Chapter 10: Wave Optics

Ray optics treats light as straight-line rays; wave optics treats light as a transverse electromagnetic wave with wavelength $\lambda$. Phenomena like interference, diffraction, and polarisation cannot be explained by rays alone — they are signatures of wave behaviour and unify with classical electrodynamics.

Concept Map

• Huygens' principle $\rightarrow$ secondary wavelets $\rightarrow$ proofs of reflection and refraction $\rightarrow$ Doppler effect.
• Superposition of coherent waves $\rightarrow$ interference $\rightarrow$ Young's experiment (fringe width $\beta = \lambda D/d$) $\rightarrow$ intensity distribution.
• Diffraction at a single slit $\rightarrow$ central maxima $\rightarrow$ angular widths $\rightarrow$ comparison with interference.
• Resolving power of microscope, telescope $\rightarrow$ Rayleigh's criterion $\rightarrow$ $\theta_{\min} = 1.22\lambda/D$.
• Polarisation $\rightarrow$ Malus's law $\rightarrow$ Brewster's law $\rightarrow$ applications.

10.1 Huygens' Principle

Definition

Each point on a wavefront acts as a source of secondary spherical wavelets spreading out in the forward direction with the same speed as the wave itself. The new wavefront at a later instant is the envelope of these wavelets.

A wavefront is a surface over which the phase of the wave is constant. From a point source we get spherical wavefronts; from a far source they are effectively plane.

Construction

Given a wavefront $W_1$ at time $t$, draw spheres of radius $vt'$ centred at every point of $W_1$ (where $v$ is wave speed and $t'$ a later time). The forward envelope of these spheres gives $W_2$, the wavefront at $t + t'$.

Note on backward wavelets

The original Huygens' construction did not explain why there is no backward wave. Fresnel and Kirchhoff later showed that the obliquity factor $(1 + \cos\theta)/2$ kills the backward propagation.

Worked Example

A point source emits in vacuum. The wavefronts are spherical with radius $ct$. At very large distance they are effectively plane wavefronts perpendicular to the direction of propagation.

Pitfalls

• Secondary wavelets are physical only as a calculational device; they are not extra sources of light.
• Huygens' construction gives shape but not amplitude variation along the wavefront — for that we need Fresnel/Kirchhoff diffraction theory.

10.2 Proof of Laws of Reflection and Refraction (Huygens')

Reflection

A plane wavefront $AB$ travelling with speed $v$ in medium 1 strikes a reflecting surface at incidence angle $i$. Point $A$ hits the surface first; point $B$ takes time $t = BC/v$ to reach the surface at $C$. In the same time a secondary wavelet from $A$ expands to a hemisphere of radius $AE = vt = BC$.

The new wavefront is the tangent from $C$ to that hemisphere, $CE$. In $\triangle ABC$ and $\triangle CEA$:

• $AB = CE$ (both perpendicular distances $= vt$).
• $BC = AE = vt$.
• $\angle ABC = \angle CEA = 90^\circ$.

Hence the triangles are congruent, so $\angle BAC = \angle ECA$, i.e. angle of incidence $i$ = angle of reflection $r$. Also, incident ray, reflected ray, and normal lie in the same plane (the plane of $AB$, $CE$ and the surface normal).

Refraction

A plane wavefront $AB$ in medium 1 (speed $v_1$) strikes a refracting interface at angle $i$. While $B$ travels to $C$ (distance $v_1 t$), a wavelet from $A$ has expanded into medium 2 to radius $AE = v_2 t$. The refracted wavefront is the tangent $CE$ from $C$.

In $\triangle ABC$: $\sin i = BC/AC = v_1 t / AC$. In $\triangle AEC$: $\sin r = AE/AC = v_2 t / AC$.

Dividing:

$\frac{\sin i}{\sin r} = \frac{v_1}{v_2} = \frac{n_2}{n_1} \quad\Rightarrow\quad n_1 \sin i = n_2 \sin r.$

Pitfalls

• The ratio $v_1/v_2$ equals $n_2/n_1$, not $n_1/n_2$ — denser medium has slower wave.
• Frequency stays the same on refraction; wavelength becomes $\lambda_2 = \lambda_1 v_2/v_1 = \lambda_1 n_1/n_2$.

10.3 Doppler Effect for Light (Qualitative)

For light, only the relative velocity of source and observer matters (no preferred medium). For non-relativistic radial velocity $v$ (positive when receding),

$\frac{\Delta \lambda}{\lambda} \approx \frac{v}{c} \quad\text{(redshift)},$

with a blueshift ($\Delta\lambda < 0$) when the source approaches. A more accurate relation:

$\frac{\nu_{\text{obs}}}{\nu_{\text{src}}} = \sqrt{\frac{1 - v/c}{1 + v/c}}\quad\text{(special-relativistic).}$

Applications

• Spectral lines from distant galaxies are redshifted, supporting an expanding universe (Hubble).
• Doppler radar; police speed guns.

Pitfalls

• "Redshift" is a shift of $\lambda$ to longer values (lower frequency). Cosmological redshift is not due to motion through space but due to the expansion of space itself.

10.4 Coherent Sources and Superposition

Definition

Two sources are coherent if they emit waves of the same frequency and maintain a constant phase difference. Ordinary light sources (sodium lamps, bulbs) emit randomly out of phase, so two such bulbs never produce stable interference.

Two coherent sources are usually obtained by dividing a single wavefront (Young's double slit, Fresnel biprism) or by amplitude division (thin films).

Superposition

If $y_1 = a \cos(\omega t)$ and $y_2 = a \cos(\omega t + \phi)$ overlap, the resultant is $y = 2a \cos(\phi/2)\cos(\omega t + \phi/2)$. Intensity ($\propto$ amplitude$^2$):

$I = 4 I_0 \cos^2\!\left(\frac{\phi}{2}\right).$

With unequal amplitudes $a_1, a_2$:

$I_{\text{res}} = I_1 + I_2 + 2\sqrt{I_1 I_2}\cos\phi.$

Conditions

Constructive (bright fringe): $\phi = 2n\pi$, path difference $\Delta = n\lambda$, $n = 0, \pm 1, \pm 2, \ldots$ Destructive (dark fringe): $\phi = (2n+1)\pi$, path difference $\Delta = (n + 1/2)\lambda$.

Worked Example

Two coherent sources of intensities $I$ and $4I$. Find $I_{\max}/I_{\min}$.

$I_{\max} = (\sqrt I + \sqrt{4I})^2 = 9I$. $I_{\min} = (\sqrt I - \sqrt{4I})^2 = I$. Ratio $= 9$.

Pitfalls

• Coherent does not mean equal amplitude; only same frequency and constant phase.
• The amplitudes add (with relative phase), then we square — never add the intensities directly.

10.5 Young's Double-Slit Experiment

Setup

A monochromatic light source illuminates a single slit $S$, then a double slit $S_1 S_2$ (separation $d$) at distance $D$ from a screen. Each slit acts as a coherent source. Path difference at a point $P$ on the screen, at distance $y$ from the central axis, is

$\Delta = S_2 P - S_1 P \approx \frac{y d}{D} \quad (D \gg d, y).$

Derivation of fringe width $\beta$

Bright fringes: $\Delta = n\lambda \Rightarrow y_n = n\lambda D/d$. Dark fringes: $\Delta = (n + 1/2)\lambda \Rightarrow y_n = (n + 1/2)\lambda D/d$.

Distance between consecutive bright (or dark) fringes:

$\boxed{\;\beta = \frac{\lambda D}{d}\;}.$

Angular fringe width $\beta/D = \lambda/d$.

Intensity distribution

At point $P$, phase difference $\phi = 2\pi \Delta/\lambda = 2\pi y d/(\lambda D)$. Intensity (equal-amplitude sources):

$I(y) = 4 I_0 \cos^2\!\left(\frac{\pi y d}{\lambda D}\right).$

The pattern is a cosine-squared with peaks of $4I_0$ at $y_n = n\lambda D/d$ and zeros midway.

Conditions for sharp fringes

• Sources must be coherent (derived from same parent wavefront).
• $D \gg d$ (small-angle approximation).
• Source slit $S$ must be narrow (else fringes wash out by superposition of patterns).
• Slit separation $d$ small (for $\beta$ to be observable).
• Monochromatic source (else different colours produce fringes of different $\beta$).

Effect of immersion in a medium of index $n$

$\lambda \to \lambda/n$, so $\beta \to \beta/n$.

Worked Example

In a YDSE, $\lambda = 500\,\text{nm}$, $d = 0.5\,\text{mm}$, $D = 1\,\text{m}$. Find $\beta$.

$\beta = (500\times 10^{-9})(1)/(0.5\times 10^{-3}) = 10^{-3}\,\text{m} = 1.0\,\text{mm}$.

Worked Example — White light

In a YDSE with white light, the central fringe is white (all wavelengths give $\Delta = 0$). The first-order fringes farther from centre are coloured (different $\beta$ per $\lambda$) and at large orders they overlap, washing out.

Pitfalls

• $\beta$ depends on $\lambda$, $D$, and $d$. Don't forget to convert units consistently.
• Adding a thin slab of thickness $t$ and index $n$ in front of one slit shifts the entire pattern by $\Delta x = (n - 1)tD/d$ towards that slit.
• Fringe pattern requires finite source size to be small but not zero; perfectly point-like sources are an idealisation.

10.6 Diffraction at a Single Slit

Definition

When a plane wave illuminates a slit of width $a$, the wavefront within the slit is partitioned into many Huygens secondary sources. Their interference on a distant screen yields a diffraction pattern with a broad central maximum and weaker secondary maxima.

Position of minima

At angle $\theta$, divide the slit into pairs of points $a/2$ apart. Their path difference is $(a/2)\sin\theta$. They cancel if this equals $\lambda/2$, i.e. $a \sin\theta = \lambda$. More generally, minima are at

$a \sin\theta_n = n\lambda,\quad n = \pm 1, \pm 2, \ldots$

Central maximum width

The first minimum is at $\sin\theta_1 \approx \lambda/a$. So angular width of the central maximum:

$\Delta\theta = 2\theta_1 = \frac{2\lambda}{a}.$

Linear width on a screen distance $D$ away: $W = 2\lambda D/a$.

Secondary maxima

Approximately at $a \sin\theta = (n + 1/2)\lambda$, with intensity dropping rapidly: relative intensities $\sim 1, 1/22, 1/61, \ldots$ of the central peak.

Intensity distribution

$I(\theta) = I_0 \left(\frac{\sin\beta}{\beta}\right)^2,\quad \beta = \frac{\pi a \sin\theta}{\lambda}.$

Comparison: Interference vs Diffraction

Worked Example

A slit of width $a = 0.2\,\text{mm}$, $\lambda = 600\,\text{nm}$, $D = 1\,\text{m}$. Find central-maximum width.

$W = 2\lambda D/a = 2(600\times 10^{-9})(1)/(0.2\times 10^{-3}) = 6.0\,\text{mm}$.

Pitfalls

• "Minima at $a \sin\theta = n\lambda$" and "maxima at $a \sin\theta = (n+1/2)\lambda$" — opposite to YDSE! Easy to confuse.
• $n = 0$ is not a minimum in single-slit diffraction; it is the central maximum.
• The central peak is twice as wide as the side peaks.

10.7 Resolving Power — Rayleigh's Criterion

Statement

Two point sources are just resolved when the central maximum of one coincides with the first minimum of the other.

Telescope

For a circular aperture of diameter $D$, the angular limit of resolution:

$\theta_{\min} = 1.22\frac{\lambda}{D}.$

Resolving power $= 1/\theta_{\min} = D/(1.22\lambda)$.

Larger aperture and shorter wavelength improve resolution.

Microscope

The smallest separation resolvable in the object plane:

$d_{\min} = \frac{1.22 \lambda}{2 n \sin\theta} = \frac{0.61 \lambda}{NA},$

where $NA = n \sin\theta$ is the numerical aperture. Resolving power $= 1/d_{\min}$. To improve, use oil immersion (raise $n$) and shorter $\lambda$ (UV, electron microscopes).

Worked Example

Resolving power of a $20\,\text{cm}$ aperture telescope at $\lambda = 500\,\text{nm}$.

$\theta_{\min} = 1.22(500\times 10^{-9})/(0.2) = 3.05\times 10^{-6}\,\text{rad}$, i.e. $\sim 0.6''$.

Pitfalls

• The $1.22$ factor comes from the first zero of the Bessel function — specific to circular apertures (a rectangular aperture has factor $1$).
• For a microscope, $d_{\min}$ refers to lateral resolution in the object plane.

10.8 Polarisation

Definition

In a transverse EM wave, the electric field $\vec E$ oscillates perpendicular to the propagation direction $\hat k$. Polarised light has $\vec E$ confined to one plane; unpolarised light has $\vec E$ rapidly fluctuating in all transverse directions.

Polarisation is a property unique to transverse waves; it does not exist for longitudinal waves like sound. The fact that light can be polarised proves that EM waves are transverse.

Polaroids

A polaroid sheet transmits the component of $\vec E$ along its transmission axis and absorbs the perpendicular component. Light passing through an ideal polaroid drops to half of the original intensity if input is unpolarised:

$I_{\text{out}} = \frac{I_0}{2}.$

Malus's law

When polarised light of intensity $I_0$ passes through a polaroid whose axis makes angle $\theta$ with $\vec E$,

$\boxed{\;I = I_0 \cos^2\theta\;}.$

Derivation: only the component $E\cos\theta$ passes through; intensity $\propto E^2$, so $I = I_0\cos^2\theta$.

Brewster's law

When light reflects from a dielectric at the Brewster angle $i_B$, the reflected ray is completely polarised in the plane parallel to the surface (i.e. perpendicular to the plane of incidence). Refracted ray is partially polarised.

Geometric condition: reflected and refracted rays are perpendicular ($i_B + r = 90^\circ$). With Snell, $n = \sin i_B/\sin r = \sin i_B/\cos i_B$, giving

$\boxed{\;\tan i_B = n\;}.$

Worked Example — Malus

Unpolarised light $I_0$ hits two polarisers with axes at $30^\circ$. Find transmitted intensity.

After first: $I_0/2$. After second: $(I_0/2)\cos^2 30^\circ = (I_0/2)(3/4) = 3I_0/8$.

Worked Example — Brewster

Brewster angle for glass ($n = 1.5$): $i_B = \arctan(1.5) = 56.3^\circ$.

Applications

• Polaroid sunglasses cut glare reflected from horizontal surfaces (which is partially polarised horizontally).
• Photoelasticity: stress patterns in transparent solids become visible between crossed polaroids.
• LCD displays: rely on polarising sheets and birefringent liquid crystals.
• Optical activity: sugar solutions rotate the plane of polarisation; used in sugar industry.
• Three-D cinema: orthogonal polarisations for left/right eye.

Pitfalls

• $I = I_0 \cos^2\theta$ applies only to polarised input. For unpolarised input through a single polaroid you get $I_0/2$.
• Brewster's law $\tan i_B = n$ is not the same as the critical angle $\sin i_c = 1/n$.
• Sunglasses use vertical transmission axes because reflected glare is horizontally polarised.

Solved Problems

1. In YDSE, $d = 1\,\text{mm}$, $D = 2\,\text{m}$, $\lambda = 600\,\text{nm}$. Find fringe width and position of the 3rd bright fringe.

$\beta = (600\times 10^{-9})(2)/(10^{-3}) = 1.2\,\text{mm}$. 3rd bright $= 3\beta = 3.6\,\text{mm}$ from centre.

2. A YDSE is immersed in water ($n = 4/3$). By what factor does $\beta$ change?

$\lambda$ in water is $\lambda/n$, so $\beta \to \beta/n = (3/4)\beta$.

3. A glass slab ($n = 1.5$, $t = 5\,\mu\text{m}$) is placed in front of one of the slits in a YDSE with $d = 0.2\,\text{mm}$, $D = 1\,\text{m}$. Find the fringe shift.

$\Delta x = (n-1)tD/d = 0.5 \cdot 5\times 10^{-6}\cdot 1 / 2\times 10^{-4} = 1.25\times 10^{-2}\,\text{m} = 1.25\,\text{cm}$ towards the slit covered.

4. Single slit of $a = 0.1\,\text{mm}$, $D = 50\,\text{cm}$, $\lambda = 500\,\text{nm}$. Find the width of the central maximum.

$W = 2\lambda D/a = 2(500\times 10^{-9})(0.5)/(10^{-4}) = 5\times 10^{-3}\,\text{m} = 5\,\text{mm}$.

5. Two polaroids cross at $90^\circ$. A third is inserted between them at $45^\circ$. Find transmitted intensity (unpolarised input $I_0$).

After 1st: $I_0/2$. After 2nd: $(I_0/2)\cos^2 45^\circ = I_0/4$. After 3rd: $(I_0/4)\cos^2 45^\circ = I_0/8$.

6. Resolving power of a telescope of aperture $5\,\text{cm}$ at $\lambda = 550\,\text{nm}$.

$\theta_{\min} = 1.22(550\times 10^{-9})/(0.05) = 1.34\times 10^{-5}\,\text{rad}$.

7. Brewster angle at an air-water interface ($n = 1.33$).

$i_B = \arctan(1.33) = 53.1^\circ$.

JEE/NEET Edge Cases

• The angular fringe width $\beta/D = \lambda/d$ is invariant under uniform scaling of $D$; useful when the screen distance is changed.
• A YDSE in white light produces a central white fringe surrounded by coloured fringes; violet is closest to the centre on each side, red farthest (because $\beta \propto \lambda$).
• Putting one slit in a slab shifts the whole pattern; fringe width does not change.
• Single-slit diffraction overlaid on YDSE: the YDSE pattern is modulated by the single-slit envelope; missing orders occur where YDSE maximum coincides with single-slit minimum, i.e. $d \sin\theta = m\lambda$ and $a \sin\theta = n\lambda$, giving $m/n = d/a$.
• Brewster angle changes with wavelength because $n$ does — strong polarisation occurs only for monochromatic light.
• The condition "$\Delta = (n + 1/2)\lambda$" gives destructive interference only in YDSE; in single-slit diffraction the minima formula is $a\sin\theta = n\lambda$ (a different counting).

Quick Recap

• Huygens: every wavefront point emits a secondary wavelet; envelope gives the next wavefront.
• $\sin i/\sin r = v_1/v_2 = n_2/n_1$.
• Coherent: same $\nu$, constant phase. Superposition: $I = I_1 + I_2 + 2\sqrt{I_1 I_2}\cos\phi$.
• YDSE fringe width $\beta = \lambda D/d$; intensity $4I_0 \cos^2(\pi y d/(\lambda D))$.
• Single-slit: minima at $a\sin\theta = n\lambda$; central width $2\lambda D/a$.
• Resolving power $\theta_{\min} = 1.22\lambda/D$.
• Malus: $I = I_0\cos^2\theta$. Brewster: $\tan i_B = n$.
• Doppler: $\Delta\lambda/\lambda \approx v/c$.

Formula Sheet