Chapter 33, Problem 1P

To determine

The phase difference which requires the minimum path length difference.

Explanation of Solution

Introduction:

A path length difference introduces a phase difference that can be observed for a monochromatic visible light.

Write the expression of phase difference

  $\Delta \varphi =\left(\frac{2\pi }{\lambda }\right)\Delta x$

Here, $\Delta \varphi$ is phase difference, $\Delta x$ is path length difference and $\lambda$ is wavelength of the monochromatic light being used.

Since the phase difference is directly proportional to the path length difference therefore the minimum path length difference corresponds to the least phase difference of $90°$ .

Conclusion:

The smallest path length difference corresponds to the minimum phase difference of $90°$ . Thus, option $\left(a\right)$ is correct.

Since a phase difference of $180°$ is greater than that of $90°$ therefore the corresponding path length difference will also be comparatively higher. Thus, option $\left(b\right)$ is incorrect.

Since a phase difference of $270°$ is larger than that of $90°$ therefore the associated path length difference will be relatively bigger. Thus, option $\left(c\right)$ is incorrect.

The phase difference is directly proportional only to the path length difference at a constant wavelength of light. Thus, option $\left(d\right)$ is incorrect.