The Wave Model of Light

Maxwell's equations gave a wave speed $c = \dfrac{1}{\sqrt{\mu_0\varepsilon_0}} \approx 3\times10^8\text{ m/s}$, matching the measured speed of light, so light was identified as an electromagnetic wave.

$\lambda = \dfrac{c}{f} = \dfrac{3.0\times10^8}{3.0\times10^6} = 100\text{ m}$.

An EM wave is transverse: $\vec{E}$ and $\vec{B}$ oscillate at right angles to each other and to the direction of propagation, and the wave is self-sustaining, needing no medium.

Hertz generated and detected radio waves and showed they reflect, refract, polarise, interfere and travel at $c$, i.e. they behave exactly like light. This confirmed Maxwell's prediction that light is an EM wave.

The cool gas absorbs exactly the wavelengths it would emit when hot, removing them from the continuous light. The result is dark absorption lines at those characteristic wavelengths.

Bright (constructive) fringes need the two waves in phase, so the path difference is a whole number of wavelengths: $d\sin\theta = m\lambda$. Dark fringes occur at $(m+\tfrac12)\lambda$.

$\Delta y = \dfrac{\lambda L}{d} = \dfrac{(600\times10^{-9})(1.5)}{0.20\times10^{-3}} = \dfrac{9.0\times10^{-7}}{2.0\times10^{-4}} = 4.5\times10^{-3}\text{ m} = 4.5\text{ mm}$.

$1.5\lambda = (1+\tfrac12)\lambda$ is an odd number of half-wavelengths, so the waves arrive exactly out of phase and cancel: the point is dark.

Diffraction is significant when the aperture is about the size of the wavelength or smaller. Through a gap much wider than $\lambda$ the wave barely spreads.

$d = \dfrac{1}{\text{lines per metre}} = \dfrac{1\times10^{-3}\text{ m}}{400} = 2.5\times10^{-6}\text{ m}$.

$\sin\theta = \dfrac{m\lambda}{d} = \dfrac{(1)(500\times10^{-9})}{2.5\times10^{-6}} = 0.20$, so $\theta = \sin^{-1}(0.20) = 11.5^\circ$.

Only a transverse wave has a plane of vibration that a filter can select, so polarisation is direct evidence light is transverse. Longitudinal waves (like sound) vibrate along the direction of travel and cannot be polarised.

Malus' law: $I = I_0\cos^2\theta = I_0\cos^2 30^\circ = I_0(0.866)^2 = 0.75\,I_0$.

For the $m^{\text{th}}$ bright fringe $y_m = \dfrac{m\lambda L}{d}$, so $\lambda = \dfrac{y_m\,d}{mL} = \dfrac{(0.024)(0.25\times10^{-3})}{(5)(2.0)} = \dfrac{6.0\times10^{-6}}{10} = 6.0\times10^{-7}\text{ m} = 600\text{ nm}$.

$\Delta y = \dfrac{\lambda L}{d}$. Doubling $L$ multiplies by $2$ and halving $d$ multiplies by $2$ again, so $\Delta y$ increases by a factor of $4$.

A maximum needs $\sin\theta = \dfrac{m\lambda}{d} \le 1$, so $m \le \dfrac{d}{\lambda} = \dfrac{2.5\times10^{-6}}{600\times10^{-9}} = 4.17$. The highest whole order is $m = 4$ ($m=5$ would need $\sin\theta > 1$).

$\sin\theta_{\text{blue}} = \dfrac{450\times10^{-9}}{2.0\times10^{-6}} = 0.225 \Rightarrow \theta_{\text{blue}} = 13.0^\circ$. $\sin\theta_{\text{red}} = \dfrac{650\times10^{-9}}{2.0\times10^{-6}} = 0.325 \Rightarrow \theta_{\text{red}} = 19.0^\circ$. The separation is about $6.0^\circ$ (red bends more).

$\dfrac{I}{I_0} = \cos^2\theta = 0.25$, so $\cos\theta = 0.5$ and $\theta = 60^\circ$.

Without the middle filter the crossed polarisers would pass nothing. The $45^\circ$ filter rotates the polarisation: $I_0 \to I_0\cos^2 45^\circ = \tfrac12 I_0 \to \tfrac12 I_0\cos^2 45^\circ = \tfrac14 I_0$. (For unpolarised incident light the first filter would halve it, giving $I_0/8$, the trap.)

Spectra reveal composition (which lines are present), temperature (the continuous peak and line ratios), radial velocity and rotation (Doppler shift and broadening), and even density and magnetic field (line shapes/splitting). Distance is not read directly from a spectrum; it needs other methods.

$d = \dfrac{m\lambda}{\sin\theta} = \dfrac{2(600\times10^{-9})}{\sin 31^\circ} = \dfrac{1.2\times10^{-6}}{0.515} = 2.33\times10^{-6}\text{ m}$. Lines per metre $= \dfrac{1}{d} = 4.3\times10^{5}$, so about $430$ lines/mm. Forgetting the order ($m=2$) gives the trap $860$.

Both obey $d\sin\theta = m\lambda$, so the maxima sit at the same angles. But many slits interfere so that the bright maxima become very narrow and intense, letting a grating resolve closely spaced wavelengths a double slit cannot.

$\dfrac{1500}{600} = 2.5$, so the path difference is $2.5\lambda = (2+\tfrac12)\lambda$, an odd number of half-wavelengths. The waves arrive out of phase and cancel, so P is dark.

At $m=0$ all colours have $\theta = 0$, so the centre is white. For $m\ge1$, $\sin\theta = \dfrac{m\lambda}{d}$ depends on $\lambda$, so each order fans into a spectrum with red bent the most (unlike a prism). At higher orders the long-wavelength end of one order can overlap the short-wavelength end of the next.

In water the frequency is unchanged but the speed drops to $c/n$, so the wavelength becomes $\lambda/n$. Since $\Delta y = \dfrac{\lambda L}{d} \propto \lambda$, the spacing shrinks by the same factor: $\Delta y_{\text{water}} = \dfrac{\Delta y}{1.33} \approx 0.75\,\Delta y$.