Wave Optics

A flat surface separates medium 1, where light travels at speed $v_1$, from medium 2, where it travels at $v_2$. A plane wavefront AB arrives at an angle of incidence $i$, tilted so that its lower end, point A, has just touched the surface, while its upper end, point B, still has a distance BC to travel (along the ray) before it reaches the surface at point C. Call the time B takes to cover that distance $\tau$:

Point A has not been idle in the meantime. The instant the wavefront touched it, A became the source of a secondary wavelet spreading into medium 2 at that medium’s own speed, $v_2$. So by the time B reaches C, A’s wavelet is a hemisphere (below the surface) of radius:

Every point of the surface between A and C launches its own wavelet too, as the wavefront sweeps across it; the closer a point is to C, the later it starts, and the smaller its wavelet has grown. The refracted wavefront is the envelope touching all of them: the plane CE drawn from C just tangent to A’s wavelet, touching it at point E. Now compare two right-angled triangles that share the same hypotenuse AC, the strip of surface between A and C.

Triangle ABC has its right angle at B (BC runs along the incident ray, which is perpendicular to the wavefront AB), and its angle at A is the angle between the incident wavefront and the surface, which by the fact above is $i$. Triangle AEC has its right angle at E (a tangent always meets the radius AE at 90°), and its angle at C is the angle between the refracted wavefront CE and the surface, which is the angle of refraction $r$. Reading off the side opposite each of those angles:

Divide the first by the second, and both $\tau$ and $AC$ cancel:

The refractive index of a medium is defined as $n = c/v$, so $v_1 = c/n_1$ and $v_2 = c/n_2$:

Notice what the wave picture predicts along the way. If the ray bends toward the normal ($r<i$), then $\sin i/\sin r$ is bigger than 1, so $v_2$ must be smaller than $v_1$: light has to travel more slowly in the denser medium. The older particle (corpuscular) model of light predicted exactly the opposite, and Foucault’s 1850 measurement of the speed of light in water settled the question in favour of waves.

One more result comes for free. The frequency $\nu$ of the light is fixed by the source: every point of the boundary is shaken by the incident wave and simply follows along at its frequency, so $\nu$ is the same on both sides. Since $v = \nu\lambda$ holds in each medium:

So light entering a denser medium slows down, and its wavelength shrinks by exactly the same factor, while its frequency stays put.

Reflection is the same construction with one change: the wavelet from A now spreads back into medium 1, the medium the light came from (above the surface, on the same side as the incoming light), so it travels at the same speed $v$ as the incident wave itself. By the time B covers $BC = v\tau$ to reach the surface at C, A’s wavelet has grown to radius $AE = v\tau$, exactly equal to BC. The reflected wavefront is again the tangent CE drawn from C to this wavelet.

The same two triangles appear: ABC, right-angled at B, with its angle at A equal to the angle of incidence $i$; and AEC, right-angled at E, with its angle at C equal to the angle of reflection $r$. Reading off the opposite sides once more:

The right-hand sides are identical, so $\sin i = \sin r$, and since both angles lie between 0° and 90°, the angles themselves must be equal:

In the language of geometry, the two triangles share the hypotenuse AC and have one other side equal (BC = AE), so they are congruent by RHS, and their matching angles $i$ and $r$ are equal. Refraction differs from reflection only because, there, the wavelet from A grows at a different speed from the incident wave, which is what breaks the symmetry between the two triangles and bends the ray.

By the superposition principle, the actual displacement at P is simply the sum $y = y_1 + y_2$. Expand the second cosine using $\cos(A+B) = \cos A\cos B - \sin A\sin B$, then group the $\cos\omega t$ terms and the $\sin\omega t$ terms together:

The two brackets are just constants; they don’t change with time. Give them names, using a new amplitude $A$ and a new phase $\delta$:

so that, running the same cosine identity backwards:

The combined disturbance is still a plain oscillation at the same frequency, just with a new amplitude $A$. To find it, square the two defining equations and add them; $\delta$ disappears from the left because $\cos^2\delta + \sin^2\delta = 1$, and the same identity tidies up the right:

Intensity is proportional to the square of the amplitude, $I = kA^2$, with the same constant $k$ for every wave here. Each wave on its own therefore gives $I_1 = ka_1^2$ and $I_2 = ka_2^2$, and the cross term can be written as $k\,a_1a_2 = \sqrt{(ka_1^2)(ka_2^2)} = \sqrt{I_1I_2}$. Multiplying the amplitude equation through by $k$:

The first two terms are what you would get by just adding the two intensities. The third, $2\sqrt{I_1I_2}\cos\phi$, is the interference term: positive where the waves arrive roughly in step, negative where they arrive roughly opposed, and the entire reason fringes exist. If the phase difference drifts randomly, as it does for the incoherent sources of §10.3, $\cos\phi$ averages out to zero and only $I = I_1 + I_2$ survives: uniform brightness, no fringes, which is exactly the washed-out pattern described there.

$\cos\phi$ can be no larger than +1, which happens when $\phi = 0, \pm2\pi, \pm4\pi, \ldots$ (the waves arrive exactly in step). The interference term is then as large as it can be, and the three terms form a perfect square:

$\cos\phi$ can be no smaller than −1, which happens when $\phi = \pm\pi, \pm3\pi, \ldots$ (crest arriving on trough). The interference term is now as negative as it can be:

To connect these phase conditions to path differences: a path difference of one whole wavelength puts the second wave exactly one full cycle, a phase of $2\pi$, behind the first. A path difference $\Delta$ is the fraction $\Delta/\lambda$ of a wavelength, so it produces the same fraction of $2\pi$:

So $\phi = 2n\pi$ is the same statement as $\Delta = n\lambda$, and $\phi = (2n+1)\pi$ is the same as $\Delta = \left(n+\tfrac12\right)\lambda$: precisely the bright and dark fringe conditions used for Young’s experiment in §10.5. Notice too that unless $I_1 = I_2$, the minimum is not zero: two unequal waves can never cancel completely, so their dark fringes are only dim, never truly black.

In the most common case, two identical slits, both waves have the same intensity, $I_1 = I_2 = I_0$, and the general result tidies up considerably:

Using the half-angle identity $1+\cos\phi = 2\cos^2(\phi/2)$:

Now $I_{\max} = 4I_0$ (at $\phi = 0$) and $I_{\min} = 0$ (at $\phi = \pi$), which the general formulas above confirm: $(\sqrt{I_0}+\sqrt{I_0})^2 = 4I_0$ and $(\sqrt{I_0}-\sqrt{I_0})^2 = 0$. The bright fringes are four times as bright as either wave alone (not twice, because it is the amplitudes that add, and the sum then gets squared), and the dark fringes are genuinely black. No energy has been created at the bright fringes, only moved there from the dark ones: $\cos^2$ averages to $\tfrac12$ across a fringe, so the average intensity is $4I_0\times\tfrac12 = 2I_0$, exactly the total the two waves bring with them.

(The current NCERT text stops at Malus’s law; this result is added here because it is a standard, frequently tested extension of the same polarization ideas.) Ordinary reflection also polarizes light, partially at most angles, but completely at one special angle of incidence, the Brewster angle $\theta_B$. At this angle, the reflected and refracted rays turn out to be exactly perpendicular to each other, $\theta_B + r = 90°$, where $r$ is the angle of refraction. Substituting $r = 90°-\theta_B$ into Snell’s law for a ray going from air into a medium of index $n$:

At this one angle, the reflected light is purely polarized, vibrating only perpendicular to the plane of incidence (the component that would vibrate within that plane cannot be reflected at all once the reflected and refracted rays are perpendicular, since a dipole cannot radiate along its own axis of oscillation). This is why polarized sunglasses, oriented to block exactly this component, cut glare from water and glass so effectively: most glare is reflected light sitting near its own Brewster angle.