Optics and Waves Domain Guide

The medium owns the speed and the source owns the frequency. Every 'what happens when the wave enters...' question resolves the moment you say that sentence: the wavelength must absorb the entire change. It is also why a standing wave on a string has a fixed set of allowed wavelengths — the boundary conditions, not the driver, decide which wavelengths fit.

Interference questions are path-difference questions in disguise. Find the extra distance one route travels, express it in wavelengths, and ask whether that is a whole number or a half-integer. Every formula in wave optics — double slit, thin film, grating, single slit — is this one procedure with a different geometry supplying the path difference, so learning the procedure is cheaper and safer than learning four formulas that superficially resemble each other.

Diffraction and interference are the same phenomenon at different scales, which is why a real double-slit pattern is an interference comb inside a diffraction envelope. The slit separation $d$ sets how closely the fringes are packed; the slit width $a$ sets how far the pattern extends before fading. Because $d > a$ always, the fringes are always finer than the envelope, and the ratio $d/a$ tells you exactly which orders go missing.

ETS lists optics and wave phenomena as a content area, and our learning model groups mechanical waves, wave optics, and geometric optics beneath it. Begin with the mechanical layer, because it fixes the vocabulary. A transverse wave on a string travels at $v = \sqrt{T/\mu}$ with $T$ the tension and $\mu$ the mass per unit length; sound in an ideal gas travels at $v = \sqrt{\gamma RT/M}$, so it speeds up with temperature and slows down in a heavier gas. Intensity falls as $1/r^{2}$ for a spherical wave and is proportional to the square of the amplitude, which is why doubling the amplitude quadruples the power.

Standing waves turn a continuous medium into a discrete spectrum. A string fixed at both ends and a pipe open at both ends both give $f_n = nv/2L$ for $n = 1,2,3,\ldots$ — all harmonics present. A pipe closed at one end must have a node at the closed end and an antinode at the open end, so only odd multiples fit: $f_n = nv/4L$ with $n = 1,3,5,\ldots$, and its fundamental is an octave below the open pipe of the same length. Two slightly detuned sources produce beats at $f_{\text{beat}} = |f_1 - f_2|$, an audible demonstration that superposition acts on amplitudes rather than intensities.

The Doppler effect for sound is one formula with four sign choices: $f' = f\dfrac{v \pm v_o}{v \mp v_s}$, where the upper signs apply when the observer moves toward the source and when the source moves toward the observer. Rather than memorising the sign table, use the physical check: any motion that closes the gap raises the observed frequency, and any motion that opens it lowers the frequency. Choose the signs that produce that result and you cannot get it wrong. Light requires the relativistic formula instead, $f' = f\sqrt{(1-\beta)/(1+\beta)}$ for a receding source, because there is no medium to measure motion against.

Geometric optics is the ray limit, valid when obstacles are much larger than the wavelength. Snell's law $n_1\sin\theta_1 = n_2\sin\theta_2$ governs refraction, and going from dense to less dense allows total internal reflection beyond the critical angle $\sin\theta_c = n_2/n_1$. Mirrors and thin lenses share the equation $\dfrac{1}{s_o}+\dfrac{1}{s_i} = \dfrac{1}{f}$ with $m = -s_i/s_o$, and a spherical mirror has $f = R/2$. The sign convention carries the physics: for a converging lens or concave mirror $f > 0$; for a diverging lens or convex mirror $f < 0$; $s_i > 0$ means a real image on the far side of a lens or the same side of a mirror; $m < 0$ means inverted. A diverging lens or convex mirror can produce only an upright, reduced, virtual image, which is worth knowing as a fact so that a whole class of items can be answered without computation.

Wave optics returns when the obstacle is comparable to the wavelength. The double slit gives maxima at $d\sin\theta = m\lambda$ with $m = 0,\pm1,\pm2,\ldots$, and on a distant screen the fringe spacing is $\Delta y = \lambda D/d$. The single slit gives minima at $a\sin\theta = m\lambda$ with $m = \pm1,\pm2,\ldots$ and no $m = 0$, because the centre is the brightest point rather than a dark one. Note carefully that these two conditions have the same algebraic form and opposite meanings, which is precisely why the exam pairs them. A grating obeys the same maxima condition as the double slit, $d\sin\theta = m\lambda$, but with thousands of slits the maxima become extremely narrow, which is what makes a grating a spectrometer rather than a curiosity.

Thin films add one more ingredient: a reflection off a medium of higher refractive index flips the phase by half a wavelength, while a reflection off a lower-index medium does not. For a film of thickness $t$ and index $n$ in air, light reflecting off the top face gets a half-wave shift and light reflecting off the bottom face does not, so the net condition for constructive reflection becomes $2nt = (m+\tfrac{1}{2})\lambda_{\text{vac}}$ — the half-integer that would otherwise mean destructive. This is why a soap film's thinnest region looks black in reflection rather than bright, and why an anti-reflection coating is designed a quarter-wavelength thick in the coating material. Polarisation completes the picture: an ideal polariser passes half of unpolarised light, then Malus's law $I = I_0\cos^{2}\theta$ governs each subsequent polariser, and light reflected at Brewster's angle $\tan\theta_B = n_2/n_1$ is completely polarised parallel to the surface.