Physics and Mathematics
Diffraction of Light
1. Concept Overview
When a wave (like light) meets an obstacle or passes through a narrow aperture comparable in size to its wavelength, it bends and spreads out — this phenomenon is called diffraction.
Think of water waves passing through a narrow gap: instead of continuing in straight lines only, they spread into the shadow region. Light behaves similarly — diffraction is most noticeable when the slit width or obstacle size is of the same order as the wavelength of light.
Why it matters: Diffraction explains why perfect images are never infinitely sharp (limits of resolution), why slits and edges produce patterns of dark and bright bands, and why instruments have finite resolving power.
2. Clear Explanation and Two-Level Mathematical Treatment
- L1: NCERT-style geometric derivation of minima (simple condition)
- L2: Full amplitude / intensity derivation leading to the ([\dfrac{\sin \beta}{\beta}]) envelope
Setup (single slit)
- Slit width = ([a])
- Monochromatic light of wavelength = ([\lambda]) normally incident on the slit
- Screen at distance = ([D]) (with ([D \gg a]))
- Point on screen at angle = ([\theta]) measured from central axis
L1 — NCERT-style Condition for Minima (Geometry / Path-Difference Method)
Divide the slit of width ([a]) into two equal halves. Consider waves from corresponding points in the two halves. For a direction ([\theta]), the path difference between corresponding points is:
[\Delta] [= \left(\dfrac{a}{2}\right)\sin\theta – \left(-\dfrac{a}{2}\right)\sin\theta = a\sin\theta]
For the first minimum, choose (\theta) such that the waves from corresponding points are exactly out of phase (path difference ([\lambda/2]) between halves). More generally, setting the slit into ([m]) equal parts yields minima at:
[\boxed{a\sin\theta = m\lambda \quad (m = 1,2,3,\dots)}]
So the minima (dark fringes) occur at angles ([\theta_m]) satisfying:
[\sin\theta_m = \dfrac{m\lambda}{a}]
The central maximum is between [\theta_1] and [-\theta_1]. This is the NCERT/qualitative minima condition.
L2 — Full Amplitude & Intensity Derivation (Fraunhofer Single-Slit Diffraction)
We treat the slit as a continuous array of sources (Huygens’ principle) and integrate contributions.
Let the slit extend from [-a/2] to [+a/2] along the aperture. Take an observation point at far-field angle ([\theta]). A point at position ([x]) inside the slit contributes a phasor proportional to ([e^{ikx\sin\theta}]) where ([k = \dfrac{2\pi}{\lambda}]).
Total complex amplitude ([E(\theta)]) at the point on the screen:
[E(\theta)] [\propto \displaystyle \int_{-a/2}^{a/2} e^{ikx\sin\theta} dx]
Integrate:
[E(\theta)] [\propto \left( \dfrac{e^{ikx\sin\theta}}{ik\sin\theta} \right)_{-a/2}^{a/2}] [= \dfrac{e^{i k (a/2)\sin\theta} – e^{-i k (a/2)\sin\theta}}{ik\sin\theta}]
Use sine identity:
[E(\theta)] [\propto \dfrac{2\sin \big( \tfrac{k a}{2}\sin\theta \big)}{k\sin\theta}]
Define:
[\beta] [\equiv \dfrac{\pi a}{\lambda}\sin\theta] [\qquad] [\text{(note: } k=\tfrac{2\pi}{\lambda}\text{)}]
Then:
[E(\theta)] [\propto \dfrac{\sin\beta}{\beta}]
Intensity is proportional to the square of amplitude:
[\boxed{I(\theta) = I_0 \left( \dfrac{\sin\beta}{\beta} \right)^2}]
where ([I_0]) is the central maximum intensity at [\theta=0] (where [\beta\to0]) and [\lim_{\beta\to0} (\sin\beta/\beta)^2 = 1)].
Minima locations: zeros of [\sin\beta] (except [\beta=0]) occur when:
[\beta = m\pi] [\quad] [(m = \pm1,\pm2,\dots)]
Substitute definition of [\beta]:
[\dfrac{\pi a}{\lambda}\sin\theta] [= m\pi] [\Rightarrow a\sin\theta = m\lambda]
Maxima: secondary maxima occur between minima, but their intensities are smaller and are given by maxima of [(\sin\beta/\beta)^2] (no simple analytic form for their positions; they are approximately near [\beta \approx (m+\tfrac12)\pi] but with reduced amplitude).
3. Dimensions and Units
| Quantity | Symbol | Dimension | SI Unit |
|---|---|---|---|
| Slit width | ([a]) | [L] | metre (m) |
| Wavelength | ([\lambda]) | [L] | metre (m) |
| Angle | ([\theta]) | dimensionless (radians) | radian |
| Wave number | ([k]) | [L(^{-1})] | metre(^{-1}) |
| Intensity | ([I]) | [Power per area] | W m(^{-2}) |
4. Key Features / Physical Insights
- Diffraction is significant when slit width ([a]) is comparable to wavelength ([\lambda]).
- The central maximum is the brightest and widest; width between first minima on either side ~ ([2\lambda/a]) in angular terms.
- Single-slit intensity follows the envelope ([I(\theta)=I_0(\sin\beta/\beta)^2]).
- Minima at ([a\sin\theta = m\lambda]) — these are exact for Fraunhofer (far-field) diffraction.
- Secondary maxima exist but are much weaker than central maximum.
- Diffraction limits the resolving power of optical systems (Rayleigh criterion builds on this).
- Diffraction patterns are symmetric about (\theta=0).
5. Important Formulas to Remember
| Topic | Formula |
|---|---|
| Minima (single slit) | ([a\sin\theta = m\lambda]) for (m = 1,2,\dots) |
| First minima angular position | ([\sin\theta_1 = \dfrac{\lambda}{a}]) |
| Intensity distribution | ([I(\theta) = I_0 \left(\dfrac{\sin\beta}{\beta}\right)^2]) where ([\beta = \dfrac{\pi a}{\lambda}\sin\theta]) |
| Small-angle approx. fringe half-angle | ([\theta \approx \dfrac{m\lambda}{a}]) when (\sin\theta \approx \theta) |
| Angular width of central maximum | ([2\theta_1 \approx 2\lambda/a]) |
6. Conceptual Questions with Solutions
1. Why does diffraction become noticeable when slit width is comparable to wavelength?
Because only then do the path differences across the slit produce significant phase differences. If [a \gg \lambda], phase variations across the slit are negligible and geometric optics applies; if [a \sim \lambda], interference between parts of the aperture causes substantial spreading.
2. Why is the central maximum the brightest?
At [\theta=0\] all contributions from the slit are in phase, adding coherently to produce maximum amplitude and hence maximum intensity.
3. Why are secondary maxima weaker?
Because partial cancellations occur across the aperture for angles away from zero; the constructive contributions are smaller and less perfectly phased.
4. How does increasing slit width [a] affect diffraction?
Increasing [a] reduces angular spread: minima satisfy ([a\sin\theta=m\lambda]), so larger [a] gives smaller [\theta] → narrower central maximum.
5. Why do we approximate [\sin\theta \approx \theta] often?
Because in Fraunhofer geometry the angles are small (screen far away, [D\gg a]), simplifying expressions and giving linear relations useful for experiments.
6. Can diffraction occur with sound or water waves?
Yes — diffraction is a general wave phenomenon, not limited to light.
7. Why don’t we see diffraction when light passes through a wide door?
A door is huge compared to [\lambda] [~500 nm], so diffraction angles are minuscule and undetectable.
8. Is the diffraction pattern dependent on polarization?
Single-slit intensity formula assumes scalar waves; for polarized light, diffraction pattern shape is essentially unchanged, though vector effects can matter in complex apertures.
9. What is meant by Fraunhofer diffraction?
Fraunhofer (far-field) diffraction refers to the pattern observed when both source and screen are effectively at infinity relative to the aperture, or when lenses are used to make plane-wave illumination and observe the Fourier-plane pattern.
10. What is Fresnel diffraction?
Fresnel (near-field) diffraction deals with finite distances; the wavefront curvature matters and calculations are more complex than Fraunhofer integrals.
11. Why are diffraction minima given by [a\sin\theta = m\lambda] and not by integer multiples of [2\lambda]?
Because minima arise when the aperture integrates to zero due to destructive phasing; the geometry yields the condition above (derived from zeros of [\sin\beta\].
12. Can a rectangular aperture have a similar pattern?
Yes; the diffraction pattern is separable: a rectangular aperture gives a product of sinc-squared patterns in orthogonal directions.
13. How does wavelength affect resolution via diffraction?
Smaller [\lambda] produces narrower central maxima, improving resolving power; hence blue/violet light gives better resolution than red.
14. Do we get interference and diffraction together?
Often yes — e.g., Young’s double-slit pattern is interference between slits modulated by single-slit diffraction envelope from each slit.
15. Why is intensity finite at [\beta\to0] even though formula has [\beta] in denominator?
Because [\sin\beta \approx \beta] for small [\beta], so [\lim_{\beta\to0} (\sin\beta/\beta)^2 = 1], rendering intensity finite and maximal at center.
7. FAQ / Common Misconceptions
1. “Diffraction only happens for light with special properties.”
No — diffraction is universal for all waves whenever the obstacle/aperture size is comparable to wavelength.
2. “Diffraction contradicts ray optics.”
No — ray optics is an approximation valid when wavelengths are much smaller than object sizes. Diffraction extends wave behaviour beyond the ray approximation.
3. “No diffraction when slit wider than wavelength.”
There is still diffraction, but the angular spread is small and often negligible.
4. “All minima are equally spaced in angle.”
No — minima satisfy [a\sin\theta=m\lambda]; in small-angle approx. they appear equally spaced in position on a distant screen, but angular spacing varies for large angles.
5. “Diffraction produces only dark fringes.”
It produces a pattern with both maxima and minima; the central maximum is bright and wide.
6. “The [\dfrac{\sin\beta}{\beta}] pattern is exact for any distance.”
This is the Fraunhofer (far-field) result. In near-field (Fresnel) region the pattern differs.
7. “Slit width affects only intensity, not angular positions.”
False. Slit width governs angular minima positions via [a\sin\theta=m\lambda].
8. “Diffraction is negligible in microscopes.”
Not true — diffraction limits the resolving power of microscopes (Rayleigh criterion).
9. “Diffraction requires coherent light.”
Diffraction from a single aperture can be observed with incoherent light, but patterns are clearer with coherent/monochromatic sources.
10. “Diffraction violates energy conservation because light spreads.”
No. Energy is conserved; spreading distributes intensity over larger area, lowering local intensity.
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