Login

Concepts

Read through concept notes with worked explanations and diagrams, chapter by chapter.

Concept 01 / 07Wave Optics

Huygens' Principle and Wavefronts: Deriving Reflection and Refraction from a Wave Picture

A wavefront is a surface of constant phase; treating every point on it as a source of secondary wavelets reproduces the laws of reflection and refraction and explains why wavelength, not frequency, changes in a medium.

Figure 1
A wavefront is the locus of all points of a wave that are vibrating in the same phase at a given instant. A source emits continuously, so there is a whole family of such surfaces, one for every value of the phase; we usually draw the ones passing through the crests, separated by one wavelength λ\lambda. The perpendicular to the wavefront at a point gives the direction in which energy travels there, and that perpendicular is exactly what geometrical optics calls a ray. Rays and wavefronts are therefore two dialects of the same physics. Rays are convenient when λ\lambda is negligible compared with the apertures and obstacles; wavefronts become indispensable the moment it is not, which is the whole of wave optics. Shapes and amplitudes. A point source in a homogeneous isotropic medium sends out spherical wavefronts; a long narrow slit source sends out cylindrical wavefronts; a source at infinity, or a point source at the focus of a converging lens, gives plane wavefronts. Energy conservation fixes how the amplitude behaves. The same power PP crosses every wavefront, so for a sphere I=P/4πr2I = P/4\pi r^2, giving I∝1/r2I \propto 1/r^2 and amplitude A∝1/rA \propto 1/r. For a cylinder the area goes as rr, so I∝1/rI \propto 1/r and A∝1/rA \propto 1/\sqrt{r}. For a plane wavefront the area does not change at all and the amplitude stays constant, which is why a well collimated laser beam does not fade with distance. Huygens' principle has three clauses. First, every point of a wavefront acts as an independent source of secondary spherical wavelets. Second, these wavelets travel forward with the speed of the wave in that medium, so in time tt each has radius vtvt. Third, the new position of the wavefront after time tt is the forward envelope, that is, the common tangent surface, of all these secondary wavelets. Law of reflection. Let a plane wavefront ABAB strike a plane mirror, with AA touching the mirror first and BB still a distance away. Let BB reach the mirror at the point CC after a time tt, so BC=vtBC = vt. During that same time the secondary wavelet born at AA has grown into a hemisphere of radius AD=vtAD = vt on the same side of the mirror. The reflected wavefront is the tangent CDCD drawn from CC to that hemisphere. Now compare the two right triangles ABCABC and ADCADC. They share the hypotenuse ACAC, the angle at BB and the angle at DD are both right angles, and BC=AD=vtBC = AD = vt. So the triangles are congruent, which forces the angle BACBAC to equal the angle DCADCA. These two angles are precisely the angle of incidence ii and the angle of reflection rr, because the angle between a wavefront and the surface always equals the angle between the corresponding ray and the normal. Hence i=ri = r. Equivalently sin⁡i=BC/AC\sin i = BC/AC and sin⁡r=AD/AC\sin r = AD/AC, and since BC=ADBC = AD we get sin⁡i=sin⁡r\sin i = \sin r. The incident ray, the normal and the reflected ray also lie in one plane because the whole construction does. Law of refraction. Repeat the construction at a boundary between two media with speeds v1v_1 and v2v_2, as in the diagram above. In the time tt taken by BB to travel to CC in medium 1, the wavelet from AA spreads into medium 2 with the smaller speed v2v_2, so its radius is only AD=v2tAD = v_2 t while BC=v1tBC = v_1 t. The refracted wavefront is the tangent CDCD. From the two right triangles sharing the hypotenuse ACAC, sin⁡i=BC/AC=v1t/AC\sin i = BC/AC = v_1 t/AC and sin⁡r=AD/AC=v2t/AC\sin r = AD/AC = v_2 t/AC. Dividing, sin⁡isin⁡r=v1v2\dfrac{\sin i}{\sin r} = \dfrac{v_1}{v_2}, a constant for the given pair of media. Writing n=c/vn = c/v for the refractive index, v1/v2=n2/n1v_1/v_2 = n_2/n_1, so n1sin⁡i=n2sin⁡rn_1 \sin i = n_2 \sin r. Snell's law drops out of the geometry with no extra assumptions, and it comes with a bonus that the ray picture could never supply: light must travel slower in the optically denser medium. Newton's corpuscular theory predicted the opposite, and Foucault's 1850 measurement in water settled the argument in favour of waves. What changes at a boundary and what does not. The boundary is driven by the incident wave and re-radiates at exactly the frequency that shakes it, so frequency ν\nu is the same in both media. Since v=νλv = \nu\lambda and vv drops by the factor nn, the wavelength must drop by the same factor, λmedium=λvacuum/n\lambda_{medium} = \lambda_{vacuum}/n. This single fact drives a large part of wave optics: it is why a slab of thickness tt and index μ\mu contains μt/λ\mu t/\lambda waves instead of t/λt/\lambda, and hence why it adds an extra optical path (μ−1)t(\mu - 1)t compared with the same thickness of vacuum. Optical path is always μ×\mu \times geometric path, and phase difference is always Δϕ=2πλ×\Delta\phi = \dfrac{2\pi}{\lambda}\times (optical path difference), with λ\lambda measured in vacuum. Total internal reflection also follows. Going from denser to rarer, sin⁡r=(n1/n2)sin⁡i\sin r = (n_1/n_2)\sin i with n1>n2n_1 > n_2. Beyond the critical angle the required sin⁡r\sin r exceeds 11, so no tangent plane to the secondary wavelets exists in the second medium: no refracted wavefront can be constructed, and all the energy goes into the reflected wavefront. Gotchas that JEE Advanced likes. First, Huygens' construction as originally stated also predicts a backward wave, because a spherical wavelet spreads in all directions. The cure is Fresnel's obliquity factor 1+cos⁡θ2\dfrac{1 + \cos\theta}{2}, which multiplies the amplitude of a wavelet emitted at an angle θ\theta to the forward direction; it equals 11 straight ahead and 00 straight back, killing the backward wave. Second, Huygens' principle is a construction for where the wavefront goes, not a dynamical law: by itself it gives directions, not intensities. The moment you let the secondary wavelets interfere with each other rather than merely taking their envelope, you get the Huygens-Fresnel principle, and out of it come interference and diffraction. Third, shapes of wavefronts after optical elements are a favourite: a plane wavefront becomes converging spherical after a converging lens, diverging spherical after a diverging lens, and stays plane but tilted after a prism, because the part of the wavefront that traverses the thicker part of the glass is delayed most. Key Takeaways and Formulae: A wavefront is a surface of constant phase; rays are its normals. Intensity I∝A2I \propto A^2. Spherical wave: A∝1/rA \propto 1/r, I∝1/r2I \propto 1/r^2. Cylindrical wave: A∝1/rA \propto 1/\sqrt{r}, I∝1/rI \propto 1/r. Plane wave: AA constant. Huygens: new wavefront = forward envelope of secondary wavelets of radius vtvt. Reflection: i=ri = r, from BC=AD=vtBC = AD = vt. Refraction: sin⁡isin⁡r=v1v2=n2n1=1n2\dfrac{\sin i}{\sin r} = \dfrac{v_1}{v_2} = \dfrac{n_2}{n_1} = {}^{1}n_2. In a medium: ν\nu unchanged, v=c/nv = c/n, λmedium=λvacuum/n\lambda_{medium} = \lambda_{vacuum}/n. Optical path =μ×= \mu \times geometric path; a slab of thickness tt adds (μ−1)t(\mu - 1)t; phase difference Δϕ=2πλΔx\Delta\phi = \dfrac{2\pi}{\lambda}\Delta x. Common mistake: writing λ\lambda inside a medium as the vacuum value. Always convert, or work with optical paths and keep λ\lambda as the vacuum wavelength. Common mistake: thinking the secondary wavelets are real new sources that can be detected individually. They are a bookkeeping device that reproduces the correct envelope. Brain Teaser 1: A plane wavefront in air strikes a still water surface at an angle of incidence of 45∘45^\circ. For water μ=4/3\mu = 4/3. Find the acute angle between the reflected wavefront and the refracted wavefront. Solution: First get the refraction angle. sin⁡r=sin⁡45∘4/3=0.7071×34=0.5303\sin r = \dfrac{\sin 45^\circ}{4/3} = \dfrac{0.7071 \times 3}{4} = 0.5303, so r=32.0∘r = 32.0^\circ. Now work with rays, because the angle between two planes equals the angle between their normals, and the rays are exactly the normals to the wavefronts. The reflected ray leaves at 45∘45^\circ on the other side of the normal, in medium 1; the refracted ray goes on at 32.0∘32.0^\circ from the same normal, into medium 2. Going round from the reflected ray to the refracted ray through the interface, the three angles at the point of incidence add to 180∘180^\circ: angle between them =180∘−45∘−32.0∘=103.0∘= 180^\circ - 45^\circ - 32.0^\circ = 103.0^\circ. Since each wavefront is perpendicular to its own ray, the two wavefront planes meet at the same pair of angles, 103.0∘103.0^\circ and its supplement 77.0∘77.0^\circ. The acute angle asked for is therefore about 77∘77^\circ. A useful check on the method: if the second medium had μ→1\mu \to 1, then r→45∘r \to 45^\circ and the angle between the rays would become 90∘90^\circ, which is right, since reflected and refracted rays then differ only by the sign of the normal component. Brain Teaser 2: Monochromatic light of vacuum wavelength 600600 nm passes through a glass slab of thickness 66 μ\mum and refractive index 1.51.5. How many complete waves does the light contain inside the slab, how many would the same 66 μ\mum of vacuum contain, and what is the resulting extra optical path? Use the result to explain why frequency, and not wavelength, is the quantity conserved at a boundary. Solution: Inside the glass the wavelength is λ′=λ/μ=600/1.5=400\lambda' = \lambda/\mu = 600/1.5 = 400 nm. Number of waves in the slab =t/λ′=6×10−6400×10−9=15= t/\lambda' = \dfrac{6 \times 10^{-6}}{400 \times 10^{-9}} = 15. In the same geometric thickness of vacuum, the number of waves is t/λ=6×10−6600×10−9=10t/\lambda = \dfrac{6 \times 10^{-6}}{600 \times 10^{-9}} = 10. The slab therefore packs 55 extra waves into the same space, corresponding to an extra optical path of 5λ=35\lambda = 3 μ\mum. Check against the formula: (μ−1)t=0.5×6=3(\mu - 1)t = 0.5 \times 6 = 3 μ\mum. They agree. Why frequency is conserved: wavefronts cannot be created or destroyed at the boundary, and they cannot pile up there, because the boundary has no room to store them. So the number of wavefronts arriving per second in medium 1 must equal the number leaving per second in medium 2, which is the statement that ν\nu is the same on both sides. Physically, the electrons at the surface are driven by the incident field and, in a linear medium, re-radiate at precisely the driving frequency. Since vv changes and ν\nu cannot, λ\lambda is forced to change, exactly as the count above shows. Brain Teaser 3: A plane wavefront falls normally on a thin converging lens of focal length ff made of glass of index μ\mu. Using only the idea that light is slowed inside glass, show that the emergent wavefront is spherical with its centre at the focus, and prove that every part of the beam arrives at the focus in phase. Solution: The lens is thickest on the axis and thins out towards the rim. Light through the centre spends the longest time in the slow medium, so the central part of the wavefront is delayed most and lags behind the edges. A plane wavefront with its middle pulled back is a wavefront curved so as to converge, so the emergent wavefront is spherical and closing in on a point on the axis. To prove that point is a genuine common focus, compare optical paths from the emergent wavefront to the axial point FF at distance ff. Take a ray at height hh from the axis. Its remaining geometric path to FF is f2+h2≈f+h22f\sqrt{f^2 + h^2} \approx f + \dfrac{h^2}{2f}, which is longer than the axial ray's path ff by h22f\dfrac{h^2}{2f}. That excess must be cancelled by the extra glass the axial ray had to cross. For a thin lens of surface radii R1R_1 and R2R_2, the thickness at height hh is t(h)=t0−h22(1R1−1R2)t(h) = t_0 - \dfrac{h^2}{2}\left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right), so the extra optical path collected by the central ray relative to the ray at height hh is (μ−1)[t0−t(h)]=(μ−1)h22(1R1−1R2)=h22f(\mu - 1)\left[t_0 - t(h)\right] = (\mu - 1)\dfrac{h^2}{2}\left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right) = \dfrac{h^2}{2f}, using the lensmaker's formula 1f=(μ−1)(1R1−1R2)\dfrac{1}{f} = (\mu - 1)\left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right). The two excesses are equal and opposite, so the total optical path from the incoming plane wavefront to FF is the same for every hh. Equal optical path means equal phase, so all the wavelets arrive at FF in step and interfere constructively. This is the wave-optics meaning of a focus, and it shows that the lensmaker's formula is exactly the condition for the lens to convert a plane wavefront into a spherical one.