Interference was observed in a chamber where air was present. Now, the chamber is evacuated and the same light is used. A careful observer will note:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → interference in which width of the fringe will be slightly increased
The fringe width in a double-slit set-up is \( \beta = \lambda D/d \), where \( \lambda \) is the wavelength of the light in the medium. The frequency of the light stays the same but its wavelength in a medium of refractive index n is \( \lambda = \lambda_0/n \).
Air has a refractive index slightly above 1 (about 1.0003); vacuum has exactly 1. On evacuating the chamber n falls, so \( \lambda \) rises slightly and so does \( \beta \).
\( \displaystyle \beta_{\mathrm{vacuum}} = \frac{\lambda_0 D}{d} \gt \beta_{\mathrm{air}} = \frac{\lambda_0 D}{n\,d} \)
The two slits still act as coherent sources, so interference continues, with the same kind of bright and dark bands; only the spacing changes.