Wave Optics — NEET notes

Huygens' Principle

Every point on a wavefront acts as a source of secondary wavelets. The envelope of these wavelets gives the new wavefront. This elegant geometric idea explains reflection, refraction, and diffraction — all without knowing what light actually is.

i r = v_1 v_2 = n_ 21

Wavefront

A wavefront is the locus of all points in a medium that are in the same phase of oscillation. Its shape depends on the source geometry and distance from the source.

A wavefront is the locus of all points that have the same phase at a given instant. The direction of wave propagation is always perpendicular to the wavefront — these perpendicular lines are called rays.

• Spherical wavefront — from a point source; surface is a sphere expanding outward
• Cylindrical wavefront — from a linear source (like a slit illuminated by a distant source); surface is a cylinder
• Plane wavefront — at very large distances from any source; the sphere's curvature becomes negligible
• Rays are always to wavefronts — they show the direction of energy flow
• Intensity 1/r^2 for spherical wavefronts (energy spreads over area 4 r^2 )

Huygens' Construction

Huygens proposed that each point on a wavefront emits secondary spherical wavelets. The common tangent (envelope) of these wavelets at a later time gives the new wavefront position.

Huygens' Principle (1678) states: (1) Every point on a wavefront serves as a source of secondary spherical wavelets that spread out at the speed of light in that medium. (2) The new wavefront at any later time is the forward envelope (common tangent) of all these wavelets.

1. Start with a known wavefront at time t
2. Treat every point on this wavefront as a new point source of secondary wavelets
3. Each wavelet expands as a sphere of radius v t in time t
4. Draw the forward envelope (common tangent) of all these wavelets
5. This envelope is the new wavefront at time t + t

• Only the forward envelope is taken — the backward wavefront is rejected (Huygens assumed it has zero amplitude)
• The construction works for any wavefront shape: plane, spherical, or cylindrical
• Speed of secondary wavelets = speed of light in that medium ( v = c/n )
• This principle does NOT require knowledge of whether light is a particle or wave — it is purely geometric

Huygens proposed this in 1678, long before Maxwell's equations (1865). It was a geometric tool that gave correct predictions for reflection, refraction, and diffraction — even without a complete wave theory.

Reflection via Huygens' Principle

Huygens' construction applied to a plane wavefront hitting a reflecting surface proves the law of reflection: angle of incidence equals angle of reflection.

Consider a plane wavefront hitting a flat mirror at angle i . When point B of the wavefront reaches the mirror, point A has already hit and started emitting secondary wavelets. By the time B reaches B', A's wavelet has expanded to radius v t . The new reflected wavefront is the tangent from B' to A's wavelet.

1. Plane wavefront AB approaches mirror at angle i
2. Point A hits the mirror first and starts emitting a secondary wavelet in the reflected medium (same medium, same speed v )
3. By the time point B reaches the mirror at B', the wavelet from A has radius v t = BB' (since the speed is the same)
4. Draw the tangent from B' to A's wavelet — this is the reflected wavefront A'B'
5. From the geometry: i = r (triangles ABB' and A'B'B are congruent)

i = r (Law of Reflection)

• The reflected wavefront is plane (if the incident wavefront is plane and the mirror is flat)
• Incident ray, reflected ray, and normal all lie in the same plane (follows from the 2D construction)
• This proof works because the speed of light is the same before and after reflection (same medium)

Refraction via Huygens' Principle

When a wavefront crosses into a medium with different speed, wavelets expand at different rates, causing the wavefront to tilt — that tilt is refraction, and it directly gives Snell's law.

A plane wavefront travels from medium 1 (speed v_1 ) to medium 2 (speed v_2 ) at angle i . Point A hits the boundary first and emits wavelets at speed v_2 . Point B is still in medium 1, traveling at v_1 . The different speeds cause the refracted wavefront to tilt.

1. Wavefront AB in medium 1 approaches boundary at angle i
2. Point A reaches boundary first, emits wavelet at speed v_2 in medium 2
3. Point B still travels in medium 1 at speed v_1 and covers distance BB' = v_1 t
4. Meanwhile, A's wavelet in medium 2 has radius AA' = v_2 t
5. The refracted wavefront is the tangent from B' to A's wavelet
6. From the geometry: i / r = v_1 / v_2 = n_2 / n_1

i r = v_1 v_2 = n_2 n_1 = n_ 21 (Snell's Law)

• Light bends toward normal when entering a denser medium ( v_2 < v_1 , so r < i )
• Light bends away from normal when entering a rarer medium ( v_2 > v_1 , so r > i )
• Frequency does NOT change during refraction — only speed and wavelength change
• _2 / _1 = v_2 / v_1 — wavelength is shorter in the denser medium
• Refractive index n = c/v — higher n means slower light

Quick recall

In Huygens' derivation of Snell's law, if v₂ < v₁, what happens to the wavelength in medium 2?

Wavelength decreases. Since = v/f and frequency stays constant, _2 = v_2/f < v_1/f = _1 . Light has shorter wavelength in a denser medium.

Interference of Light

When two coherent light waves overlap, they add up (constructive) or cancel out (destructive) depending on their path difference. This creates bright and dark bands — visible proof that light is a wave, not a particle.

I = I_1 + I_2 + 2 I_1 I_2

Coherent Sources

Two sources are coherent if they emit waves with a constant phase difference over time. Without coherence, bright and dark fringes would flicker and average out — no stable pattern.

For a stable interference pattern, the two sources must be coherent: they must have the same frequency and a constant phase difference. Two independent light bulbs are NOT coherent — each atom emits light independently with random phases.

• Coherent sources: same frequency + constant phase difference
• Two independent sources (two bulbs, two candles) are NEVER coherent — random atomic emissions
• Coherent sources are obtained by splitting one source into two: Young's double slit, Lloyd's mirror, Fresnel biprism
• Laser light is coherent — all photons have the same frequency and phase (stimulated emission)
• Incoherent sources produce uniform average illumination — no fringe pattern visible

Path Difference & Phase Difference

Path difference is the extra distance one wave travels compared to the other. Phase difference is the corresponding angular mismatch. They are directly proportional — converting between them is essential.

When two waves reach a point P, one may have traveled a longer path than the other. This extra distance is the path difference ( x ). The corresponding phase difference ( ) tells you whether the waves add up or cancel.

= 2 x

A path difference of one full wavelength ( ) corresponds to a phase difference of 2 (waves back in step). Half a wavelength ( /2 ) gives a phase difference of (waves completely out of step).

• Path difference x = 0, , 2 , 3 , → phase difference = 0, 2 , 4 , → constructive (bright)
• Path difference x = /2, 3 /2, 5 /2, → phase difference = , 3 , 5 , → destructive (dark)
• General: x = n → constructive; x = (2n-1) /2 → destructive
• One complete wavelength of path difference = one complete cycle of phase ( 2 )

Quick recall

Two coherent waves have a path difference of 1.5λ. Is the interference constructive or destructive?

Destructive. x = 1.5 = 3 /2 → this is a half-odd multiple of → = 3 → waves cancel. Dark fringe.

Constructive & Destructive Interference

Constructive interference gives maximum intensity (bright). Destructive gives minimum intensity (dark). The resultant intensity depends on the phase difference and the individual amplitudes.

When two waves of amplitudes a_1 and a_2 superpose with phase difference , the resultant amplitude is:

A = a_1^2 + a_2^2 + 2a_1 a_2

Since intensity (amplitude) ^2 , the resultant intensity is:

I = I_1 + I_2 + 2 I_1 I_2

• Constructive ( = 0, 2 , 4 , ): I_ = ( I_1 + I_2 )^2 = I_1 + I_2 + 2 I_1 I_2
• Destructive ( = , 3 , 5 , ): I_ = ( I_1 - I_2 )^2 = I_1 + I_2 - 2 I_1 I_2
• If I_1 = I_2 = I_0 : I_ = 4I_0 and I_ = 0 → perfect cancellation
• Energy is NOT destroyed in destructive interference — it is redistributed from dark fringes to bright fringes

Young's Double Slit Experiment

Thomas Young's 1801 experiment shattered Newton's particle theory of light. Two narrow slits produce an interference pattern of bright and dark fringes on a screen. The fringe width formula β = λD/d is one of the most important equations in wave optics.

= D d

Experimental Setup

Monochromatic light passes through a single slit S (for coherence) and then through two narrow slits S₁ and S₂. The overlapping waves create alternating bright and dark bands on a distant screen.

The single slit S acts as a coherent line source. Light from S reaches both S₁ and S₂ simultaneously, so S₁ and S₂ become coherent sources (same frequency, constant phase difference). Waves from S₁ and S₂ overlap on the screen, and the path difference at each point determines whether you see a bright or dark fringe.

• d = slit separation (distance between S₁ and S₂), typically 0.1 to 1 mm
• D = distance from slits to screen, typically 1 m
• D d — this is the far-field (Fraunhofer) condition, necessary for the standard formula
• Slits must be narrow compared to — otherwise diffraction patterns from each slit become too narrow to overlap
• Central bright fringe is at the centre of the screen (equidistant from both slits → zero path difference)

Path Difference at Point P

For a point P at distance y from the centre of the screen, the path difference between waves from S₁ and S₂ determines whether P is bright or dark.

Let point P be at distance y from the central point O on the screen. The path from S₂ to P is longer than from S₁ to P (assuming P is above O and S₂ is the lower slit). The path difference is:

x = S_2P - S_1P = yd D

This approximation uses D d and D y . It comes from the geometry: the path difference equals d d = yd/D for small angles.

• Path difference: x = yd/D = d
• Bright fringe (constructive): x = n → y_n = n D d , where n = 0, 1, 2,
• Dark fringe (destructive): x = (2n-1) 2 → y_n = (2n-1) D 2d , where n = 1, 2, 3,
• Central maximum at n = 0 : y = 0 (zero path difference, always bright)
• Pattern is symmetric about the central maximum

Fringe Width

Fringe width β is the distance between two consecutive bright (or dark) fringes. It is uniform, depends on λ, D, and d, and is the most tested formula from this chapter.

= D d

The fringe width is the spacing between any two consecutive bright fringes (or any two consecutive dark fringes). It is the same everywhere on the screen — the pattern is equally spaced.

• = D / d — valid for both bright and dark fringes
• Fringes are equally spaced (unlike diffraction, where they are not)
• : red light ( = 700 nm) gives wider fringes than blue ( = 400 nm)
• D : move the screen farther → fringes spread out
• 1/d : increase slit separation → fringes get closer together
• Angular fringe width: = / d (independent of D )

Quick recall

In YDSE, if the slit separation is doubled and the screen distance is halved, what happens to the fringe width?

' = (D/2)/(2d) = D/(4d) = /4 . Fringe width becomes one-quarter of original. Both changes reduce .

Intensity Distribution

With two identical slits, the intensity at any point on the screen follows a cos² pattern. The central maximum has 4 times the single-slit intensity.

For two identical coherent sources (equal intensity I_0 ), the intensity at angle (or position y ) on the screen is:

I = 4I_0 ^2 ( 2 ) = 4I_0 ^2 ( d )

where = 2 d / is the phase difference. At the central maximum ( = 0 ), I = 4I_0 . At the first dark fringe, I = 0 .

• Maxima ( I = 4I_0 ) occur at = 0, 2 , 4 , → path difference = n
• Minima ( I = 0 ) occur at = , 3 , 5 , → path difference = (n - 1 2 )
• All bright fringes have the same intensity 4I_0 (in ideal YDSE with point-source slits)
• In practice, the single-slit diffraction envelope modulates the pattern — central fringes are brighter
• Average intensity over the screen: 2I_0 (energy conservation — same as sum of individual intensities)

YDSE Modifications

NEET frequently tests what happens when wavelength, medium, slit separation, or screen distance changes. A quick checklist covers all variants.

• White light YDSE: Central fringe is white (all wavelengths have zero path difference there). Away from centre, each wavelength has a different , so fringes overlap and colours appear. After a few orders, the pattern washes out.
• Slit widths unequal: Amplitudes differ → I_ 0 → fringes have reduced contrast but do not disappear
• One slit closed: No interference — you get a single-slit diffraction pattern from the open slit
• Source moved closer to slits: Fringe pattern unchanged (as long as coherence is maintained)

Diffraction of Light

When light passes through a narrow slit (comparable to its wavelength), it bends and spreads — forming a characteristic pattern with a bright central maximum flanked by weaker secondary maxima. Diffraction proves light does not always travel in straight lines.

a = n (minima)

What is Diffraction?

Diffraction is the bending of waves around obstacles or through narrow openings. It becomes significant when the slit width is comparable to the wavelength of light.

Diffraction is the phenomenon of bending of light around the edges of an obstacle or aperture. It is a consequence of Huygens' principle — secondary wavelets from different parts of the aperture interfere with each other. Diffraction is significant only when the obstacle/slit size a .

• Diffraction is essentially interference of wavelets from a single aperture
• Pronounced when slit width a (comparable to wavelength)
• If a : light travels in straight lines (ray optics regime — negligible diffraction)
• If a : wave spreads in all directions (complete diffraction)
• Fraunhofer diffraction: source and screen at infinity (parallel rays) — the standard NCERT case
• Fresnel diffraction: source or screen at finite distance — NOT in NEET syllabus

NCERT states: 'The single slit diffraction pattern is actually the interference pattern of the secondary wavelets from the same slit.' This is a key conceptual point — diffraction is not a separate phenomenon from interference.

Single Slit Diffraction Pattern

A single slit of width a produces a broad central maximum flanked by alternating minima and weaker secondary maxima. The central max is twice as wide as the side maxima.

In Fraunhofer diffraction at a single slit of width a , the diffraction pattern on a distant screen shows a broad central maximum flanked by alternating dark and bright fringes. The central maximum is the brightest and widest.

Minima (dark fringes): a = n , n = 1, 2, 3,

Secondary maxima (approx): a = (n + 1 2 ) , n = 1, 2,

• Condition for minima: a = n (where n = 1, 2, — NOT zero)
• Central maximum: between first minima on either side (from - /a to + /a )
• Width of central maximum: angular width = 2 /a ; linear width = 2 D/a
• Width of secondary maxima: D/a — half the central maximum width
• Intensity of secondary maxima drops rapidly: 1st secondary 4.5 % of central, 2nd 1.6 %
• n = 0 gives the central maximum, NOT a minimum (unlike the formula appearance)

Central Maximum Width

The central maximum width is the most important parameter in single-slit diffraction. It is twice the width of secondary maxima and depends inversely on slit width.

Angular width of central max = 2 a

Linear width of central max = 2 D a

• Central max width : longer wavelength → wider central maximum
• Central max width 1/a : narrower slit → wider central maximum (counter-intuitive!)
• Central max width D : farther screen → wider pattern
• Central maximum is exactly twice the width of each secondary maximum
• If a is very large ( a ): central max shrinks to a point → ray optics (no visible diffraction)

Quick recall

A slit of width 0.1 mm is illuminated by light of wavelength 600 nm. Screen is at 2 m. Find the width of the central maximum.

Width = 2 D/a = 2 600 10^ -9 2 / (0.1 10^ -3 ) = 24 10^ -3 m = 24 mm. The central bright band is 2.4 cm wide!

Polarization of Light

Unpolarized light vibrates in all directions perpendicular to propagation. Polarization restricts it to one plane — proving light is a transverse wave. Malus's law, Brewster's angle, and Polaroid applications are NEET essentials.

I = I_0 ^2 (Malus's Law)

Unpolarized vs Polarized Light

Ordinary light has electric field oscillations in all directions perpendicular to propagation. Polarized light oscillates in just one plane. Only transverse waves can be polarized — this proves light is transverse.

In unpolarized light, the electric field vector E oscillates in all possible directions perpendicular to the direction of propagation, with no preferred direction. In plane-polarized (or linearly polarized) light, E oscillates in only one fixed plane.

• Ordinary light (from bulb, Sun, candle) is unpolarized — E vibrates randomly in all directions
• Polarized light has E restricted to one plane — the plane of vibration
• The plane of polarization is perpendicular to the plane of vibration (NCERT convention)
• Only transverse waves can be polarized; longitudinal waves (sound) cannot
• Polarization is the definitive proof that light is a transverse wave

Malus's Law

When polarized light passes through an analyser tilted at angle θ to its polarization direction, the transmitted intensity follows cos²θ. At 90°, nothing gets through.

When plane-polarized light of intensity I_0 passes through a second polarizer (analyser) whose transmission axis makes angle with the polarization direction, the transmitted intensity is:

I = I_0 ^2 (Malus's Law)

• = 0° : I = I_0 (polarizer and analyser parallel — maximum transmission)
• = 90° : I = 0 (crossed polarizers — complete blocking, zero transmission)
• = 45° : I = I_0/2 (half the intensity passes through)
• = 60° : I = I_0/4
• When unpolarized light of intensity I_0 passes through a single polarizer: transmitted intensity = I_0/2 (half is absorbed)

Quick recall

Unpolarized light of intensity I₀ passes through two polaroids with axes at 60° to each other. What is the transmitted intensity?

After first polaroid: I_1 = I_0/2 (unpolarized → polarized). After second: I_2 = (I_0/2) ^2 60° = (I_0/2)(1/4) = I_0/8 .

Brewster's Angle

When light hits a surface at Brewster's angle, the reflected light is completely polarized. The reflected and refracted rays are perpendicular to each other. This is nature's own polarizer.

When unpolarized light hits a transparent surface (like glass) at a specific angle called Brewster's angle ( i_B ), the reflected light is completely plane-polarized with E parallel to the surface. At this angle, the reflected and refracted rays are exactly 90° apart.

i_B = n (Brewster's Law)

At Brewster's angle: i_B + r = 90°

• i_B = n where n is the refractive index of the reflecting medium
• At Brewster's angle: reflected ray refracted ray ( i_B + r = 90° )
• Reflected light is completely polarized (vibrations parallel to surface only)
• Refracted light is partially polarized (not fully — some parallel component passes through)
• For glass ( n = 1.5 ): i_B = ^ -1 (1.5) 56.3°
• For water ( n = 1.33 ): i_B = ^ -1 (1.33) 53.1°

Polaroid Applications

Polaroids are synthetic sheets that polarize light by selective absorption. They have wide-ranging practical applications from reducing glare to liquid crystal displays.

A Polaroid is a thin commercial sheet that produces plane-polarized light by selective absorption. It transmits the component of E parallel to its pass axis and absorbs the perpendicular component. Invented by Edwin Land in 1932.

• Sunglasses — Polaroid lenses with vertical pass axis block horizontally polarized glare from roads and water
• Photography — Polarizing filters reduce reflections from glass/water, enhance sky contrast
• LCD screens — Liquid crystals rotate polarization direction; two crossed polaroids act as a light valve
• 3D movies — Two images projected with orthogonal polarizations; viewers wear polaroid glasses (one eye sees each image)
• Window panes in trains/aircraft — Adjustable polaroids control light intensity
• Stress analysis — Photoelasticity uses crossed polaroids to visualize stress in transparent materials

Quick recall

Two polaroids are placed with their pass axes at 90° (crossed). Can any light pass through?

No — I = I_0 ^2 90° = 0 . But if a third polaroid is inserted between them at 45° to each, light DOES pass: I = I_0/2 ^2 45° ^2 45° = I_0/8 .

Resolving Power of Optical Instruments

Diffraction limits how fine a detail any optical instrument can see. Rayleigh's criterion sets the benchmark: two point sources are just resolved when one's central maximum falls on the other's first minimum. Larger apertures and shorter wavelengths give better resolution.

_ = 1.22 D

Rayleigh Criterion

Two point sources are considered just resolved when the central maximum of one's diffraction pattern coincides with the first minimum of the other. Below this limit, the images merge into a blur.

Every point source of light, when imaged through a circular aperture (like a lens), produces a diffraction pattern called an Airy disc — a bright central spot surrounded by faint rings. Two nearby point sources produce overlapping Airy discs.

_ = 1.22 D

where _ is the minimum angular separation for two sources to be just resolved, is the wavelength, and D is the diameter of the aperture (lens or mirror).

• Rayleigh criterion: two images are just resolved when the central maximum of one falls on the first minimum of the other
• _ = 1.22 /D — the limit of resolution (smaller _ → better resolution)
• Larger aperture D → smaller _ → better resolution (this is why telescopes have large mirrors)
• Shorter wavelength → smaller _ → better resolution (electron microscopes use electrons with very short )
• The factor 1.22 comes from the first zero of the Bessel function for a circular aperture

Resolving Power of a Telescope

A telescope's resolving power depends on its objective lens/mirror diameter D and the wavelength. Magnification helps your eye see the detail, but only aperture determines what detail actually exists in the image.

Resolving Power of Telescope = 1 _ = D 1.22

• Resolving power D (objective diameter) — larger objective → better resolution
• Resolving power 1/ — shorter wavelength → better resolution
• Magnification does NOT improve resolving power — it only enlarges an already-resolved or unresolved image
• Limit of resolution _ = 1.22 /D — the smallest angle between two stars that can be distinguished
• For visible light ( 500 nm) and D = 1 m: _ 6 10^ -7 rad

Resolving Power of a Microscope

A microscope resolves fine structures by minimizing the distance between two just-resolvable points. Shorter wavelength and larger numerical aperture improve resolution.

d_ = 1.22 2n = 1.22 2 NA

where d_ is the minimum distance between two points that can be resolved, n is the refractive index of the medium between the object and objective, is the half-angle of the cone of light entering the objective, and NA = n is the numerical aperture.

• Resolving power of microscope = 1/d_ = 2n /(1.22 )
• Smaller d_ → better resolution → can see finer details
• d_ : shorter wavelength → better resolution
• Oil immersion ( n 1.5 ) reduces d_ by factor n → better resolution
• Electron microscopes use electrons ( 0.01 nm) → resolve atomic-level structures
• Visible light microscopes: max resolution 200 nm (cannot see viruses or atoms)

Interference in Thin Films

The beautiful colours of soap bubbles, oil slicks, and anti-reflection coatings all arise from interference between light reflected from the top and bottom surfaces of a thin transparent film. The film thickness determines which wavelengths constructively interfere — and therefore which colours you see.

2 t r = n (constructive, with phase change)

How Thin Film Colours Arise

When light hits a thin transparent film, part reflects from the top surface and part from the bottom. These two reflected beams interfere — the film thickness and refractive index determine whether they add up or cancel for each colour.

Consider a thin film of thickness t , refractive index , with air on both sides. When white light hits the film at angle i , two reflected beams form: one from the top surface (ray 1) and one from the bottom surface (ray 2). The path difference between them determines which wavelengths interfere constructively.

• Two reflected beams interfere: one from the top surface, one from the bottom
• Optical path difference between them: 2 t r (where r is the angle of refraction inside the film)
• At near-normal incidence ( r 0 ): path difference 2 t
• The film must be thin (few hundred nm to a few μm) for visible light interference
• Thick films: the two beams separate too far to overlap → no colour seen

Phase Change on Reflection

When light reflects off a denser medium, it undergoes an additional phase change of π (equivalent to an extra λ/2 path). This phase shift is critical for getting the thin-film interference conditions right.

When light traveling in a rarer medium reflects from a denser medium (e.g., air to glass), the reflected wave undergoes a phase change of (equivalent to an additional path of /2 ). When light reflects from a rarer medium (glass to air), there is no phase change.

• Reflection at denser medium → phase change of (extra /2 path)
• Reflection at rarer medium → NO phase change
• For a film in air: top surface reflection (air → film) has phase change; bottom surface reflection (film → air) has no phase change
• Net additional phase difference between the two reflected beams = (half-wavelength)
• This extra swaps the constructive and destructive conditions compared to the no-phase-change case

Conditions for Constructive & Destructive Interference in Thin Films

Including the phase change from reflection, the conditions for bright and dark colours in reflected light from a thin film in air are the opposite of what naive path difference would suggest.

For a thin film of refractive index and thickness t in air, with light incident at angle i (refracted at angle r inside the film):

The geometric path difference is 2 t r . The top-surface reflection adds an extra /2 . So the effective path difference is 2 t r /2 .

Constructive (bright in reflection): 2 t r = (2n - 1 ) 2 , n = 1, 2, 3,

Destructive (dark in reflection): 2 t r = n , n = 0, 1, 2,

• These conditions are for reflected light (film in air, one phase change)
• For transmitted light, conditions are reversed (constructive and destructive swap)
• At near-normal incidence: r 1 , so 2 t (2n-1) /2 for bright
• If both surfaces have the same phase change (e.g., film between two denser media), the conditions swap again — always count the net phase changes
• For white light: only wavelengths satisfying the constructive condition appear in reflection → coloured appearance

Quick recall

A soap film (μ = 1.33) in air has thickness 300 nm. For near-normal incidence, which visible wavelength appears brightest in reflected light?

For constructive: 2 t = (2n-1) /2 . So = 4 t/(2n-1) = 4(1.33)(300)/(2n-1) = 1596/(2n-1) nm. For n=1 : = 1596 nm (IR, not visible). For n=2 : = 532 nm (green — visible!). For n=3 : = 319 nm (UV). So green light at 532 nm dominates.

Anti-Reflection Coatings

A thin film of the right thickness and refractive index can make destructive interference eliminate reflections. This is why camera lenses look purplish — the coating cancels green/yellow reflection.

An anti-reflection (AR) coating is a thin film applied to a lens surface to minimize reflection. The film thickness is chosen so that reflected beams from the top and bottom of the coating destructively interfere — cancelling the reflection for a specific wavelength.

t = 4 _f (minimum thickness for destructive reflection)

• Coating thickness = /(4 _f) for minimum reflection at wavelength
• Ideal coating refractive index: _f = _ glass — gives equal reflection from both surfaces → perfect cancellation
• Common coating: MgF₂ ( = 1.38 ) on glass ( = 1.5 ) — optimized for green light (~550 nm)
• Coated lenses reflect less green/yellow → remaining reflection is reddish-blue → lenses appear purplish
• AR coatings increase transmission from ~96% (bare glass) to ~99.5% per surface — critical for multi-element camera lenses