CLASS 12 PHYSICS · CHAPTER 10 · STUDY NOTES
Wave Optics
An exploration of the wave nature of light, covering Huygens' principle, interference, diffraction, and polarisation, which explain phenomena beyond the scope of ray optics.
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Introduction to Wave Theory
Use wave optics when ray directions alone cannot explain the observed brightness or field direction.One experiment can distinguish two predictions
Read the assumptions and limits
The electromagnetic description needs no material ether. Ray and wave descriptions are useful in different limits; photon behaviour is introduced separately. Wavefronts show equal phase, not necessarily equal amplitude.
Original teaching models · NCERT Physics Part II, chapter 10, PDF pages 1–2 · no recorded completion
- Wave optics
- A description that uses wave phase and superposition to explain effects such as interference, diffraction and polarisation.
Refraction gives a test of competing models. To explain bending towards the normal, the old corpuscular model predicted faster light in water. Huygens’ wave model predicted slower light. Measurements supported the wave prediction.
Light is an electromagnetic wave: its electric and magnetic fields vary across space and time. It needs no material ether. Ray optics remains a useful approximation when the relevant dimensions are large compared with wavelength.
Let’s look a little closer.
Go deeper: Keep the models in their useful limits
An interference pattern depends on relative phase, which a single geometric ray does not encode. Diffraction describes spreading from a finite aperture; polarisation concerns the direction of a transverse field.
The electromagnetic wave model does not by itself describe every exchange of energy with matter. Photon behaviour is introduced in the next chapter. A useful model is chosen for the observation being explained.
Take a moment. What do you think?
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Question
Light of vacuum wavelength 450 nm enters glass with n = 1.50. Find its phase speed and wavelength there. Does its frequency change?
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Huygens' Principle and Wavefronts
Build the next front from an envelope, not by joining wavelet centres.The new front is the envelope of the wavelets
Read the assumptions and limits
Wavelets are a mathematical construction. Their forward envelope is selected; a complete wave treatment explains why this does not predict an unwanted freely propagating backward front. Circles show two-dimensional sections of spherical wavelets in an isotropic medium.
Original teaching models · NCERT Physics Part II, chapter 10, PDF pages 3–4 · no recorded completion
- Wavefront
- A surface joining points at the same phase of a wave. In an isotropic medium, rays are normal to it.
Each point on an existing wavefront acts as a source of secondary wavelets. After time Δt, their radius is vΔt. The forward common tangent surface is the new wavefront in Huygens’ construction.
A point source gives spherical fronts. Far from the source, a small portion is approximately plane. A plane wavefront in a uniform medium advances without changing its plane shape.
Go deeper: What the construction represents
The circles in the visual are two-dimensional sections of spherical wavelets. Their radii encode travel distance in the same time interval. A wavefront records equal phase, not necessarily equal amplitude at every point.
Selecting the forward envelope is part of this elementary construction. A complete wave treatment explains why it does not predict an unwanted freely propagating backward front. The construction is not a picture of material pellets emitted into space.
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Question
A wave travels at 2.00 × 10⁸ m/s in a medium. What is the secondary-wavelet radius after 2.00 femtoseconds?
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Sign inNew here? Sign up freeNCERT reference: chapter PDF pages 3, 4.
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Refraction and Reflection using Huygens' Principle
Give both wavelets the same travel time, then compare their radii.Unequal wavelet radii tilt the new front
Read the assumptions and limits
The electromagnetic description needs no material ether. Ray and wave descriptions are useful in different limits; photon behaviour is introduced separately. Wavefronts show equal phase, not necessarily equal amplitude.
Original teaching models · NCERT Physics Part II, chapter 10, PDF pages 4–7 · no recorded completion
- Wavefront refraction
- For a stationary boundary, sin i / sin r = v₁/v₂ = n₂/n₁; frequency is unchanged.
| Medium Change | Speed (v) | Wavelength (λ) | Frequency (ν) |
|---|---|---|---|
| Rarer to Denser | Decreases | Decreases | Remains Constant |
| Denser to Rarer | Increases | Increases | Remains Constant |
One end of an incident plane front reaches the interface first. While the other end travels BC = v₁Δt, a wavelet in the second medium travels AE = v₂Δt. Drawing the tangent from C gives the refracted front.
If v₂ < v₁, the refracted ray bends towards the normal; if v₂ > v₁ and transmission is possible, it bends away. For reflection the speeds are equal, giving i = r. All angles use the normal.
Go deeper: Frequency, phase and a lens
The source fixes the temporal frequency across a stationary interface. The wavelength changes with speed: λ₂/λ₁ = v₂/v₁. Reflection and transmission can divide the intensity, so unchanged frequency does not imply unchanged intensity.
A converging lens delays the central portion of an incident plane front more than its edges. The emerging front is curved towards a focus. A finite real lens still has diffraction and may have aberrations; the sketch uses ideal paraxial optics.
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Question
A plane wave enters a medium of index 1.60 from air at 30°. Find r and the ratio v₂/v₁.
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Sign inNew here? Sign up freeNCERT reference: chapter PDF pages 4, 5, 6, 7.
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Superposition and Coherence
Add electric-field components before calculating brightness.Add fields first; square the resulting amplitude
Read the assumptions and limits
These are scalar components of same-frequency, same-polarisation fields in a linear medium. Coherence requires a stable relative phase over the observation time. Intensity is proportional to squared amplitude in the same medium. The phase-average illustration samples uniformly, not by assuming instantaneous fields stop interfering.
Original teaching models · NCERT Physics Part II, chapter 10, PDF pages 8–11 · no recorded completion
- Coherent sources
- Sources with a stable relative phase over the observation time; for a stationary interference pattern they have the same frequency.
In a linear medium the instantaneous fields add. For same-frequency, matched-polarisation waves, I = I₁ + I₂ + 2√(I₁I₂) cos φ. Equal amplitudes reinforce at φ = 0 and cancel at φ = π.
Different amplitudes do not cancel completely: I_min = (√I₁−√I₂)² and I_max = (√I₁+√I₂)². Independent light sources usually have rapidly changing relative phase, making their interference term average to zero during observation.
Go deeper: Where the bright energy comes from
Interference redistributes intensity across a pattern. Dark regions do not destroy energy: brighter regions account for the redistribution when the complete pattern and boundaries are considered.
Coherence is about phase stability, not simply whether two lamps emit the same nominal colour. The visual uses scalar components of fields with matched polarisation and the same frequency. Orthogonal polarisations require a vector treatment and do not give this scalar interference term.
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Question
Two matched waves have field-amplitude ratio A₂/A₁ = 2. Find I/I₁ at phase differences 0°, 90° and 180°.
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Sign inNew here? Sign up freeNCERT reference: chapter PDF pages 8, 9, 10, 11, 14.
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Young's Double Slit Experiment
Compare the two paths to one screen point.Two paths create a repeating intensity pattern
Read the assumptions and limits
Fraunhofer and small-angle approximations: screen distance is large compared with slit dimensions and y ≪ D. The diffraction curve is (sin α/α)² with α = πay/(λD), not a sine-wave brightness curve. Double-slit modes assume equal coherent amplitudes and matched polarisation; the finite-width envelope uses a = 0.10 mm. Plot brightness is normalised, so narrowing a slit here does not imply unchanged transmitted power. Colours are a display convention.
Original teaching models · NCERT Physics Part II, chapter 10, PDF pages 11–12 · no recorded completion
- Fringe spacing
- In the small-angle, distant-screen model, β = λD/d, with wavelength λ, screen distance D and slit separation d.
A common source illuminates two narrow slits, providing coherent contributions. At screen coordinate y, path difference Δ ≈ dy/D. Bright fringes occur when Δ = mλ; dark fringes for equal amplitudes occur when Δ = (m+½)λ.
The screen pattern repeats with spacing β = λD/d. Increasing wavelength or screen distance widens the spacing; increasing slit separation narrows it. The centre at y = 0 is a bright fringe when the slit contributions start in phase.
Go deeper: Geometry and a finite slit width
The approximation uses y ≪ D and slit dimensions much smaller than D. A diagram of two slits and a nearby screen is not to scale; use the stated distances in the calculation.
Actual finite-width slits add a diffraction envelope. The visual’s envelope mode uses width a = 0.10 mm; this controls the broad envelope while separation d controls the fine fringe spacing. A missing interference order can coincide with a diffraction minimum.
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Question
A 500 nm source illuminates slits separated by 0.50 mm, with a screen 1.60 m away. Find fringe spacing and the position of the third bright fringe from the centre.
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Diffraction of Light
Compare contributions from across one opening.A narrower slit spreads the central maximum
Read the assumptions and limits
Fraunhofer and small-angle approximations: screen distance is large compared with slit dimensions and y ≪ D. The diffraction curve is (sin α/α)² with α = πay/(λD), not a sine-wave brightness curve. Double-slit modes assume equal coherent amplitudes and matched polarisation; the finite-width envelope uses a = 0.10 mm. Plot brightness is normalised, so narrowing a slit here does not imply unchanged transmitted power. Colours are a display convention.
Original teaching models · NCERT Physics Part II, chapter 10, PDF pages 12–14 · no recorded completion
- Single-slit minimum
- For a slit of width a, far-field minima satisfy a sin θ = mλ, where m = ±1, ±2, …; m = 0 is the central maximum.
| Feature | Interference (Double Slit) | Diffraction (Single Slit) |
|---|---|---|
| Source | Superposition of two wavefronts | Superposition from parts of same wavefront |
| Fringe Width | All fringes are of equal width | Central maximum is twice as wide |
| Intensity | All bright fringes have equal intensity | Intensity decreases for higher orders |
At the first minimum, pair each point in the upper half with a point a/2 below it. When a sin θ = λ, each pair has path difference λ/2 and cancels. At θ = 0 the contributions are in phase, giving the central maximum.
For a distant screen and small angles, the first minima are at y ≈ ±λD/a. The central width is approximately 2λD/a. A narrower slit spreads the pattern more widely; secondary maxima are weaker than the central peak.
Go deeper: What the intensity graph adds
The far-field normalised intensity is (sin α/α)², with α = πa sin θ/λ ≈ πay/(λD). Its value tends to one at α = 0. The secondary maxima are not exactly halfway between successive minima; the elementary midpoint estimate is approximate.
Finite apertures also limit how closely an optical instrument can distinguish two directions. Increasing magnification alone cannot overcome diffraction. The plotted intensity is normalised: widening the pattern by narrowing the slit does not imply constant transmitted power.
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Question
A 0.25 mm slit is illuminated by 500 nm light, with a screen 1.50 m away. Find the first-minimum distance from the centre and the central width. What if a is halved?
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Polarisation of Light
Project the field direction; square the projection for intensity.Project the field onto the analyser’s axis
Read the assumptions and limits
The circles show pass-axis directions as seen looking along propagation, not sequential positions in a single transverse plane. Ideal polarisers are assumed. Unpolarised input loses half its intensity at the first one; Malus’ law applies to the already polarised beam. The transmission axis is perpendicular to the aligned absorbing molecular direction in the usual polaroid model. Longitudinal sound waves do not have this transverse polarisation.
Original teaching models · NCERT Physics Part II, chapter 10, PDF pages 15–17 · no recorded completion
- Malus’ law
- For linearly polarised light incident on an ideal analyser, I = I_p cos²θ, where θ is the angle between the incoming electric field and its transmission axis.
Light is transverse. In linearly polarised light, the electric field oscillates along one fixed transverse direction. An ideal polariser transmits the field component along its pass-axis.
Unpolarised light has no fixed transverse direction; an ideal first polariser transmits half its time-averaged intensity. An analyser then follows Malus’ law. Crossed ideal axes block the already polarised beam, while inserting a tilted middle polariser can allow some light through.
Go deeper: Axis direction and common confusions
The usual polaroid material absorbs electric fields along the aligned molecular direction; its transmission axis is perpendicular to that direction. Do not confuse the absorbing molecular alignment with the pass-axis.
Malus’ law uses the intensity of the polarised beam arriving at the analyser, not the unpolarised intensity before the first polariser. With three polarisers, apply one projection at each successive axis. Longitudinal sound does not have this transverse polarisation.
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Question
Unpolarised light of intensity 80 W/m² passes through an ideal polariser and an analyser at 60°. Find the output. If the last axis is crossed and a middle axis at 30° is inserted, what is the output then?
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Sign inNew here? Sign up freeNCERT reference: chapter PDF pages 15, 16, 17.