Step 1: Understanding the Concept:
This question compares the fringe widths in two key wave optics phenomena: interference (typically from a double-slit) and diffraction (typically from a single-slit). Fringe width refers to the separation between consecutive bright or dark bands in the pattern.
Step 2: Detailed Explanation:
Interference Fringes (e.g., Young's Double-Slit Experiment):
In the interference pattern produced by two coherent sources, all the bright and dark fringes have the same width. The fringe width (\(\beta\)) is given by the formula \(\beta = \frac{\lambda D}{d}\), where \(\lambda\) is the wavelength, \(D\) is the distance to the screen, and \(d\) is the slit separation. This width is constant across the pattern. The intensity of all bright fringes is also the same.
Diffraction Fringes (e.g., Single-Slit Diffraction):
In the diffraction pattern produced by a single slit, the fringes are not of equal width. The central bright fringe (central maximum) is much wider and more intense than the other fringes (secondary maxima). The width of the central maximum is twice the width of any of the secondary maxima. The width of the central maximum is given by \(\frac{2\lambda D}{a}\), while the width of the secondary maxima is \(\frac{\lambda D}{a}\), where \(a\) is the slit width. Also, the intensity of the secondary maxima decreases rapidly as we move away from the center.
Comparison:
Interference fringes are of equal width and intensity.
Diffraction fringes are of unequal width and intensity.
Therefore, the width of diffraction fringes is unequal when compared to each other, and fundamentally different from the uniform width of interference fringes.
Step 3: Final Answer:
The statement implies a general comparison. Since diffraction fringes are not of uniform width, while interference fringes are, the correct description is that they are unequal. Thus, option (B) is the most appropriate answer.