If two coherent sources are infinitely close to each other, do you expect an interference pattern?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → No. The fringe width becomes very large, so no interference pattern is obtained.
The bright fringes in Young’s experiment lie at \( x_n=n\lambda D/d \), so the fringe width is
\( \begin{aligned} \beta &= x_{n+1}-x_n \\ &= \frac{\lambda D}{d} \end{aligned} \)
It is inversely proportional to the separation \( d \) of the sources. Sources infinitely close to each other mean \( d\rightarrow0 \), and then \( \beta\rightarrow\infty \).
Option (1) Yes. We can always expect to obtain an interference pattern. — Incorrect. Fringes are seen only when their spacing \( \lambda D/d \) is small enough for bright and dark bands to fit on the screen. For \( d\rightarrow0 \) this fails.
Option (2) Yes. Depending on the wavelength of light, we may obtain an interference pattern. — Incorrect. For any finite wavelength, \( \lambda D/d \) grows without limit as \( d\rightarrow0 \), so no choice of wavelength brings the fringes back.
Option (3) No. The fringe width becomes very large, so no interference pattern is obtained. — Correct. With \( d\rightarrow0 \) the fringe width becomes very large, so a single bright fringe spreads over the whole screen. The screen looks uniformly lit, and no pattern of bright and dark bands appears.
Option (4) No. The fringe width becomes very small, so no interference pattern is obtained. — Incorrect. Since \( \beta \) varies as \( 1/d \), bringing the sources closer makes the fringes wider. They do not become narrower.
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,
\( \displaystyle x=x_n=\frac{n\lambda D}{d} \)\( ;\ n=0,\pm1,\pm2,\ldots \qquad (10.13) \)”