Question

Interference was observed in interference chamber where air was present, now the chamber is evacuated, and if the same light is used, a careful observer will see

• A. no interference
• B. interference with brighter bands
• C. interference with dark bands
• D. interference with larger width

Answer 1

The Correct Option is D

 To solve this problem, we need to understand the effect of changing the medium in an interference setup. Let's explore the steps involved:

Understanding Interference: In an interference setup, when light waves overlap, they either constructively or destructively interfere, producing patterns of bright and dark bands. The key factor in determining the pattern is the path difference between the interfering waves.

Effect of Medium on Wavelength: The wavelength of light in a medium is different from its wavelength in a vacuum. The wavelength \((\lambda)\) in a medium is given by \(\lambda = \frac{\lambda_0}{n}\), where \(\lambda_0\) is the wavelength in a vacuum and \(n\) is the refractive index of the medium.

Path Difference and Fringe Width: The fringe width \((w)\) is related to the wavelength and the geometry of the setup by the formula \(w = \frac{\lambda \cdot D}{d}\), where \(D\) is the distance to the screen and \(d\) is the separation between the slits or sources.

Changes when Chamber is Evacuated: When the chamber is evacuated, the medium changes from air to vacuum. The refractive index of air is approximately 1, but still greater than the refractive index of vacuum (which is exactly 1). Thus, the wavelength of light will increase when the chamber is evacuated, since \(\lambda_0 > \lambda\) in air.

Conclusion on Fringe Width: As the wavelength increases (larger in vacuum compared to air), the fringe width will also increase because it is directly proportional to the wavelength. Therefore, the observer will see interference with larger width.

Thus, the correct answer is: interference with larger width.