A full lesson: a cartoon parade to start, then Snell's law built from wavelets, why frequency never changes, the critical angle from the same picture, the law of reflection, how lenses, prisms and mirrors reshape wavefronts, NEET shortcuts, problems to solve yourself, and 12 MCQs.
The cadets who step onto the sand first slow down, so the line swings round and marches closer to the normal. A wavefront entering glass does exactly the same: that is refraction.
On Republic Day, a line of NCC cadets marches arm in arm across the school road. The road ends in the sandy sports ground, and the line meets it at a slant. Nobody gives an order, yet the line turns. The cadet who reaches the sand first starts taking shorter steps; the others are still on the road taking long steps, so they catch up and the line swings round. Kabir keeps the drum beat the same all the time.
Light does the same thing at the surface of glass or water. The line of cadets is a wavefront, each cadet is a point on it, the drum beat is the frequency and the step length is the wavelength. In this lesson Huygens' wavelets turn that picture into Snell's law, the critical angle and the law of reflection.
Take a toy car (or a rolling pin) and roll it at a slant from the smooth floor onto a doormat. Which way does it turn, towards the line perpendicular to the mat's edge, or away from it?
Towards the perpendicular (the normal). The wheel that hits the mat first slows down, the other wheel keeps going, and the axle swings round. Rolled from the mat onto the floor, it turns away from the normal. Same as light going from glass into air.
"Remember three words from the parade: the line is the wavefront, the drum beat is the frequency, the step length is the wavelength. The drum beat never changes."
A plane wavefront AB meets the boundary at angle of incidence i. Medium 1 has speed v₁, medium 2 has speed v₂. Play the construction (or step through it), change i with the slider or by dragging the incoming ray, and pick medium 2.
The end B of the wavefront is still in medium 1.
A is already in medium 2, so its wavelet uses the speed v₂. The tangent from C to this wavelet is the refracted wavefront CE.
Angle ABC and angle AEC are both 90° (a wavefront is tangent to the wavelet).
Using n = c/v. If v₂ < v₁, then r < i: the ray bends towards the normal.
"The whole proof is one sentence: in the same time, B covers v₁t in the first medium and A's wavelet covers v₂t in the second. Write those two sines and divide."
In the Huygens figure, BC is one wavelength in medium 1 if t is one period, and AE is one wavelength in medium 2. So λ₁/λ₂ = v₁/v₂. Watch a wave cross into the second medium. Two floats count the crests passing each side.
Crests cannot pile up or vanish at the boundary, so as many leave per second as arrive. The source fixes ν.
In glass (n = 1.5), 600 nm light has λ = 400 nm and v = 2 × 10⁸ m/s.
A red ribbon looks red under water. Photon energy E = hν also stays the same; only the speed changes.
"Speed and wavelength both shrink by the same factor n. Frequency? Never. In the parade, the drum beat never changes, only the step length."
"If light is slower in glass, does the light lose energy when it enters glass?"
No (apart from a little reflection and absorption). NCERT: the energy carried by a wave depends on its amplitude, not on its speed. And photon energy hν is fixed because ν is fixed.
Now the wave goes from a denser medium (top) into a rarer one (bottom), where it is faster. The wavelet from A is now bigger than BC. Raise i and watch what happens when the wavelet becomes so big that C is inside it.
So sin r = AE/AC > sin i: the ray bends away from the normal.
When AE = AC, the tangent from C touches the boundary itself: r = 90°.
Rarer over denser. For i > i_c there is no refracted wavefront and all the light is reflected: total internal reflection.
| Pair | n₁ → n₂ | sin i_c | i_c |
|---|---|---|---|
| Glass → air | 1.5 → 1 | 2/3 | 41.8° |
| Water → air | 4/3 → 1 | 3/4 | 48.6° |
| Diamond → air | 2.42 → 1 | 0.413 | 24.4° |
| Glass → water | 1.5 → 4/3 | 8/9 | 62.7° |
"The geometry tells you why TIR happens: there is simply no tangent line from C. Rare over dense, always less than one."
A plane wavefront AB falls on a mirror. This time the wavelet from A stays in the same medium, so it grows at the same speed v. Play it, step through it, or change i.
Reflection: same medium, same speed. That is the whole trick.
Both are right-angled (at E and at B), share the hypotenuse AC, and have AE = BC (RHS).
Angle between AB and the mirror is i; between CE and the mirror is r. Congruent triangles make them equal.
"Refraction: two different radii, v₁t and v₂t. Reflection: the same radius v t. Same construction, one change."
Glass slows light down, so the part of a wavefront that crosses more glass is held back more. Pick an object and watch a plane wavefront go through it. Change the refractive index to see the effect.
The base is thicker, so the lower part lags: rays bend towards the base. δ ≈ (n − 1)A.
The centre is thickest and lags most. The wavefront curves round and shrinks to the focus F.
The edges of the mirror are nearer, so the edges reflect first and get ahead: converging to F.
"NCERT's key sentence: the total time taken from a point on the object to the corresponding point on the image is the same along every ray. The thick middle of the lens is shorter in distance but slower; the edges are longer but faster."
Speeds and wavelengths go on top in the same order; refractive indices go upside down.
600 nm in glass (1.5): 400 nm. In water (4/3): 450 nm.
Glass 41.8°, water 48.6°, diamond 24.4°. Bigger n, smaller i_c.
Glass → water: n₂/n₁ = (4/3)/(3/2) = 8/9, less than 1, so water is rarer than glass.
Optical path n t versus air path t. Used again in Young's experiment (Lesson 4).
Wave model. The corpuscular model said the opposite (faster in denser).
Doppler effect for light. Older NCERT editions (§10.3.4) add: if a source moves away from you with speed v ≪ c, the frequency you receive drops: Δν/ν ≈ −v_radial/c (red shift); approaching gives a blue shift. This section is not in the rationalised NCERT, so it is only for awareness.
"Most NEET questions here take ten seconds if you write the chain sin i over sin r equals v₁ over v₂ equals λ₁ over λ₂ equals n₂ over n₁, and remember: frequency never changes."
Cover the solution, write the formula, and solve on paper. Open the hint only if you are stuck for more than a minute.
Yellow light of wavelength 589 nm in air enters water (n = 4/3). Find its speed, wavelength and frequency in water.
v = c/n, λ = λ₀/n, and the frequency is the same as in air: ν = c/λ₀.
A plane wavefront in air meets a glass surface at i = 60°. The glass has n = √3. Find the angle of refraction and the speed of light in the glass.
sin r = sin i / n; v = c/n.
In the Huygens construction for air → glass (n = 1.5), AC = 10 cm and i = 30°. Find BC, the wavelet radius AE, the angle r and the time t taken by B to reach C.
BC = AC sin i; AE = BC × v₂/v₁ = BC/n; sin r = AE/AC; t = BC/c.
Find the critical angle for light going from glass (n = 1.5) into water (n = 4/3).
sin i_c = n_rare / n_dense.
A glass plate 6 mm thick (n = 1.5) is placed across part of a plane wavefront. How much later does that part of the wavefront come out, compared with the part that went through air? What extra air distance is this equal to?
Time in glass n t/c, time in the same thickness of air t/c.
Take c = 3 × 10⁸ m/s. Write the formula first, then click.
Score: 0 / 12
| If the question says… | Write this first | The trap |
|---|---|---|
| "prove Snell's law using Huygens principle" | BC = v₁t, AE = v₂t, sin i = BC/AC, sin r = AE/AC | Using v₁t for the wavelet in medium 2 |
| "wavelength / speed in the medium" | λ = λ₀/n, v = c/n | Multiplying by n instead of dividing |
| "which quantity is unchanged" | frequency (and colour, photon energy) | Choosing wavelength because "colour depends on wavelength" |
| "critical angle" | sin i_c = n_rare / n_dense | Writing n_dense / n_rare (greater than 1, impossible) |
| "light goes from glass to water" | relative index 8/9, i_c = 62.7° | Using 1/1.5 as if the second medium were air |
| "proof of the law of reflection" | AE = BC = v t, ΔEAC ≅ ΔBCA | Forgetting to say the speed is the same (same medium) |
| "wavefront after a convex lens / concave mirror" | spherical, converging to F | Saying plane (that is for a source at F) |
| "sin i / sin r = 1.5, find v₂" | v₂ = v₁ / 1.5 | Corpuscular answer v₂ = 1.5 v₁ |
Lesson 3 lets two wavefronts overlap: superposition, coherent sources and interference. The idea of extra optical path, (n − 1)t, comes back when a glass sheet shifts Young's fringes in Lesson 4.
Values used: c = 3 × 10⁸ m/s, n(water) = 4/3, n(glass) = 1.5, n(diamond) = 2.42, visible λ ≈ 400–700 nm.
Follow the four steps: understand it, remember it, try it, then prove it with past papers.