The Workbook

1. Snell's law: \(\sin i = \mu\sin r\). \(\sin r = \dfrac{\sin 45^\circ}{\sqrt{2}} = \dfrac{1}{2}\), so \(r = 30^\circ\).

2. The ray bends towards the normal by \(i - r = 45^\circ - 30^\circ = 15^\circ\).

1. \(v = \dfrac{c}{\mu} = \dfrac{3 \times 10^8}{1.5} = 2 \times 10^8\) m/s.

2. \(\lambda_g = \dfrac{\lambda}{\mu} = \dfrac{600}{1.5} = 400\) nm.

3. The frequency is fixed by the source and does not change: \(f = \dfrac{c}{\lambda} = \dfrac{3 \times 10^8}{6 \times 10^{-7}} = 5 \times 10^{14}\) Hz in both media.

1. \(_w\mu_g = \dfrac{\mu_g}{\mu_w} = \dfrac{3/2}{4/3} = \dfrac{9}{8}\) and \(_g\mu_w = \dfrac{\mu_w}{\mu_g} = \dfrac{8}{9}\).

2. \(\mu_g\sin i = \mu_w\sin r\), so \(\sin r = \dfrac{\mu_g}{\mu_w}\sin i = \dfrac{9}{8} \times 0.8 = 0.9\). Water is rarer than glass, so the ray bends away from the normal (\(r > i\)).

1. \(\mu_1\sin 60^\circ = \mu_2\sin 30^\circ\), so \(\dfrac{\mu_2}{\mu_1} = \dfrac{\sin 60^\circ}{\sin 30^\circ} = \dfrac{\sqrt{3}/2}{1/2} = \sqrt{3}\).

2. Medium 2. The ray bends towards the normal on entering it, which happens when light enters a denser medium.

3. \(\dfrac{v_1}{v_2} = \dfrac{c/\mu_1}{c/\mu_2} = \dfrac{\mu_2}{\mu_1} = \sqrt{3}\). Light is slower in the denser medium.

At the first face, \(\sin i = \mu\sin r\). The two faces are parallel, so the normals are parallel and the angle of incidence at the second face is the same \(r\).

At the second face, \(\mu\sin r = \sin e\). So \(\sin e = \sin i\) and \(e = i\). The emergent ray makes the same angle with the parallel normal, so it is parallel to the incident ray (but shifted sideways).

1. \(v = \dfrac{c}{\mu} = 2 \times 10^8\) m/s. \(t = \dfrac{0.03}{2 \times 10^8} = 1.5 \times 10^{-10}\) s.

2. Optical path \(= \mu \times\) distance \(= 1.5 \times 3 = 4.5\) cm. Check: in \(1.5 \times 10^{-10}\) s light in air covers \(3 \times 10^8 \times 1.5 \times 10^{-10} = 0.045\) m.

1. Apparent depth \(= \dfrac{\text{real depth}}{\mu} = \dfrac{4}{4/3} = 3\) m.

2. Normal shift \(= 4 - 3 = 1\) m. The stone appears \(1\) m higher than it is.

Apparent depth \(= \dfrac{t}{\mu} = \dfrac{6}{1.5} = 4\) cm. The letters appear raised by \(t - \dfrac{t}{\mu} = t\left(1 - \dfrac{1}{\mu}\right) = 6 \times \dfrac{1}{3} = 2\) cm.

1. Each layer contributes its own apparent depth: \(\dfrac{8}{4/3} + \dfrac{6}{1.5} = 6 + 4 = 10\) cm.

2. Real depth \(= 8 + 6 = 14\) cm. Shift \(= 14 - 10 = 4\) cm. Check: \(8 \times \dfrac{1}{4} + 6 \times \dfrac{1}{3} = 2 + 2 = 4\) cm.

1. The bird is in the rarer medium, viewed from the denser one. Apparent height \(= \mu \times 3 = 4\) m above the surface. Distance from the fish \(= 2 + 4 = 6\) m.

2. The fish is in the denser medium, viewed from the rarer one. Apparent depth \(= \dfrac{2}{\mu} = 1.5\) m below the surface. Distance from the bird \(= 3 + 1.5 = 4.5\) m.

1. The apparent depth of the fish is \(\dfrac{d}{\mu}\), so it changes at \(\dfrac{4}{4/3} = 3\) m/s. The bird itself dives at \(3\) m/s. The bird sees the fish approach at \(3 + 3 = 6\) m/s.

2. The apparent height of the bird is \(\mu h\), so it changes at \(\dfrac{4}{3} \times 3 = 4\) m/s. The fish itself rises at \(4\) m/s. The fish sees the bird approach at \(4 + 4 = 8\) m/s.

The slab raises the image of the dot by \(t\left(1 - \dfrac{1}{\mu}\right) = 3 \times \dfrac{1}{3} = 1\) cm. The microscope must be raised by \(1\) cm.

1. \(\sin r = \dfrac{\sin 45^\circ}{\sqrt{2}} = \dfrac{1}{2}\), so \(r = 30^\circ\).

2. \(d = \dfrac{10\sin(45^\circ - 30^\circ)}{\cos 30^\circ} = \dfrac{10 \times 0.26}{0.87} \approx 3\) cm.

Let the ray enter at \(A\) and leave at \(B\). Inside the slab the path \(AB\) makes angle \(r\) with the normal, so \(AB = \dfrac{t}{\cos r}\).

The emergent ray is parallel to the incident ray. The lateral shift \(d\) is the perpendicular distance from \(B\) to the line of the incident ray. The angle between \(AB\) and that line is \(i - r\), so \(d = AB\sin(i - r) = \dfrac{t\sin(i - r)}{\cos r}\).

1. For small angles \(\sin(i - r) \approx i - r\) and \(\cos r \approx 1\), so \(d \approx t(i - r)\). Snell's law becomes \(i = \mu r\), so \(r = \dfrac{i}{\mu}\) and \(d \approx t\,i\left(1 - \dfrac{1}{\mu}\right)\).

2. \(d \approx 6 \times 0.1 \times \left(1 - \dfrac{1}{1.5}\right) = 0.6 \times \dfrac{1}{3} = 0.2\) cm.

\(\sin r = \dfrac{\sin 90^\circ}{\sqrt{2}} = \dfrac{1}{\sqrt{2}}\), so \(r = 45^\circ\). \(d = \dfrac{t\sin(90^\circ - 45^\circ)}{\cos 45^\circ} = t = 10\) cm. At grazing incidence the shift is largest.

Let the object be at distance \(x\) in front of the slab and the eye beyond it. Rays from the object cross the slab and are shifted by \(t\left(1 - \dfrac{1}{\mu}\right)\) towards the eye, whatever the value of \(x\).

Reason: the shift is fixed by the refraction at the two faces, which depends only on \(t\) and \(\mu\). The air paths before and after the slab do not bend the ray. So the image position does not depend on where the slab sits.

1. \(\mu = \dfrac{\sin 60^\circ}{\sin 30^\circ} = \sqrt{3}\).

2. \(d = \dfrac{t\sin(i - r)}{\cos r} = \dfrac{6\sin 30^\circ}{\cos 30^\circ} = \dfrac{6 \times 1/2}{\sqrt{3}/2} = \dfrac{6}{\sqrt{3}} = 2\sqrt{3} \approx 3.46\) cm.

3. The faces are parallel, so the ray leaves at the angle of incidence, \(60^\circ\) to the normal.

1. All the faces are parallel, so the angle of refraction at one face equals the angle of incidence at the next. Snell's law at each face gives \(\sin i = \mu_1\sin r_1 = \mu_2\sin r_2 = \mu_3\sin r_3 = \sin e\). So \(e = i\) and the emergent ray is parallel to the incident ray.

2. From the chain, \(\sin r_2 = \dfrac{\sin i}{\mu_2} = \dfrac{\sin 45^\circ}{\sqrt{2}} = \dfrac{1}{2}\), so \(r_2 = 30^\circ\). \(\mu_1\) drops out: the angle in any slab depends only on that slab's index and the angle in air.

Shift \(= t\left(1 - \dfrac{1}{\mu}\right)\): \(3 = 12\left(1 - \dfrac{1}{\mu}\right)\), so \(1 - \dfrac{1}{\mu} = \dfrac{1}{4}\) and \(\mu = \dfrac{4}{3}\).

1. Shift \(= 9\left(1 - \dfrac{1}{1.5}\right) = 9 \times \dfrac{1}{3} = 3\) cm.

2. Under water the index that matters is the relative one, \(_w\mu_g = \dfrac{1.5}{4/3} = \dfrac{9}{8}\). Shift \(= 9\left(1 - \dfrac{8}{9}\right) = 1\) cm. The slab does less in water because it is only slightly denser than water.

Zero. The image of the letters is raised by \(t\left(1 - \dfrac{1}{\mu}\right) = 2\) cm, and this shift does not depend on the position of the slab (Question 17). The image stays \(2\) cm above the page while the slab moves.

The reflected ray makes \(60^\circ\) with the normal. If the refracted ray is perpendicular to it, then \(60^\circ + 90^\circ + r = 180^\circ\), so \(r = 30^\circ\).

\(\mu = \dfrac{\sin 60^\circ}{\sin 30^\circ} = \sqrt{3} \approx 1.73\).

1. Light crosses the slab once on the way down and once on the way up. Seen through the slab, the mirror appears raised by the normal shift \(t\left(1 - \dfrac{1}{\mu}\right) = 6 \times \dfrac{1}{3} = 2\) cm.

2. Replace the real mirror by an effective mirror \(2\) cm higher, \(18\) cm below \(O\). The plane mirror forms the image \(18\) cm behind it, that is \(18 + 18 = 36\) cm below \(O\).

   Check by steps: \(O\) is \(14\) cm above the glass. From inside the glass it appears \(14 \times 1.5 = 21\) cm above the top face, or \(27\) cm above the mirror. The mirror image is \(27\) cm below the mirror, \(33\) cm below the top face in glass. From air it appears at \(\dfrac{33}{1.5} = 22\) cm below the top face, that is \(14 + 22 = 36\) cm below \(O\).

1. \(\sin C = \dfrac{1}{\sqrt{2}}\), so \(C = 45^\circ\).

2. \(\sin C = \dfrac{1}{2}\), so \(C = 30^\circ\). A denser medium has a smaller critical angle.

3. \(\sin C = \dfrac{2}{3}\), so \(C \approx 42^\circ\).

First, the ray must travel from a denser medium towards a rarer one. Second, the angle of incidence must be larger than the critical angle, \(i > C\).

From air into glass the ray goes towards the denser medium. It bends towards the normal and always finds an angle of refraction, so it is never totally reflected.

At the critical angle the refracted ray grazes the surface: \(\mu_g\sin C = \mu_w\sin 90^\circ\), so \(\sin C = \dfrac{\mu_w}{\mu_g} = \dfrac{4/3}{3/2} = \dfrac{8}{9}\).

\(\dfrac{8}{9} > \dfrac{2}{3}\), so \(C\) is larger than for glass in air. Water is closer to glass in optical density, so the ray needs a larger angle before it is totally reflected.

1. \(\sin C = \dfrac{1}{\sqrt{2}}\), \(C = 45^\circ\). The angle of incidence \(60^\circ > C\), so the ray is totally reflected and does not emerge.

2. \(\sqrt{2}\sin 30^\circ = \sin r\) gives \(\sin r = \dfrac{\sqrt{2}}{2} = \dfrac{1}{\sqrt{2}}\), so \(r = 45^\circ\).

3. Grazing means \(r = 90^\circ\), which happens at \(i = C = 45^\circ\).

1. Total reflection needs \(45^\circ \ge C\), that is \(\sin 45^\circ \ge \dfrac{1}{\mu}\), so \(\mu \ge \sqrt{2} \approx 1.41\).

2. For \(\mu = 1.5\), \(C \approx 42^\circ < 45^\circ\), so the ray is totally reflected. It was moving horizontally and leaves vertically: it is turned through \(90^\circ\). It meets the bottom face normally and passes out without bending.

1. \(\sin C = \dfrac{\mu_{\text{clad}}}{\mu_{\text{core}}} = \dfrac{1.2}{1.5} = 0.8\), so \(C \approx 53^\circ\).

2. The angle of incidence at the wall is \(90^\circ\) minus the angle with the axis. Guiding needs \(i \ge 53^\circ\), so the angle with the axis is at most \(90^\circ - 53^\circ = 37^\circ\).

3. At the flat end the ray refracts from air into the core: \(\sin i_{\max} = 1.5\sin 37^\circ = 1.5 \times 0.6 = 0.9\). Any ray entering at less than about \(64^\circ\) to the axis is guided.

1. \(\sin C = \dfrac{4}{5}\), so \(\cos C = \dfrac{3}{5}\) and \(\tan C = \dfrac{4}{3}\).

2. \(r = h\tan C = 3 \times \dfrac{4}{3} = 4\) m. Area \(= \pi r^2 = 16\pi \approx 50\) m\(^2\).

\(r = \dfrac{h}{\sqrt{\mu^2 - 1}}\). \(\mu^2 - 1 = \dfrac{16}{9} - 1 = \dfrac{7}{9}\), so \(\sqrt{\mu^2 - 1} = \dfrac{\sqrt{7}}{3}\) and \(r = \sqrt{7} \times \dfrac{3}{\sqrt{7}} = 3\) m.

1. Rays from every direction above the water enter the eye of the fish within angle \(C\) of the vertical. The cone has vertex angle \(2C \approx 98^\circ\).

2. Rays from the bottom that meet the surface at more than \(C\) are totally reflected. Outside the cone the surface acts as a mirror, and the fish sees the reflection of the bottom of the lake.

Rays from the coin that reach the surface within radius \(r = h\tan C\) leave the liquid; the rest are totally reflected. The disc must cover that circle: \(r = 12 \times \dfrac{4}{3} = 16\) cm, so the diameter is \(32\) cm.

A ray from the source at depth \(h\) reaches the surface at distance \(r\) from the point directly above the source with \(\tan i = \dfrac{r}{h}\). Light escapes only while \(i \le C\), so the edge of the lit circle has \(r = h\tan C\).

\(\sin C = \dfrac{1}{\mu}\) gives \(\cos C = \dfrac{\sqrt{\mu^2 - 1}}{\mu}\) and \(\tan C = \dfrac{1}{\sqrt{\mu^2 - 1}}\). So \(r = \dfrac{h}{\sqrt{\mu^2 - 1}}\).

\(\sin C = \dfrac{1}{2.4} = \dfrac{5}{12} \approx 0.42\), so \(C \approx 25^\circ\). Light that enters a cut diamond meets most of the inside faces at more than \(25^\circ\) and is totally reflected many times. It comes out through a few faces only, in bright flashes.

The ray enters the water as long as \(i \le C\), where \(\sin C = \dfrac{\mu_w}{\mu_g} = \dfrac{8}{9}\). So the largest \(\sin i\) is \(\dfrac{8}{9}\) (about \(63^\circ\)).

\(\sin 90^\circ = \mu\sin r\) gives \(\sin r = \dfrac{1}{\mu} = \sin C\), so \(r = C\).

Every ray from air has \(i \le 90^\circ\), so its angle of refraction is at most \(C\). A ray inside the glass at an angle larger than \(C\) cannot have come from the air. Reversing the path, such a ray cannot get out into the air either: it is totally reflected.

The air near the hot road is hotter and less dense, so its refractive index is lower than that of the cooler air above. A ray from the sky coming down at a small angle passes into rarer and rarer air and bends away from the vertical. Near the road it is turned back upward. The eye traces the ray back to the road, so it sees an image of the sky there, which looks like a pool of water.

1. The boundaries are parallel, so the angle of refraction at one boundary is the angle of incidence at the next. Snell's law at each boundary gives \(\mu_1\sin\theta_1 = \mu_2\sin\theta_2 = \mu_3\sin\theta_3 = \ldots\)

2. \(1 \times \sin 60^\circ = \sqrt{3}\sin\theta\), so \(\sin\theta = \dfrac{\sqrt{3}/2}{\sqrt{3}} = \dfrac{1}{2}\) and \(\theta = 30^\circ\).

\(\mu\sin\theta = \sqrt{2}\sin 45^\circ = 1\). The ray is horizontal when \(\theta = 90^\circ\), that is \(\mu \times 1 = 1\), so \(\mu = 1\).

Beyond that layer \(\mu\sin\theta\) would have to exceed \(\mu\), which is impossible. The ray is totally reflected at that level and comes back down, retracing the same angles in reverse.

1. \(\sin\theta_2 = \dfrac{\sin 60^\circ}{\sqrt{1.5}} = \dfrac{\sqrt{3}/2}{\sqrt{3}/\sqrt{2}} = \dfrac{1}{\sqrt{2}}\), so \(\theta_2 = 45^\circ\). \(\sin\theta_3 = \dfrac{\sin 60^\circ}{\sqrt{3}} = \dfrac{1}{2}\), so \(\theta_3 = 30^\circ\).

2. Towards the vertical. Each layer is denser than the one above, so the angle with the normal keeps decreasing.

1. At the origin \(\mu = \mu_0\) and \(\theta = 90^\circ\), so \(\mu\sin\theta = \mu_0\) everywhere on the path.

2. \(\sin\theta = \dfrac{\mu_0}{\mu} = \dfrac{1}{\sqrt{1 + y/a}}\). Then \(\cos\theta = \sqrt{1 - \sin^2\theta} = \sqrt{\dfrac{y/a}{1 + y/a}}\), and \(\tan\theta = \dfrac{\sin\theta}{\cos\theta} = \dfrac{1}{\sqrt{y/a}} = \sqrt{\dfrac{a}{y}}\).

1. \(\dfrac{dx}{dy} = \sqrt{\dfrac{a}{y}}\), so \(dx = \sqrt{a}\,y^{-1/2}\,dy\). Integrating from the origin: \(x = \sqrt{a} \times 2\sqrt{y} = 2\sqrt{ay}\). Squaring, \(x^2 = 4ay\), that is \(y = \dfrac{x^2}{4a}\), a parabola.

2. \(y = \dfrac{16}{4} = 4\) m at \(x = 4\) m, which is where the curve in the figure reaches \(y = 4\).

Slope \(\dfrac{dy}{dx} = \dfrac{x}{2a} = \dfrac{2}{2} = 1\) at \(x = 2\) m. The ray makes \(45^\circ\) with the \(x\) axis (and \(45^\circ\) with the \(y\) axis).

Check: \(y = 1\) m there, and \(\tan\theta = \sqrt{a/y} = 1\), so \(\theta = 45^\circ\) with the \(y\) axis.

\(\dfrac{d}{dy}(\mu\sin\theta) = \dfrac{d\mu}{dy}\sin\theta + \mu\cos\theta\dfrac{d\theta}{dy} = 0\). So \(\dfrac{d\theta}{dy} = -\dfrac{\sin\theta}{\mu\cos\theta}\dfrac{d\mu}{dy} = -\dfrac{\tan\theta}{\mu}\dfrac{d\mu}{dy}\).

When \(\dfrac{d\mu}{dy} > 0\), \(\dfrac{d\theta}{dy} < 0\): the angle with the \(y\) axis decreases, so the ray turns towards the \(y\) axis (towards the denser region).

\(\mu\sin\theta = \mu_0\sin 30^\circ = \dfrac{\mu_0}{2}\). The ray is horizontal when \(\sin\theta = 1\), that is \(\mu(y) = \dfrac{\mu_0}{2}\).

\(\sqrt{1 - \dfrac{y}{a}} = \dfrac{1}{2}\) gives \(1 - \dfrac{y}{a} = \dfrac{1}{4}\), so \(y = \dfrac{3a}{4}\). Above this height the ray cannot go; it turns back down.

1. The angle with the vertical at the eye is \(90^\circ - \alpha\), so \(\mu_e\sin(90^\circ - \alpha) = \mu_e\cos\alpha\). Grazing the ground means \(\theta = 90^\circ\) there: \(\mu_e\cos\alpha = \mu_g\sin 90^\circ\), so \(\cos\alpha = \dfrac{\mu_g}{\mu_e}\).

2. \(1 - \dfrac{\alpha^2}{2} = \dfrac{1}{1.0003} \approx 1 - 0.0003\), so \(\alpha^2 = 0.0006\) and \(\alpha \approx 0.024\) rad \(\approx 1.4^\circ\). Only rays within about \(1.4^\circ\) of the horizontal turn back, which is why the mirage is seen far down the road.

1. At minimum deviation the ray passes symmetrically, so \(r_1 = r_2 = \dfrac{A}{2} = 30^\circ\).

2. \(\sin i = \mu\sin r_1 = \sqrt{2} \times \dfrac{1}{2} = \dfrac{1}{\sqrt{2}}\), so \(i = 45^\circ\) (and \(e = 45^\circ\)).

3. \(\delta_{\min} = 45^\circ + 45^\circ - 60^\circ = 30^\circ\).

1. \(\sin r_1 = \dfrac{\sin 60^\circ}{\sqrt{3}} = \dfrac{1}{2}\), so \(r_1 = 30^\circ\). \(r_2 = A - r_1 = 30^\circ\). \(\sin e = \sqrt{3}\sin 30^\circ = \dfrac{\sqrt{3}}{2}\), so \(e = 60^\circ\).

2. \(\delta = i + e - A = 60^\circ + 60^\circ - 60^\circ = 60^\circ\).

3. Yes. \(i = e\) (and \(r_1 = r_2\)), which is the condition for minimum deviation.

\(\mu = \dfrac{\sin\frac{60^\circ + 30^\circ}{2}}{\sin 30^\circ} = \dfrac{\sin 45^\circ}{\sin 30^\circ} = \dfrac{1/\sqrt{2}}{1/2} = \sqrt{2} \approx 1.41\).

\(A = 60^\circ\) and \(\delta_{\min} = 60^\circ\): \(\mu = \dfrac{\sin 60^\circ}{\sin 30^\circ} = \dfrac{\sqrt{3}/2}{1/2} = \sqrt{3} \approx 1.73\).

1. \(\sin r_1 = \dfrac{\sin 90^\circ}{\sqrt{2}}\), so \(r_1 = 45^\circ\) (the critical angle). \(r_2 = 90^\circ - 45^\circ = 45^\circ\).

2. \(\sin e = \sqrt{2}\sin 45^\circ = 1\), so \(e = 90^\circ\): the ray leaves grazing the second face. \(\delta = 90^\circ + 90^\circ - 90^\circ = 90^\circ\).

1. \(\sin C = \dfrac{1}{2}\), so \(C = 30^\circ\).

2. \(A \le 2C = 60^\circ\). For \(A = 60^\circ\) only the grazing ray gets through, leaving grazing. For \(A > 60^\circ\), \(r_2 = A - r_1 > C\) for every ray, since \(r_1 \le C\); every ray is totally reflected at the second face.

1. \(\delta = i + e - A = 45^\circ + 45^\circ - 60^\circ = 30^\circ\).

2. The ray passes symmetrically, \(i = e\). Deviation is least exactly when the path is symmetric, so this is \(\delta_{\min}\).

3. \(r_1 = r_2 = 30^\circ\) and \(\mu = \dfrac{\sin 45^\circ}{\sin 30^\circ} = \sqrt{2}\).

\(\delta = (1.5 - 1) \times 4^\circ = 2^\circ\).

1. \(\delta_1 = 0.5 \times 6^\circ = 3^\circ\), \(\delta_2 = 0.6 \times 4^\circ = 2.4^\circ\). Same way round, the deviations add: \(5.4^\circ\).

2. Opposite ways round, they subtract: \(3^\circ - 2.4^\circ = 0.6^\circ\), towards the base of the first prism.

For small angles \(\sin\theta \approx \theta\), so Snell's law gives \(i = \mu r_1\) and \(e = \mu r_2\).

\(\delta = i + e - A = \mu(r_1 + r_2) - A = \mu A - A = (\mu - 1)A\). The result does not depend on \(i\): a thin prism deviates every ray by the same small angle.

1. The lowest point of the curve is \(\delta_{\min} = 30^\circ\) at \(i = 45^\circ\).

2. At minimum deviation \(e = i = 45^\circ\), so \(A = i + e - \delta_{\min} = 90^\circ - 30^\circ = 60^\circ\).

3. \(\mu = \dfrac{\sin 45^\circ}{\sin 30^\circ} = \sqrt{2}\).

4. At small \(i\), \(r_1\) is small and \(r_2 = A - r_1\) is large. Below about \(28^\circ\), \(r_2\) exceeds the critical angle (\(45^\circ\)) and the ray is totally reflected inside. No ray emerges, so there is no deviation to plot.

\(\omega = \dfrac{1.52 - 1.48}{1.50 - 1} = \dfrac{0.04}{0.5} = 0.08\).

1. \(\delta_y = (\mu_y - 1)A = 0.5 \times 5^\circ = 2.5^\circ\).

2. \(\delta_v - \delta_r = (\mu_v - \mu_r)A = 0.04 \times 5^\circ = 0.2^\circ\).

3. \(\dfrac{0.2}{2.5} = 0.08 = \omega\). Dispersive power is angular dispersion per unit mean deviation.

1. Zero mean deviation: \((\mu_1 - 1)A_1 = (\mu_2 - 1)A_2\), so \(0.5 \times 6^\circ = 0.6 \times A_2\) and \(A_2 = 5^\circ\).

2. Each prism has mean deviation \(3^\circ\). Angular dispersion of each is \(\omega\delta\): crown \(0.08 \times 3^\circ = 0.24^\circ\), flint \(0.1 \times 3^\circ = 0.3^\circ\), in opposite senses. Net dispersion \(= 0.3^\circ - 0.24^\circ = 0.06^\circ\), with no mean deviation.

1. \(\omega_1\delta_1 = \omega_2\delta_2\): \(0.08 \times 0.5A_1 = 0.1 \times 0.6 \times 4^\circ\), so \(0.04A_1 = 0.24^\circ\) and \(A_1 = 6^\circ\).

2. \(\delta_1 = 3^\circ\) and \(\delta_2 = 2.4^\circ\) in opposite senses. Net deviation \(= 0.6^\circ\), and the colours stay together.

1. \(P\) is red and \(Q\) is violet. Violet has the larger refractive index, so it is bent more at each face and deviates most, towards the base of the prism.

2. Red. \(v = \dfrac{c}{\mu}\) and red has the smallest \(\mu\).

3. In a slab every colour comes out parallel to the incident ray (only shifted sideways), so the colours overlap again and the beam stays white.

\(\dfrac{b}{\lambda^2}\) is larger for the smaller wavelength, so violet has the larger \(\mu\). A larger \(\mu\) means a larger deviation, so violet is deviated more.

1. Red: \(\mu = \sqrt{2}\) gives \(\delta_{\min} = 30^\circ\) (Question 1). Violet: \(\mu = \sqrt{3}\) gives \(\delta_{\min} = 60^\circ\) (Question 4).

2. Angular dispersion \(= 60^\circ - 30^\circ = 30^\circ\).

1. Normal incidence means \(i = 0\) and \(r_1 = 0\). Then \(r_2 = A - r_1 = 30^\circ\).

2. \(\sin e = \sqrt{2}\sin 30^\circ = \dfrac{1}{\sqrt{2}}\), so \(e = 45^\circ\). \(\delta = i + e - A = 0 + 45^\circ - 30^\circ = 15^\circ\).

\(\sin C = \dfrac{1}{2}\), so \(C = 30^\circ\). At the second face \(r_2 = 30^\circ = C\), so the emergent ray grazes the face (\(e = 90^\circ\)). Any larger prism angle would give total internal reflection there.

Under water the relative index is \(\dfrac{1.5}{4/3} = \dfrac{9}{8}\). \(\delta = \left(\dfrac{9}{8} - 1\right) \times 8^\circ = 1^\circ\). The prism deviates four times less because glass is only slightly denser than water.

Suppose the deviation were least at some angle of incidence \(i_0\) with \(e_0 \ne i_0\). Reversing the ray gives a path with angle of incidence \(e_0\) and the same deviation. So the same minimum would occur at two different angles of incidence.

A smooth curve of \(\delta\) against \(i\) with one minimum cannot do this. So the minimum must be at the one path that is its own reverse: \(i = e\).

1. \(\sin r_1 = \dfrac{\sin 45^\circ}{\sqrt{3}} = \dfrac{1}{\sqrt{6}} \approx 0.41\), so \(r_1 \approx 24^\circ\). \(r_2 = 60^\circ - 24^\circ = 36^\circ\).

2. \(\sin C = \dfrac{1}{\sqrt{3}}\), so \(C \approx 35^\circ\).

3. \(r_2 = 36^\circ > C\), so the ray is totally reflected at the second face and does not emerge. Compare Question 2: at \(i = 60^\circ\) the same prism let the ray through.

Each face acts as a plane mirror. The two faces are inclined at \(A\), so their normals are also inclined at \(A\). The incident direction lies between the two normals, making angles \(\theta_1\) and \(\theta_2\) with them, so \(\theta_1 + \theta_2 = A\).

A mirror turns a ray away from the normal by twice the angle of incidence. Beam 1 is turned by \(2\theta_1\) to one side of the incident direction, and beam 2 by \(2\theta_2\) to the other side. The angle between them is \(2\theta_1 + 2\theta_2 = 2A\).