Q.Draw a graph showing the intensity distribution of fringes due to diffraction at single slit.

The key to this question is understanding what happens to light when it passes through a narrow slit and is observed on a distant screen — that’s Fraunhofer diffraction. Unlike geometrical optics, where a slit would just cast a sharp image of itself, diffraction spreads the light into a pattern of alternating bright and dark regions because different parts of the wavefront interfere with each other.

The central region is always the brightest and widest. Why? Because all the secondary wavelets from across the slit arrive in phase at the centre of the screen — they travel equal distances, so they add constructively. That gives a central bright band, not a dark one. This immediately eliminates options (A) and (C), which claim a central dark band.

Now, as you move away from the centre, the path difference between wavelets from opposite edges of the slit becomes significant. When that path difference equals exactly one wavelength, the wavelets cancel in pairs, producing the first dark band. Beyond that, a partial cancellation gives a weaker bright band, then another dark band, and so on. The intensity of these successive bright bands drops sharply — the first side maximum is only about 5% as bright as the central maximum, and the next is even fainter.

So the pattern is: central bright, then alternating dark and bright bands, with decreasing intensity. That matches option (D) exactly.

Let’s walk through the reasoning step by step.

1. Set up the physics. In Fraunhofer diffraction, the slit width $a$ is comparable to the wavelength $\lambda$, and the screen is far away. The condition for a dark band (destructive interference) is:

Here $\theta$ is the angle from the centre. There is no $m=0$ dark band — $m=0$ gives $\theta=0$, which is the centre, and that’s where all wavelets are in phase.

2. Identify the central feature. At $\theta=0$, the path difference between any two wavelets is zero, so they all interfere constructively. This produces the central maximum — a bright band. Its intensity is the highest in the pattern. …

1. The physical structure of a CD

The information on a CD is stored as a spiral of microscopic pits on a reflecting aluminium layer. Adjacent turns of the spiral are separated by about

Visible light has $\lambda \approx 0.4$–$0.7\ \mu\text{m}$. Because $d$ is of the same order as $\lambda$ (only a few times larger), the surface is exactly the regime in which wave effects dominate — the rows of pits form a reflection diffraction grating.

2. Why that produces colours

For a grating, constructive interference of the light reflected from successive rows occurs when

Rearranged:

The angle $\theta$ therefore depends on $\lambda$. White light contains all visible wavelengths, so each colour is thrown off in a slightly different direction — red (large $\lambda$) at the largest angle, violet at the smallest. The eye, looking at one point on the disc, receives only one wavelength strongly, and the disc appears to be painted in shifting rainbow bands as you tilt it. This is dispersion by diffraction, exactly as with a grating or the colours on a butterfly wing.

3. Reject the alternatives …