\pm2 \qquad (10.14) \)
… Equations (10.13) and (10.14) show that dark and bright fringes are equally spaced.”