Diffraction at a Single Slit

Light Bends Around Edges

Look closely at a sharp shadow's edge: instead of a clean line, there are faint bright and dark bands. Light has spread into the geometric shadow — diffraction. It is a general property of all waves (sound, water, light, matter waves). We rarely notice it for light because $\lambda$ is far smaller than everyday obstacles — but it sets the ultimate resolution of every telescope and microscope, and paints the colours flashing off a CD.

Replace Young's double slit with a single narrow slit of width a (monochromatic, normal incidence) and a new pattern appears: a broad central bright maximum flanked by weaker secondary maxima, fading outwards.

Huygens-Fresnel idea: treat each strip of the slit's wavefront as a secondary source; add their contributions at angle $\theta$ with the right phase differences. The slit-edge-to-edge path difference is $a\sin\theta$.

The Single-Slit Conditions

Minima (zero intensity): when the slit divides into pairs that exactly cancel —

$$\boxed{a\sin\theta = n\lambda, \quad n = \pm1, \pm2, \pm3, ...}$$

(note: $n = 0$ is NOT a minimum — it's the central maximum.)

Secondary maxima (weak, between the minima):

$$\boxed{a\sin\theta \approx \left(n+\tfrac12\right)\lambda, \quad n = 1, 2, 3, ...}$$

Central maximum: stretches between the first minima at $\theta = \pm\lambda/a$. Its angular width is $2\lambda/a$ — twice that of any other fringe; its linear width on a screen at distance D is $\frac{2\lambda D}{a}$. Narrower slit → wider central spread (the wave 'fans out' more).

Key Point — interference vs diffraction: in Young's double slit the bright fringe condition is $d\sin\theta = n\lambda$; in single-slit diffraction the minima are at $a\sin\theta = n\lambda$. Same-looking equation, opposite meaning ($d$ = slit separation vs $a$ = slit width; bright vs dark). The single most punished confusion in this chapter.

Feynman's caution (NCERT quotes it): 'No one has been able to define the difference between interference and diffraction satisfactorily… when there are a few sources it is called interference, with a large number, diffraction.' The double-slit pattern is genuinely a single-slit diffraction envelope times the double-slit interference fringes.

[NEET Important] Central maximum is twice as wide and far brighter than the secondary maxima; minima at $a\sin\theta = n\lambda$ (not bright!). The razor-blade-and-bulb home experiment (red fringes wider than blue, since $\beta \propto \lambda$) is a Board favourite.

Solved Examples

Example 1: First minimum [NEET Numerical]

Monochromatic 600 nm light passes through a 0.2 mm slit. Find the angle of the first diffraction minimum.

Solution:

1. $a\sin\theta = \lambda$ (first minimum, n = 1).
2. $\sin\theta = \frac{600\times10^{-9}}{0.2\times10^{-3}} = 3\times10^{-3}$.
3. $\theta \approx 3\times10^{-3}$ rad $\approx 0.17^{\circ}$ — a small fan, as $\lambda \ll a$.

Example 2: Width of the central maximum [JEE Numerical]

For that slit, find the linear width of the central maximum on a screen 1.5 m away.

Solution:

1. Angular width $= 2\lambda/a = 6\times10^{-3}$ rad.
2. Linear width $= \frac{2\lambda D}{a} = 6\times10^{-3}\times1.5$.
3. $= 9\times10^{-3}$ m = 9 mm — and the central band is twice as wide as the others.

Example 3: Narrower slit spreads more [NEET Numerical]

If the slit width a is halved, what happens to the central maximum's width?

Solution:

1. Central width $\propto 1/a$.
2. Halving a doubles the central maximum's width.
3. The narrower the aperture, the more the wave fans out — the heart of the diffraction limit.

Example 4: Interference vs diffraction equation trap [JEE Numerical]

For light of 500 nm: (a) a double slit with d = 0.1 mm — where is the 1st bright fringe angle? (b) a single slit with a = 0.1 mm — where is the 1st dark fringe angle?

Solution:

1. (a) Bright: $d\sin\theta = \lambda \Rightarrow \sin\theta = \frac{500\times10^{-9}}{10^{-4}} = 5\times10^{-3}$ rad — a maximum.
2. (b) Dark: $a\sin\theta = \lambda \Rightarrow$ the same angle $5\times10^{-3}$ rad — but a minimum.
3. Identical equation, opposite physics: d-bright vs a-dark. Read the geometry, never the formula alone.

Example 5: Why doesn't a doorway diffract light visibly?

Explain why sound diffracts around a doorway but light seems not to.

Solution:

1. Diffraction is significant when $\lambda$ is comparable to the obstacle/aperture size.
2. Sound's wavelength (~1 m) is comparable to a doorway → strong spreading (you hear around corners).
3. Light's $\lambda$ (~$10^{-6}$ m) is a million times smaller than the doorway → negligible spreading — hence sharp shadows and the success of ray optics.

Example 6: Second minimum [NEET Numerical]

For a = 0.25 mm and $\lambda$ = 500 nm, find the angle of the 2nd minimum.

Solution:

1. $a\sin\theta = 2\lambda$.
2. $\sin\theta = \frac{2\times500\times10^{-9}}{0.25\times10^{-3}} = 4\times10^{-3}$.
3. $\theta \approx 4\times10^{-3}$ rad. Minima march out at $n\lambda/a$; the maxima sit between them.

Example 7: The CD's colours

Why does a CD flash rainbow colours in white light?

Solution:

1. A CD's track is a fine periodic structure (a reflection grating) — its spacing is comparable to light's wavelength.
2. Different wavelengths diffract to different angles, separating white light into colours.
3. NCERT cites exactly this as everyday diffraction; the same physics as the single slit, periodically repeated.

Example 8: Red vs blue in the home experiment [NEET pattern]

In the two-razor-blade single slit, why are the red fringes wider than the blue?

Solution:

1. Fringe positions scale with $\lambda$ ($\sin\theta = n\lambda/a$).
2. Red ($\lambda \approx 650$ nm) > blue ($\approx 450$ nm), so red's fringes sit at larger angles — wider.
3. NCERT's exact observation with red/blue filters; longer wavelength, broader pattern.

Example 9: Double slit as envelope times fringes

Explain why a real double-slit pattern has fringes of varying brightness.

Solution:

1. Each slit (width a) diffracts, producing a single-slit intensity envelope.
2. The two slits (separation d) interfere, producing closely spaced fringes.
3. The actual pattern is the product: interference fringes modulated by the diffraction envelope — so fringes near the envelope's minima are dim or missing (Feynman's point that the two phenomena are one).

Example 10: Resolution hint [JEE pattern]

Using the central-maximum width, explain why a larger telescope aperture sees finer detail.

Solution:

1. A point source images not as a point but as a diffraction disc of angular size $\sim \lambda/a$ (a = aperture).
2. Two stars closer than this angle blur together — the diffraction limit.
3. Larger a → smaller $\lambda/a$ → finer resolvable detail — the deep reason giant telescope mirrors matter (Chapter 9's aperture argument, now explained).