AP Exams (Advanced Placement) · Physics (JEE & NEET)
Ray and Wave Optics
Reflection and refraction at mirrors and lenses, optical instruments, interference, diffraction and polarisation of light.
Fourteen concepts, split down the middle: nine where light is a ray you can draw, five where it is a wave you have to add. Almost every mark lost here is a sign, a power of ten, or a condition applied in the wrong direction.
- AP Exams (Advanced Placement)
- Medium level
- 14 concepts
- 5 practice questions
1Spherical mirrors and the sign convention
One formula covers every mirror: 1/v + 1/u = 1/f, with f = R/2 and m = −v/u. What makes it hard is not the algebra but the bookkeeping. Measure every distance from the pole, take the direction of the incoming light as positive, and heights above the axis as positive. A real object then always has u negative; a concave mirror, whose centre of curvature sits on the incoming side, always has R and f negative.
Figure. Concave-mirror landmarks from 1/v + 1/u = 1/f: a distant object images near f, and an object at the centre images at the centre. Magnitudes shown; signs (real image v < 0) stay in the worked example.
How it works
- Fix the originPole at zero, positive along the light. Nothing is measured from the focus.
- Sign, then substitutePut in u, f and R with their signs already attached, then solve for v.
- Read the sign backv negative means the image is in front of the mirror, where light really goes: real.
Object between F and C
An object stands 15 cm in front of a concave mirror of radius of curvature 20 cm. Locate the image and find its magnification.
- Concave: R = −20 cm, so f = R/2−10 cm
- 1/v = 1/f − 1/u = −1/10 + 1/15−1/30 cm⁻¹
- v−30 cm, in front of the mirror
- m = −v/u = −(−30)/(−15)−2, real and inverted
Pro tip. The object sat between F and C, so the image had to land beyond C, real, inverted and enlarged — and 30 cm is beyond 20 cm. Sketch the qualitative answer before you compute, and a sign slip announces itself immediately.
A concave mirror has a radius of curvature of 20 cm. In the Cartesian convention its focal length is
- +10 cm
- −10 cm
- −20 cm
R is measured from the pole to the centre of curvature, which for a concave mirror lies on the same side as the incoming light — so R = −20 cm and f = R/2 = −10 cm. Dropping that minus is the single most expensive habit in ray optics.
2Refraction and Snell's law
At a boundary, n₁sinθ₁ = n₂sinθ₂. Going into a denser medium the ray bends toward the normal; coming out, away from it. What actually changes inside the medium is the speed, v = c/n, and with it the wavelength, λ′ = λ/n. The frequency is set by the source and never changes, which is why light does not change colour on entering glass.
Figure. The incident and refracted rays are drawn at their true angles to the normal. Notice how much the ray straightens on entering the glass — a 60° approach becomes a 35° one, and that compression is the whole of Snell's law.
How it works
- Draw the normalBoth angles are measured from the normal to the surface, never from the surface itself.
- Denser bends inwardLarger n on the far side means a smaller angle there — the ray hugs the normal.
- Divide by nSpeed and wavelength both drop by a factor n; frequency is untouched.
Into a glass block
A 600 nm beam in air strikes a glass block (n = 1.5) at 60° to the normal. Find the refraction angle, the speed inside, and the wavelength inside.
- sinθ₂ = sin 60° / 1.5 = 0.866/1.50.577
- θ₂ = arcsin(0.577)35.3°
- v = c/n = 3×10⁸ / 1.52 × 10⁸ m s⁻¹
- λ′ = λ/n = 600 nm / 1.5400 nm
Pro tip. 400 nm is violet's wavelength in air, yet the beam inside the glass still looks orange — because the eye responds to frequency, and frequency did not move. Never convert a wavelength to a colour without asking which medium it is in.
3Total internal reflection
Push the refraction angle to 90° and the refracted ray disappears along the surface. The incidence angle that does this is the critical angle, sinθ_c = 1/n, where n is the denser medium's index relative to the rarer one. Beyond it, none of the light escapes: it is all reflected back, and reflected perfectly, with no metal coating to tarnish. Two conditions, both required — the light must start in the denser medium, and the angle must exceed θ_c.
Figure. Three rays leave the same point in the glass at 25°, 41.8° and 60°. The first escapes, bent away from the normal. The second is already grazing the surface — that is the definition of θ_c. The third never leaves.
How it works
- Denser to rarerOnly then can the refraction angle be the larger of the two and reach 90° first.
- At θ_c it grazesThe refracted ray runs along the boundary; the transmitted beam is already vanishing.
- Beyond θ_c, nothing leavesAll the energy goes into the reflected ray, obeying the ordinary law of reflection.
The 45° prism
A right-angled isosceles prism of glass (n = 1.5) has light entering one short face normally and striking the hypotenuse. Does it emerge through the hypotenuse?
- sinθ_c = 1/n = 1/1.50.667
- θ_c = arcsin(0.667)41.8°
- Angle of incidence on the hypotenuse45°
- 45° > 41.8°, so the ray istotally reflected, through 90°
Pro tip. This is why binoculars and periscopes use prisms rather than mirrors: reflection is 100%, not the 90-odd percent a silvered surface manages, and there is no coating to degrade. The same 41.8° is why a diamond, with θ_c near 24°, traps light through many internal bounces before letting it out.
Light travels from air into glass. Total internal reflection at that air–glass surface
- Occurs once the incidence angle exceeds the critical angle
- Cannot occur — the light is going from rarer to denser
- Occurs only if the glass is thicker than the wavelength
Going rarer to denser, the refraction angle is always the smaller one, so it can never reach 90°. TIR needs the light to start in the denser medium. The critical angle exists for the air–glass pair, but only for rays approaching from the glass side.
4Refraction at a single spherical surface
Before a lens has two surfaces it has one, and one curved boundary between media of index n₁ and n₂ obeys n₂/v − n₁/u = (n₂ − n₁)/R. Same origin as always — the pole of the surface, distances positive along the light — with n₂ the index of the medium the refracted ray ends up in. It is n₂ and n₁ that appear at a single boundary, not the (n − 1) of the whole lens; that familiar factor only emerges when you apply this relation at both faces of a thin lens and subtract.
Flatten the surface, R → ∞, and the right-hand side dies, leaving n₂/v = n₁/u — the apparent-depth rule. An object at real depth d under a medium appears at d × (n₂/n₁): a pool looks shallower from above (n₂ < n₁), and a bird looks higher seen from underwater (n₂ > n₁).
Figure. Single-surface formula: n₂/v − n₁/u = (n₂ − n₁)/R. The curvature term is the surface power; rearrange and the image term is the object term plus that power. No arc is drawn — the algebra is the figure.
How it works
- Pole is the originMeasure u, v and R from the vertex of the surface, positive along the incoming light — never from the centre of curvature.
- n₂ sits over vThe refracted ray ends in medium n₂, so n₂/v pairs with the far side and n₁/u with the object side.
- Flatten to checkSet R → ∞ and the surface is a plane: n₂/v = n₁/u, which is apparent depth = real depth × n₂/n₁.
A point object beyond a glass rod
A long glass rod (n = 1.5) ends in a convex surface of radius 10 cm. A point object sits in air on the axis, 30 cm from the vertex. Where is the image?
- n₁ = 1, n₂ = 1.5, R = +10 cm, u = −30 cmsigns fixed
- 1.5/v − 1/(−30) = (1.5 − 1)/101.5/v + 1/30 = 0.05
- 1.5/v = 0.05 − 0.03330.0167 cm⁻¹
- v = 1.5 / 0.0167+90 cm, real, inside the glass
Pro tip. The image is 90 cm inside the glass and real, because v came out positive on the side the light travels into. Notice (n − 1) never appeared: at one surface it is the two indices themselves that enter, and the 0.5 you see is n₂ − n₁, not n − 1. The two coincide here only because n₁ = 1.
A fish in water (n = 4/3) looks straight up at a bird hovering 4 m above the surface. The bird appears to be at a height of about
- 3.0 m — nearer than it really is
- 5.3 m — farther than it really is
- 4.0 m — refraction changes the direction, not the height
Viewed from the denser medium, apparent height = real height × n₂/n₁ with n₂ = 4/3 (water, where the ray ends) and n₁ = 1 (air): 4 × 4/3 = 5.33 m. The bird looks higher, the mirror image of the coin-in-a-pool case where an object in water looks shallower.
5Thin lenses and image formation
The thin lens obeys 1/v − 1/u = 1/f with m = v/u. Note the two differences from a mirror: a minus sign in the formula and no minus sign in the magnification. A converging lens has f positive, a diverging lens f negative. Solve, then read the signs: v positive means the image is on the far side where light genuinely arrives, so it is real; m negative means inverted.
Figure. The lens is drawn as a plane, the thin-lens idealisation. One ray goes in parallel and leaves through F′; the other cuts the optical centre and carries straight on. They meet below the axis at 60 cm, which is why the image is real, inverted and twice as tall.
The two rays that fix the image
- Parallel in, focus outA ray arriving parallel to the axis leaves through the far focus F′.
- Straight through the centreA ray aimed at the optical centre passes on undeviated — no bend at all.
- Where they crossThat intersection is the image tip; the axis fixes the foot.
Object beyond the focus
A 5 mm tall object stands 30 cm in front of a convex lens of focal length 20 cm. Find the image distance, the magnification and the image height.
- 1/v = 1/f + 1/u = 1/20 − 1/301/60 cm⁻¹
- v+60 cm, on the far side
- m = v/u = 60/(−30)−2
- Image height = |m| × 5 mm10 mm, inverted
Pro tip. The object sat between f and 2f, so the image had to be beyond 2f, real, inverted and magnified — and 60 cm is beyond 40 cm. Slide the object to exactly 2f and object and image are the same size at the same distance; that symmetric point is worth memorising as a check.
An object is placed 10 cm from a convex lens of focal length 20 cm. The image is
- Real and inverted, 20 cm beyond the lens
- Virtual and erect, 20 cm on the object's side
- Formed at infinity
1/v = 1/20 + (−1/10) = −1/20, so v = −20 cm and m = (−20)/(−10) = +2. Inside the focus, a converging lens acts as a magnifying glass: virtual, erect, twice the size, on the same side as the object. Only an object exactly at f throws the image to infinity.
6What sets f — lens maker's formula and power
A lens has the focal length its glass and its two curvatures give it: 1/f = (n − 1)(1/R₁ − 1/R₂), with the same sign convention as everything else. Two consequences do most of the work in exams. First, the factor is (n − 1), so it is the contrast with the surroundings that matters — immerse the lens and n becomes n_lens/n_medium. Second, define power as P = 1/f in metres, in dioptres, and lenses in contact simply add: P = P₁ + P₂.
Figure. Lens maker: 1/f = (n − 1)(1/R₁ − 1/R₂). Each curved surface contributes a term; equal double-convex radii add. Power in dioptres is 1/f with f in metres.
How it works
- Sign each radiusR is positive when the centre of curvature lies beyond the surface, along the light.
- Multiply by (n − 1)More curvature or more index contrast, shorter f, stronger lens.
- Add in dioptresPowers add for lenses in contact. Focal lengths do not.
An equiconvex lens, then a doublet
An equiconvex lens of glass (n = 1.5) has both faces of radius 20 cm. Find its focal length and power, then its power when a −2.5 D lens is placed in contact with it.
- 1/f = (n − 1)(1/R₁ − 1/R₂) = 0.5 × (1/20 + 1/20)1/20 cm⁻¹
- f+20 cm
- P = 1/f(m) = 1/0.20+5 D
- In contact: P = 5 − 2.5+2.5 D, i.e. F = 40 cm
Pro tip. R₂ is negative for the second face of an equiconvex lens, which is why the two terms add rather than cancel. Now dip that same lens in water: (n − 1) = 0.5 becomes (1.5/1.33 − 1) = 0.125, so f goes from 20 cm to about 80 cm — the lens is four times weaker without changing shape at all.
A thin converging glass lens (n = 1.5) of focal length 20 cm in air is immersed in water (n = 4/3). Its focal length becomes about
- 20 cm — f depends only on the glass and the radii
- 80 cm — the lens is roughly four times weaker
- 15 cm — the lens is stronger in a denser medium
The prefactor is (n_lens/n_medium − 1). In air that is 0.5; in water it is 1.5/1.333 − 1 = 0.125, a quarter as large, so f quadruples. A lens works by contrast, and water removes most of it — which is exactly why your eyes will not focus underwater.
7The prism: deviation and minimum deviation
A ray crossing a prism refracts twice, and the two bends add to a deviation δ = i₁ + i₂ − A, where A is the prism angle and the internal angles always satisfy r₁ + r₂ = A. Sweep the angle of incidence from grazing to grazing and δ traces a shallow valley with a single minimum. At that minimum the path runs symmetrically through the prism — i₁ = i₂ and r₁ = r₂ = A/2 — so the minimum sits where the path is symmetric, not where the ray goes straight through.
At minimum deviation the index follows from geometry alone: μ = sin((A + δ_m)/2) / sin(A/2), which is exactly how a spectrometer measures the refractive index of a glass. For a thin prism (A only a few degrees) the whole relation collapses to δ = (μ − 1)A, independent of the angle of incidence — the form used all through dispersion.
Figure. Deviation against angle of incidence is a shallow valley — the curve is drawn screen-down, so its lowest point on the page is the smallest deviation. δ falls to a single minimum and climbs again towards grazing incidence at either end. The minimum sits at the symmetric passage i₁ = i₂, and the same δ_m can be reached from two different angles of incidence on the two arms of the valley, which is why a spectrometer reading is taken right at the turning point.
How it works
- Two refractions, one identityr₁ + r₂ = A always, and δ = i₁ + i₂ − A. Snell's law at each face ties the i's to the r's.
- Symmetry marks the minimumδ is least when i₁ = i₂; the ray then crosses parallel to the base and r₁ = r₂ = A/2.
- Read the index off δ_mμ = sin((A + δ_m)/2)/sin(A/2). For a thin prism this reduces to δ = (μ − 1)A.
Minimum deviation of a 60° glass prism
An equilateral prism (A = 60°) is made of glass with n = 1.5. Find the angle of minimum deviation and the angle of incidence that produces it.
- At δ_m: r₁ = r₂ = A/230°
- sin i₁ = n sin r₁ = 1.5 × sin 30°0.75
- i₁ = arcsin 0.7548.6°
- δ_m = 2i₁ − A = 2(48.6°) − 60°37.2°
Pro tip. Check it the other way: μ = sin((60° + 37.2°)/2)/sin 30° = sin 48.6°/0.5 = 0.75/0.5 = 1.5 ✓. The thin-prism shortcut δ = (μ − 1)A would give (0.5)(60°) = 30° — a real underestimate, because 60° is nowhere near thin. Keep δ = (μ − 1)A for the few-degree prisms of dispersion problems only.
A thin prism of refracting angle 5° is made of glass of index 1.5. It deviates a ray of light by
- 2.5°
- 5.0°
- 7.5°
For a thin prism δ = (μ − 1)A = (1.5 − 1)(5°) = 2.5°, independent of the angle of incidence. Forgetting the (μ − 1) factor and reading 5° or adding A back in are the usual slips.
8Dispersion and dispersive power
A glass has a slightly larger index for violet light than for red, so through a prism violet bends more, not less, and white light fans into a spectrum. For a thin prism each colour deviates by δ = (μ − 1)A, so the angular spread between the extremes is the angular dispersion θ = (μ_v − μ_r)A, while the mean ray deviates by δ_y = (μ_y − 1)A.
Divide one by the other and the prism angle cancels: the dispersive power ω = (μ_v − μ_r)/(μ_y − 1) is a pure number belonging to the glass alone, typically about 0.02–0.03. Because ω is a material constant, two different glasses can be combined to cancel dispersion while keeping deviation (an achromatic prism) or to cancel deviation while keeping dispersion (a direct-vision prism).
Figure. Crown glass from the worked example: violet sits above red by 0.010 in μ. Angular fan gaps are unresolvable on a ray diagram; the index bars are the differences that enter ω = (μ_v − μ_r)/(μ_y − 1).
How it works
- Split the deviation by colourδ = (μ − 1)A, and μ_violet > μ_red, so violet is deviated most and red least.
- Spread over meanω = angular dispersion / mean deviation = (μ_v − μ_r)/(μ_y − 1). The prism angle A cancels.
- Pair two glassesChoose the second prism's angle so the two dispersions cancel (achromatic) or the two deviations cancel (direct vision).
Dispersive power of a crown-glass prism
A thin crown-glass prism of angle 5° has indices μ_red = 1.513, μ_yellow = 1.517 and μ_violet = 1.523. Find its mean deviation, its angular dispersion and its dispersive power.
- Mean deviation δ_y = (μ_y − 1)A = 0.517 × 5°2.59°
- Angular dispersion θ = (μ_v − μ_r)A = 0.010 × 5°0.05°
- Dispersive power ω = θ/δ_y = 0.05/2.590.0193
- Check: ω = (μ_v − μ_r)/(μ_y − 1) = 0.010/0.5170.0193 ✓
Pro tip. ω came out the same by both routes because the prism angle A cancels — 0.0193 is a fingerprint of crown glass and would be identical for a 2° or a 10° prism. That A-independence is exactly what lets an achromatic doublet remove colour from a lens: pair this crown with a flint of larger ω and matching angles, and the two spreads cancel while a net bend survives.
White light passes through a prism. Compared with red light, violet light is
- deviated more, because the glass has a larger index for violet
- deviated less, because violet has the shorter wavelength
- deviated by the same angle — only the colour differs
δ = (μ − 1)A and μ rises towards the blue end, so μ_violet > μ_red and violet deviates most. The trap is to reason from wavelength directly; it is the index, larger for the shorter wavelength in ordinary glass, that sets the bend.
9Microscopes and telescopes
Put two converging lenses in a row and you have either a microscope or a telescope; which one depends on the focal lengths, and the figure of merit is always the same. What is magnified is the angle the image subtends at the eye, not its linear size. A compound microscope multiplies two stages: the objective forms a real, enlarged image (linear magnification ≈ L/f_o, with L the tube length) and the eyepiece views that as a simple magnifier (angular gain D/f_e, D = 25 cm), so M ≈ (L/f_o)(D/f_e).
A telescope looks at objects so distant that linear size is meaningless, so it works purely in angles. In normal adjustment — final image at infinity — its angular magnification is M = f_o/f_e and its tube length is f_o + f_e. The two instruments pull the objective in opposite directions: a microscope wants both focal lengths as short as possible, a telescope wants a long objective and a short eyepiece.
Figure. Both instruments magnify angle, not linear size. A typical school telescope sits near M = f_o/f_e ≈ 10; a compound microscope stacks two factors and reaches ~100. The comparison table names the formulas; the bars name the scale.
How it works
- Microscope: two magnifiersObjective gives ≈ L/f_o, eyepiece gives D/f_e; multiply for the total M ≈ (L/f_o)(D/f_e).
- Telescope: compare anglesDistant object, so M = f_o/f_e in normal adjustment and the tube is f_o + f_e long.
- Which focal lengthsShort and short for a microscope; a long objective and a short eyepiece for a telescope.
Both magnify an angle, but a microscope multiplies two linear stages while a telescope takes a ratio of focal lengths.
| Instrument | Angular magnification (normal adjustment) | Length |
|---|---|---|
| Simple microscope (magnifier) | M = D/f (image at ∞); 1 + D/f (image at 25 cm) | — |
| Compound microscope | M ≈ (L/f_o)(D/f_e) | ≈ v_o + f_e |
| Astronomical telescope | M = f_o/f_e | f_o + f_e |
A telescope in normal adjustment
An astronomical telescope has an objective of focal length 100 cm and an eyepiece of focal length 5 cm. Find its magnifying power and tube length in normal adjustment.
- M = f_o/f_e = 100/520
- Tube length L = f_o + f_e = 100 + 5105 cm
- Eyepiece alone as a magnifier: D/f_e = 25/55
- So the objective contributes a further factor f_o/D = 100/254, and 4 × 5 = 20 ✓
Pro tip. ×20 out of a metre of tube. To double the power you either double the objective's focal length — a two-metre tube — or halve the eyepiece's. A microscope is the mirror image: there you drive f_o down to a few millimetres, because for a microscope short focal lengths magnify while for a telescope it is a long objective that does.
An astronomical telescope has objective and eyepiece focal lengths of 100 cm and 2 cm. Its magnifying power in normal adjustment is
- 50
- 200
- 20
M = f_o/f_e = 100/2 = 50, with a tube length of 102 cm. Multiplying the focal lengths (200) or dividing the wrong way are the two slips; the long focal length goes on top.
10Huygens' principle and wavefronts
Wave optics begins with a rule for advancing a wave. A wavefront is a surface of constant phase; Huygens' principle says every point on it acts as a source of secondary spherical wavelets, and the wavefront a moment later is the forward surface just touching all of them. The rays are simply the perpendiculars to the wavefronts, so this one construction reproduces the whole of geometrical optics.
It earns its place in two ways. Crossing into a slower medium the wavelets advance less far, so the front tilts, and working the geometry through gives sin θ₁/sin θ₂ = v₁/v₂ = n₂/n₁ — Snell's law, derived rather than assumed. And it pins down what interference will need: two sources are coherent when they keep a fixed phase relationship, which is why the next concepts split a single wavefront in two rather than using two independent lamps.
Figure. Wavefronts are surfaces of constant phase, drawn here as vertical lines, and the ray is their perpendicular. At normal incidence the fronts stay parallel to the boundary but crowd closer on entering the glass — the spacing here drops by the index n = 1.5, from λ to λ/n, because the wave slows while its frequency, set by the source, does not. The secondary spherical wavelets that generate each new front are arcs and cannot be drawn with this vocabulary.
How it works
- Every point reseedsTreat each point of the current wavefront as a source of a secondary spherical wavelet travelling at the local wave speed.
- Take the envelopeThe new wavefront is the forward surface tangent to all those wavelets one instant later.
- Rays are the normalsDraw the perpendiculars; where the medium slows the wavelets, the front tilts, and that tilt is Snell's law.
Snell's law from wavefronts
A plane wavefront in air (v₁ = 3 × 10⁸ m s⁻¹) meets glass (n = 1.5) at 60° to the surface normal. Use the wavefront speeds to find the refraction angle, and confirm the ratio.
- v₂ = c/n = 3 × 10⁸ / 1.52 × 10⁸ m s⁻¹
- sin θ₂ = (v₂/v₁) sin θ₁ = (1/1.5) sin 60°0.577
- θ₂ = arcsin 0.57735.3°
- sin θ₁/sin θ₂ = 0.866/0.5771.5 = n₂/n₁ ✓
Pro tip. The wavefront picture gives the same 35.3° as ray-based Snell's law because it is Snell's law — the tilt of the front and the bend of the ray are one fact seen two ways. The wavefronts crowd closer inside the glass (λ falls by n) while their frequency is untouched; that crowding is what forces the tilt.
A plane wavefront in air passes into glass at normal incidence. Inside the glass the wavefronts are
- closer together, because the wavelength has shrunk
- farther apart, because the wave has slowed
- spaced exactly as before — only the ray direction can change
Consecutive wavefronts are one wavelength apart. The speed drops by n and the frequency is fixed by the source, so λ = v/f falls by n and the fronts crowd closer. At normal incidence there is no tilt, but the spacing still changes.
11Young's double slit and fringe width
Two slits cut from the same wavefront stay in step, so their light can interfere. At a point y up the screen the path difference is Δ = dy/D. Bright fringes sit where Δ = nλ, dark ones where Δ = (n − ½)λ, and consecutive fringes of the same kind are β = λD/d apart. Everything in the pattern is that one length scale: β widens if you lengthen λ or D, and narrows if you separate the slits.
Figure. The two gaps in the barrier are the slits, separated by d. Both send light to the same point P on the screen, but the lower one travels dy/D further. Whether P is bright or dark is decided entirely by how many wavelengths fit into that small extra length.
How it works
- Split one sourceBoth slits are fed by the same wavefront, so their phase difference is fixed in time.
- Path difference dy/DFor small angles the extra distance from the far slit is d sinθ ≈ dy/D.
- Count in wavelengthsWhole numbers of λ give brightness, half-odd numbers give darkness, spaced β apart.
Locating a named fringe
In a double-slit experiment λ = 600 nm, the slit separation is d = 0.3 mm and the screen is D = 1.5 m away. Find the fringe width, and the positions of the 4th bright and 3rd dark fringes from the centre.
- β = λD/d = (600×10⁻⁹ × 1.5)/(0.3×10⁻³)3 × 10⁻³ m = 3 mm
- 4th bright: y = 4β12 mm
- 3rd dark: y = (3 − ½)β7.5 mm
Pro tip. Convert nm, mm and m to SI before dividing — one wrong power of ten is the usual failure here. And note the half: the nth dark fringe sits at (n − ½)β, not at nβ. Counting dark fringes as if they were bright is worth a mark on its own.
The whole double-slit apparatus, slits and screen alike, is immersed in water (n = 4/3). The fringe width
- Grows by a factor 4/3
- Falls to three quarters of its former value
- Is unchanged, since d and D are unchanged
β = λD/d, and it is the wavelength in the medium that counts: λ′ = λ/n. With d and D fixed, β scales down by n, so 3 mm becomes 2.25 mm. The geometry is a red herring; the wavelength is what moved.
12Intensity, phase and shifting the pattern
Fringes are not on-or-off. Convert the path difference to a phase, φ = 2πΔ/λ, and the intensity of two equal slits varies smoothly as I = 4I₀cos²(φ/2) — four times one slit's intensity at the maxima, zero at the minima, and the average still 2I₀, so nothing is lost. Now slide a thin sheet of index n and thickness t over one slit. It adds (n − 1)t to that path, and the entire pattern slides by (n − 1)tD/d toward the covered slit. The spacing β does not change, because β never depended on the path difference at the centre.
Figure. Solid and dashed curves have identical shape and identical spacing; the dashed one is simply displaced. That is what a slab over one slit does — it re-centres the pattern without rescaling it.
How it works
- Path to phaseφ = 2πΔ/λ. One whole wavelength of path is one full 2π of phase.
- Add the amplitudesEqual slits give I = 4I₀cos²(φ/2), peaking at four times a single slit.
- A slab re-centres itExtra path (n − 1)t moves the zero-order fringe; every fringe follows, β intact.
A sheet over one slit
In the same apparatus (λ = 600 nm, d = 0.3 mm, D = 1.5 m, β = 3 mm) a sheet of thickness 10 μm and index 1.5 is placed over one slit. Find the shift of the pattern.
- Extra path = (n − 1)t = 0.5 × 10 μm5 μm
- Fringes shifted = (n − 1)t/λ = 5/0.68.33
- Shift = 8.33 × β = 8.33 × 3 mm25 mm
- Fringe width βstill 3 mm
Pro tip. Read the last row again. The pattern moves 25 mm — more than eight fringes — and yet a student measuring the spacing would see nothing at all changed. Shift and spacing are independent, and questions are built on the assumption you will confuse them.
A thin transparent sheet is placed over one slit of a double-slit apparatus. The pattern
- Shifts toward the covered slit, with the spacing unchanged
- Shifts away from the covered slit, and the fringes widen
- Loses contrast until no fringes remain
The sheet lengthens one optical path by (n − 1)t, so the point of zero path difference must move toward the slit that was slowed. β = λD/d contains no term for the sheet, so the spacing is untouched. Contrast survives because both slits still carry light.
13Single-slit diffraction and resolution
A single slit of width a spreads light into a broad central band flanked by faint side bands. The dark bands sit at a sinθ = nλ with n = 1, 2, 3… — and n = 0 is excluded, because straight ahead every part of the slit arrives in phase. That condition looks exactly like the double-slit bright-fringe condition and means the opposite; nothing here costs more marks. The central maximum spans 2λ/a in angle, twice the width of any other band, and the same 1/a scaling sets the resolving limit of a circular aperture, θ ≈ 1.22λ/D.
Figure. The side bands are barely visible — the first carries under 5% of the central peak — which is why the pattern looks like a single bright band with faint echoes. Halve the slit and this whole picture stretches to twice the width.
How it works
- Pair up the slitSplit the slit in two halves; at a sinθ = λ every point cancels its partner.
- Minima, not maximaSo a sinθ = nλ locates darkness, with n = 0 excluded by construction.
- Narrow slit, wide patternWidth ∝ 1/a. Squeeze the slit and the light spreads further, not less.
How wide is the central band?
A slit of width 0.2 mm is lit by 600 nm light and the pattern is caught on a screen 2 m away. Find the width of the central maximum, and compare it with the fringe width a double slit of the same 0.2 mm separation would give.
- First minimum: sinθ = λ/a = 600×10⁻⁹/0.2×10⁻³3 × 10⁻³
- y₁ = D tanθ ≈ 2 × 3×10⁻³6 mm
- Central maximum width = 2y₁12 mm
- Double slit with d = 0.2 mm: β = λD/d6 mm
Pro tip. The central diffraction band is exactly twice a double-slit fringe of the same spacing — and that is not a coincidence: both scale as λD divided by a length, but the central band is bounded by minima on both sides at ±λ/a. In a real double-slit experiment the fringes sit inside this envelope, which is why the outer fringes fade.
In a single-slit pattern the condition a sinθ = λ locates
- The first bright band beside the centre
- The first dark band
- The centre of the pattern
Divide the slit into two halves: at this angle every point in the upper half is exactly half a wavelength out of step with its partner below, so the two halves cancel completely. The identical-looking d sinθ = λ in a double slit is a maximum, because there the two paths differ by a whole wavelength rather than being paired for cancellation.
14Polarisation — Malus and Brewster
Unpolarised light is an even mixture of every transverse orientation, so the first polariser always transmits exactly half, whatever its angle. After that the light has a definite direction and Malus's law takes over: a second polariser at θ to the first passes I = I₀cos²θ. Reflection polarises too — at Brewster's angle, tanθ_B = n, the reflected beam is completely plane-polarised, and the reflected and refracted rays leave at right angles to each other.
Figure. The curve is flat near 0° and near 180° and steepest at 45°, so a polariser is least sensitive to misalignment where it passes most light. Note that it touches zero only at 90° — crossed sheets, and nothing else, give darkness.
How it works
- First sheet halves itAveraging cos²θ over all orientations gives ½ — the angle of the first sheet is irrelevant.
- Then MalusEvery sheet after the first transmits cos²θ of what reaches it, θ measured to the previous axis.
- Reflection at θ_Btanθ_B = n; the reflected ray is fully polarised perpendicular to the plane of incidence.
Three sheets beat two
Unpolarised light of intensity I₀ falls on polariser P₁. P₃ is crossed with P₁ at 90°, and P₂ is inserted between them at 45° to P₁. Find the emergent intensity, then remove P₂ and find it again.
- After P₁ (unpolarised light halves)I₀/2
- After P₂ at 45°: (I₀/2) cos²45°I₀/4
- After P₃, 45° from P₂: (I₀/4) cos²45°I₀/8
- Remove P₂, leaving P₁ ⟂ P₃0
Pro tip. Taking a sheet away leaves you with less light, not more. The middle sheet does not just filter — it rotates the transmitted polarisation to 45°, so P₃ is no longer perpendicular to what arrives at it. Nothing in the calculation is unusual; the surprise comes from thinking of polarisers as filters rather than as projections.
Light reflected from a glass surface (n = 1.5) is completely plane-polarised when the angle of incidence is about
- 0°, i.e. normal incidence
- 56°
- 34°
tanθ_B = n = 1.5 gives θ_B = 56.3°, measured from the normal. 34° is its complement — the refraction angle at that incidence, which is what makes the reflected and refracted rays perpendicular. At normal incidence there is no preferred transverse direction, so nothing is polarised.
Notes
- Mirrors and lenses: The mirror formula \dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f} and lens formula \dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f} (Cartesian sign convention) give image positions, with magnification m=-\dfrac{v}{u} for lenses. Focal length of a mirror is f=R/2.
- Refraction and the lens maker's formula: Snell's law is n_1\sin\theta_1=n_2\sin\theta_2. Total internal reflection occurs beyond the critical angle \sin\theta_c=\dfrac{1}{n}. A thin lens obeys \dfrac{1}{f}=(n-1)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right).
- Young's double slit: Coherent slits separated by d give bright fringes where the path difference is n\lambda, with fringe width \beta=\dfrac{\lambda D}{d}. Intensity varies as I=I_0\cos^2\left(\dfrac{\phi}{2}\right) with \phi the phase difference.
- Diffraction: A single slit of width a produces minima at a\sin\theta=n\lambda; the central maximum has angular width \dfrac{2\lambda}{a}. The resolving limit of an aperture is set by \theta\approx\dfrac{1.22\lambda}{D}.
- Polarisation: Unpolarised light passing through a polariser drops to half intensity; a second polariser at angle \theta transmits I=I_0\cos^2\theta (Malus's law). Reflected light is fully polarised at Brewster's angle \tan\theta_B=n.
Formulas
- Mirror / lens: \dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f} (mirror), \dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f} (lens)
- Snell / critical angle: n_1\sin\theta_1=n_2\sin\theta_2,\quad \sin\theta_c=\dfrac{1}{n}
- Lens maker: \dfrac{1}{f}=(n-1)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)
- Fringe width: \beta=\dfrac{\lambda D}{d}
- Single-slit minima: a\sin\theta=n\lambda
- Malus / Brewster: I=I_0\cos^2\theta,\quad \tan\theta_B=n
Exam traps & shortcuts
- Power of a lens in dioptres is P=\dfrac{1}{f(\text{m})}, and powers of lenses in contact add: P=P_1+P_2 - faster than combining focal lengths.
- In YDSE, inserting a thin sheet of thickness t and index n in one path shifts the pattern by \dfrac{(n-1)tD}{d}; the fringe width \beta itself is unchanged.
- For crossed polarisers (\theta=90^\circ) no light passes, but inserting a third polariser between them at 45^\circ lets some light through - a classic conceptual trap.
Reference tables
Every line here should be reconstructible from the concept above it, not merely recalled.
| Quantity | Relation | Watch for |
|---|---|---|
| Mirror | 1/v + 1/u = 1/f, f = R/2 | m = −v/u, mirrors only |
| Thin lens | 1/v − 1/u = 1/f | m = v/u — no minus sign |
| Refraction | n₁sinθ₁ = n₂sinθ₂ | v and λ divide by n; f is fixed |
| Critical angle | sinθ_c = 1/n | Denser to rarer only |
| Lens maker | 1/f = (n − 1)(1/R₁ − 1/R₂) | (n − 1) → (n_lens/n_medium − 1) |
| Power | P = 1/f(m), P = P₁ + P₂ | Dioptres add, focal lengths do not |
| Spherical surface | n₂/v − n₁/u = (n₂ − n₁)/R | Apparent depth = real × n₂/n₁ |
| Prism | δ = i₁ + i₂ − A, r₁ + r₂ = A | Minimum when i₁ = i₂ |
| Minimum deviation | μ = sin((A + δ_m)/2) / sin(A/2) | Thin prism: δ = (μ − 1)A |
| Dispersive power | ω = (μ_v − μ_r)/(μ_y − 1) | Property of the glass, not of A |
| Telescope / microscope | M = f_o/f_e ; M ≈ (L/f_o)(D/f_e) | Angle at the eye, not linear size |
| Fringe width | β = λD/d | Divides by n if immersed |
| Fringe shift | (n − 1)tD/d | β itself is unchanged |
| Two-slit intensity | I = 4I₀cos²(φ/2), φ = 2πΔ/λ | 4I₀ at the peaks, average 2I₀ |
| Single-slit minima | a sinθ = nλ, n ≠ 0 | Dark, not bright |
| Central maximum | Angular width 2λ/a | Twice as wide as the rest |
| Resolution | θ ≈ 1.22λ/D | Larger aperture, finer detail |
| Malus | I = I₀cos²θ | Unpolarised light halves first |
| Brewster | tanθ_B = n | Reflected ⟂ refracted at θ_B |
One origin, one positive direction. Nearly every wrong answer in ray optics is a sign that was assumed rather than assigned.
| Quantity | Convention | Worked value |
|---|---|---|
| Origin | Pole of the mirror, optical centre of the lens | Never the focus |
| Positive direction | The direction the incident light travels | Usually left to right |
| Object distance u | Negative for any real object | −30 cm |
| Image distance v — lens | Positive on the far side, where the light genuinely arrives | +60 cm, so real |
| Image distance v — mirror | Negative in front of the mirror; reflected light travels back along the negative direction | −30 cm, so real |
| Converging lens f | Positive | +20 cm |
| Diverging lens f | Negative | −20 cm |
| Concave mirror f | Negative — C sits on the incoming side | −10 cm for R = 20 cm |
| Convex mirror f | Positive | +10 cm for R = 20 cm |
| Radius R of a lens face | Positive if its centre lies beyond the surface | R₁ = +20, R₂ = −20 cm |
| Heights | Positive above the axis | m > 0 means erect |
Recap
Read only this the night before.
- Signs
- Measure from the pole or the optical centre, against the light is negative. m = v/u for a lens, −v/u for a mirror — the one asymmetry worth memorising.
- Refraction
- Frequency never changes; speed and wavelength both divide by n. TIR only from denser to rarer, only beyond sinθ_c = 1/n — 41.8° for glass, 48.6° for water.
- Lenses
- 1/f = (n − 1)(1/R₁ − 1/R₂). Dioptres add for lenses in contact. Immersed in water a glass lens goes about four times weaker.
- Interference
- β = λD/d. Bright at nλ, dark at (n − ½)λ. A slab of optical thickness (n − 1)t slides the whole pattern toward the covered slit and leaves β alone.
- Diffraction
- a sinθ = nλ is a minimum, and n = 0 is excluded. The central maximum is 2λ/a wide, twice everything else. Narrow the slit and the light spreads.
- Polarisation
- The first sheet halves it, then Malus. Crossed sheets pass nothing — until a third at 45° goes between them and I₀/8 comes out.
- Spherical surface
- n₂/v − n₁/u = (n₂ − n₁)/R at one curved boundary; apply it twice for the lens maker. Flatten it (R → ∞) for apparent depth = real depth × n₂/n₁.
- Prism
- δ = i₁ + i₂ − A, least when i₁ = i₂ — then μ = sin((A + δ_m)/2)/sin(A/2). Thin prism δ = (μ − 1)A; dispersive power ω = (μ_v − μ_r)/(μ_y − 1) belongs to the glass.
- Instruments & wavefronts
- Telescope M = f_o/f_e, tube f_o + f_e; microscope M ≈ (L/f_o)(D/f_e). Huygens: every point of a wavefront reseeds spherical wavelets, their envelope is the next front, rays are its perpendiculars, and the tilt on slowing is Snell's law.
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