5.P — Worked Problems: Optics

Twelve problems across Part 5, climbing from standard to genuinely awkward. Every solution shows the arithmetic. Attempt each before opening it.

Problem 1 — The glass slab

A ray strikes a parallel-sided glass slab (n = 1.50, thickness 4.0 cm) at 50° to the normal. Find the angle inside, the angle on leaving, and the sideways displacement of the emerging ray.

Solution

Entering:

1.00\sin 50° = 1.50\sin r \quad\Longrightarrow\quad \sin r = \frac{0.7660}{1.50} = 0.5107 \quad\Longrightarrow\quad r = 30.7°

Leaving. The second surface is parallel to the first, so the ray inside meets it at 30.7°:

1.50\sin 30.7° = 1.00\sin\theta \quad\Longrightarrow\quad \sin\theta = 1.50\times0.5107 = 0.7660 \quad\Longrightarrow\quad \theta = 50°

The emerging ray is parallel to the incoming one, always, for any parallel-sided slab.

Displacement. The path length inside the glass is:

L = \frac{t}{\cos r} = \frac{4.0}{\cos 30.7°} = \frac{4.0}{0.8599} = 4.652\ \text{cm}

The perpendicular offset between the actual ray and where it would have gone is:

d = L\sin(50° - 30.7°) = 4.652\times\sin 19.3° = 4.652\times0.3305 = 1.54\ \text{cm}

What to notice. The direction is restored but the position is not. This is why a thick window shifts the view slightly and why looking through a car windscreen at a steep angle displaces everything you see by a centimetre or so. It is also the working principle of a beam-shifting plate in optical instruments, where a rotatable slab is used to move a beam sideways by a controlled amount without changing its direction.

Problem 2 — The coin in the cup

A coin sits at the bottom of an opaque cup of radius 4.0 cm and depth 6.0 cm. From a position where you can just not see the coin over the rim, water is poured in until the cup is full. Now you can see it. Show this, and find how far in from the far wall the coin must be.

Solution

Set up the geometry. Your eye is positioned so the line of sight just grazes the rim on the near side and strikes the bottom at the far wall. With the cup empty, that line of sight hits the bottom at the far edge:

\tan\theta_{\text{air}} = \frac{8.0}{6.0} = 1.333 \quad\Longrightarrow\quad \theta_{\text{air}} = 53.1°

measured from the vertical (the normal to the water surface).

With water, the ray bends at the surface. It enters the water at 53.1° from the normal and refracts towards the normal:

1.00\sin 53.1° = 1.333\sin\theta_w \quad\Longrightarrow\quad \sin\theta_w = \frac{0.7999}{1.333} = 0.6000 \quad\Longrightarrow\quad \theta_w = 36.9°

The refracted ray now travels down 6.0 cm at 36.9° from vertical, covering a horizontal distance of:

x = 6.0\tan 36.9° = 6.0\times0.7508 = 4.50\ \text{cm}

from the near rim. Since the cup is 8.0 cm wide, the visible point is:

8.0 - 4.50 = 3.50\ \text{cm}

in from the far wall.

What to notice. Without water the sight line reaches the far wall at the bottom; with water it reaches a point 3.5 cm short of it, meaning 3.5 cm more of the bottom has come into view. This is the classic demonstration attributed to Ptolemy in the second century, and it is exactly the same effect that makes a pool look shallower — light from the bottom bends away from the normal on leaving the water, so it reaches your eye from a steeper apparent direction than it truly came from.

Problem 3 — Critical angle with a coating

A glass fibre core has n = 1.480 and the cladding n = 1.465. Find the critical angle, and the maximum angle from the axis at which a ray can enter the end face from air and still be guided.

Solution

Critical angle at the core–cladding boundary:

\sin\theta_c = \frac{1.465}{1.480} = 0.98986 \quad\Longrightarrow\quad \theta_c = 81.83°

measured from the normal to the boundary — that is, only 8.17° from the fibre axis.

Acceptance angle. A ray entering the flat end face at angle \alpha from the axis refracts to angle \beta inside:

1.00\sin\alpha = 1.480\sin\beta

That ray then strikes the side wall at (90° - \beta) from its normal, and for guiding we need this to be at least \theta_c:

90° - \beta \geq 81.83° \quad\Longrightarrow\quad \beta \leq 8.17°

\sin\alpha_{\max} = 1.480\sin 8.17° = 1.480\times0.14211 = 0.2103 \quad\Longrightarrow\quad \alpha_{\max} = 12.1°

The quantity \sin\alpha_{\max} is called the numerical aperture, here 0.21.

What to notice. A refractive index difference of only 1 % gives an acceptance cone of just 12°. That looks restrictive and it is exactly what makes long-haul fibre work: a narrow acceptance cone means all guided rays travel nearly parallel to the axis, so their path lengths differ very little and a pulse arrives almost as sharp as it left. A large index difference would accept much more light and smear every pulse. There is a genuine trade-off here between how much light you can get in and how far you can send it.

Problem 4 — Two lenses in series

A converging lens of f_1 = +10 cm and a diverging lens of f_2 = -15 cm are 5.0 cm apart. An object sits 20 cm before the first lens. Find the final image.

Solution

First lens:

\frac{1}{v_1} = \frac{1}{10}-\frac{1}{20} = \frac{2-1}{20} = \frac{1}{20} \quad\Longrightarrow\quad v_1 = 20\ \text{cm}

m_1 = -\frac{20}{20} = -1.0

Real, inverted, same size, 20 cm past the first lens.

Second lens. That image is 20 cm past lens 1, which is 20 - 5 = 15 cm past lens 2. Since it lies beyond lens 2, the light is converging towards it when it reaches lens 2, so it acts as a virtual object with u_2 = -15 cm.

\frac{1}{v_2} = \frac{1}{-15}-\frac{1}{-15} = -\frac{1}{15}+\frac{1}{15} = 0 \quad\Longrightarrow\quad v_2 = \infty

The light emerges parallel. The final image is at infinity.

m_2 = -\frac{v_2}{u_2} \to \infty

What to notice. This is not a failure of the calculation; it is a telescope. The first lens forms an image at its focal plane region and the second lens, with the image sitting exactly at its own focal point, converts it into parallel light — which is what a relaxed eye wants. Note also the sign convention doing real work: the object for the second lens is virtual, with negative u, because the light was already converging. Getting that sign wrong is the single most common error in multi-lens problems, and the check is always to ask whether the light arriving at the lens is converging or diverging.

Problem 5 — The prescription

A short-sighted person has a far point of 40 cm and a near point of 12 cm. Find the corrective lens power for distance vision, and where their near point moves to when wearing it.

Solution

Distance correction. The lens must take an object at infinity and form a virtual image at the far point, 40 cm in front:

\frac{1}{\infty}+\frac{1}{-0.40} = \frac{1}{f} \quad\Longrightarrow\quad f = -0.40\ \text{m}

P = \frac{1}{f} = -2.50\ \text{D}

New near point. Wearing the lens, the eye can still only accommodate down to 12 cm. So the lens must produce a virtual image at 12 cm from an object at some distance u:

\frac{1}{u}+\frac{1}{-0.12} = \frac{1}{-0.40}

\frac{1}{u} = -2.50 + 8.333 = 5.833 \quad\Longrightarrow\quad u = 0.171\ \text{m} = 17.1\ \text{cm}

What to notice. The glasses fixed distance vision and pushed the near point out from 12 cm to 17 cm. For a young person that is harmless. For someone in their fifties whose unaided near point is already 40 cm, the same correction pushes it beyond arm's length, which is exactly why myopic people over about 45 need bifocals or reading glasses in addition — the distance correction makes their close vision worse. It is also why many short-sighted people find they can read comfortably by simply taking their glasses off.

Problem 6 — Double slit with a twist

In a double-slit experiment with d = 0.30 mm and L = 1.5 m, light of 550 nm gives a certain fringe spacing. A thin sheet of plastic (n = 1.58, thickness 12 μm) is placed over one slit. By how many fringes does the pattern shift?

Solution

Fringe spacing without the plastic:

\Delta y = \frac{\lambda L}{d} = \frac{(550\times10^{-9})(1.5)}{0.30\times10^{-3}} = \frac{8.25\times10^{-7}}{3.0\times10^{-4}} = 2.75\times10^{-3}\ \text{m} = 2.75\ \text{mm}

Extra optical path from the plastic. Light passing through thickness t of index n takes as long as it would to travel nt in vacuum. Replacing t of air with t of plastic adds:

\Delta = (n-1)t = (0.58)(12\times10^{-6}) = 6.96\times10^{-6}\ \text{m}

In wavelengths:

N = \frac{\Delta}{\lambda} = \frac{6.96\times10^{-6}}{550\times10^{-9}} = 12.65

The pattern shifts by 12.65 fringes, towards the covered slit — because that path is now effectively longer, so the point of zero path difference must move to compensate.

What to notice. A 12 μm sheet — a sixth the thickness of a human hair — shifts the pattern by more than twelve whole fringes, a displacement of 35 mm on the screen. This is interference acting as an amplifier, and it is the basis of every interferometric measurement: a change far too small to see directly becomes a large, countable displacement. Note also that the shift is not a whole number, which means the central fringe no longer lands where a fringe used to be. Measuring the fractional part is how these instruments achieve nanometre precision.

Problem 7 — The anti-reflective coating that fails

A coating of magnesium fluoride (n = 1.38) is put on glass (n = 1.52) with thickness chosen to eliminate reflection at 550 nm. What happens at 400 nm and at 700 nm?

Solution

Thickness:

t = \frac{\lambda}{4n} = \frac{550\times10^{-9}}{4\times1.38} = 99.6\ \text{nm}

At any wavelength, the phase difference between the two reflections, in wavelengths, is 2nt/\lambda. Both reflections flip (air→MgF₂ and MgF₂→glass both go up in index), so the flips cancel and destructive interference in reflection needs 2nt/\lambda = (m+\frac{1}{2}).

2nt = 2(1.38)(99.6\times10^{-9}) = 2.749\times10^{-7}\ \text{m} = 274.9\ \text{nm}

The two reflected waves have a phase difference \delta = 2\pi(2nt)/\lambda, and adding two equal-amplitude waves gives a reflected intensity proportional to \cos^2(\delta/2), exactly as in Chapter 5.3. So the reflectance as a fraction of its own worst case is:

\frac{R(\lambda)}{R_{\max}} = \cos^2\left(\frac{\pi\times 274.9\ \text{nm}}{\lambda}\right)

At 550 nm: 274.9/550 = 0.500, so the angle is 90° and \cos^2 90° = 0. Perfect cancellation.

At 400 nm: 274.9/400 = 0.687, angle = 123.7°, and \cos^2(123.7°) = (-0.555)^2 = 0.308. About 31 % of the peak reflection is back.

At 700 nm: 274.9/700 = 0.393, angle = 70.7°, and \cos^2(70.7°) = (0.330)^2 = 0.109. About 11 % of the peak.

What to notice. A single-layer coating can only null one wavelength exactly, and the reflection climbs at both ends of the spectrum. Since it climbs more in the blue than the red, and the residual is a mixture of the two, the reflected light looks purple — which is precisely the colour you see glinting off a coated lens. Multi-layer coatings stack several quarter-wave films of different indices, each nulling a different wavelength, and modern ones hold reflection below 0.3 % across the whole visible band. That is the entire reason a modern zoom lens with eighteen elements is usable.

Problem 8 — Resolving a car's headlights

At what distance can the naked eye just separate a car's two headlights, 1.4 m apart? Take pupil 4.0 mm (night vision) and \lambda = 550 nm. Then repeat for a 200 mm telescope.

Solution

Eye:

\theta_{\min} = 1.22\frac{\lambda}{D} = 1.22\frac{550\times10^{-9}}{4.0\times10^{-3}} = 1.678\times10^{-4}\ \text{rad}

L = \frac{s}{\theta} = \frac{1.4}{1.678\times10^{-4}} = 8.34\times10^{3}\ \text{m} = 8.3\ \text{km}

Telescope:

\theta_{\min} = 1.22\frac{550\times10^{-9}}{0.200} = 3.355\times10^{-6}\ \text{rad}

L = \frac{1.4}{3.355\times10^{-6}} = 4.17\times10^{5}\ \text{m} = 417\ \text{km}

What to notice. The theoretical 8.3 km for the eye is far beyond what anyone actually achieves — in practice headlights merge at around 3 km. The gap is because retinal cone spacing, eye aberrations, atmospheric shimmer and the sheer brightness of the sources all degrade the ideal. The diffraction limit is a ceiling, not a promise. Note also that the pupil widens at night, which improves the theoretical resolution while the switch to rod-dominated vision makes actual acuity much worse — the optics get better and the sensor gets worse.

Problem 9 — Grating with two close lines

A grating 4.0 cm wide with 500 lines/mm is used to examine a source with lines at 588.995 nm and 589.592 nm. Find the angle of the first-order lines, whether they are resolved in first order, and how many orders are visible.

Solution

Grating spacing:

d = \frac{1}{500\ \text{mm}^{-1}} = 2.0\times10^{-6}\ \text{m}

First-order angle, for 589 nm:

\sin\theta = \frac{m\lambda}{d} = \frac{589\times10^{-9}}{2.0\times10^{-6}} = 0.2945 \quad\Longrightarrow\quad \theta = 17.1°

Resolving power needed:

\frac{\lambda}{\Delta\lambda} = \frac{589.0}{0.597} = 987

Resolving power available. Number of lines illuminated:

N = 500\ \text{mm}^{-1}\times40\ \text{mm} = 20{,}000

\frac{\lambda}{\Delta\lambda} = mN = 1\times20000 = 20{,}000

Twenty times more than needed — comfortably resolved in first order.

Highest order. The maximum possible is at \sin\theta = 1:

m_{\max} = \frac{d}{\lambda} = \frac{2.0\times10^{-6}}{589\times10^{-9}} = 3.40

So orders 1, 2 and 3 are visible; the fourth would require \sin\theta > 1.

What to notice. These are the sodium D lines, and separating them is the standard test of a spectroscope. They are 0.6 nm apart out of 589 — one part in a thousand — and a 4 cm grating separates them twenty times over. This is the resolution that lets astronomers measure a star's radial velocity to metres per second by watching such lines shift, which is how exoplanets are detected (Chapter 2.5).

Problem 10 — Three polarisers, optimised

Unpolarised light of intensity I_0 passes through two crossed polarisers with a third inserted between them at angle \theta to the first. Find the \theta that maximises transmission, and the maximum.

Solution

After the first polariser: I_1 = I_0/2.

After the middle one at \theta: I_2 = \frac{I_0}{2}\cos^2\theta.

After the last one, which is at (90° - \theta) from the middle one:

I_3 = \frac{I_0}{2}\cos^2\theta\cos^2(90°-\theta) = \frac{I_0}{2}\cos^2\theta\sin^2\theta

Use \sin\theta\cos\theta = \frac{1}{2}\sin 2\theta:

I_3 = \frac{I_0}{2}\cdot\frac{1}{4}\sin^2 2\theta = \frac{I_0}{8}\sin^2 2\theta

Maximum when \sin 2\theta = 1, so 2\theta = 90°:

\theta = 45°, \qquad I_3 = \frac{I_0}{8}

What to notice. With N polarisers each rotated by 90°/N, the transmission is \frac{1}{2}\cos^{2(N-1)}(90°/N), which tends towards I_0/2 as N grows large. With 45 filters you recover about 45 % of the input through a 90° rotation. This is not a trick — each filter really does rotate the polarisation a little — and the same mathematics in quantum mechanics is the quantum Zeno effect, where frequent measurement drags a system along a path it would not otherwise take.

Problem 11 — The rainbow's width

Using the rainbow formula from Chapter 5.1, find the angular width of the primary bow, and then find the deviation for the secondary bow and its angular position.

Solution

Primary. From Chapter 5.1, red (n = 1.331) emerges at 42.5° from the antisolar point and violet (n = 1.344) at 40.7°.

\text{width} = 42.5 - 40.7 = 1.8°

About three and a half times the Sun's angular diameter of 0.53°.

Secondary. Two internal reflections, so the deviation is:

D_2 = 2(i-r) + 2(180° - 2r) = 360° + 2i - 6r

Minimum when dD/di = 0:

2 - 6\frac{dr}{di} = 0 \quad\Longrightarrow\quad \frac{dr}{di} = \frac{1}{3}

From Snell, \cos i = n\cos r\,(dr/di) = n\cos r/3. Squaring and substituting as before:

9(1-\sin^2 i) = n^2 - \sin^2 i \quad\Longrightarrow\quad 8\sin^2 i = 9 - n^2

\sin i = \sqrt{\frac{9-n^2}{8}}

For red, n = 1.331: \sin i = \sqrt{(9-1.772)/8} = \sqrt{0.9035} = 0.9505, so i = 71.9°.

\sin r = \frac{0.9505}{1.331} = 0.7141 \quad\Longrightarrow\quad r = 45.6°

D_2 = 360 + 2(71.9) - 6(45.6) = 360 + 143.8 - 273.6 = 230.2°

Which is equivalent to 360 - 230.2 = 129.8° of turning, putting the bow at 180 - 129.8 = 50.2° from the antisolar point.

What to notice. The secondary sits at about 50°, outside the primary's 42°, and its colours are reversed — red innermost — because the extra reflection flips the ordering. The gap between them, from 42° to 50°, is Alexander's dark band, which is genuinely darker than the sky both inside the primary and outside the secondary. And the secondary is much fainter: each internal reflection transmits most of the light out of the drop rather than reflecting it, so only a few percent survives two bounces.

Problem 12 — Laser beam to the Moon

A laser with \lambda = 532 nm is expanded to a 1.0 m diameter beam and aimed at a retroreflector on the Moon, 384,400 km away. Find the spot size on the Moon. The reflector is 0.5 m across; estimate the fraction of photons returned, ignoring atmospheric losses.

Solution

Divergence of a beam of diameter D:

\theta \approx 1.22\frac{\lambda}{D} = 1.22\frac{532\times10^{-9}}{1.0} = 6.49\times10^{-7}\ \text{rad}

Spot radius on the Moon:

r = \theta L = (6.49\times10^{-7})(3.844\times10^{8}) = 249\ \text{m}

Spot diameter about 500 m.

Fraction hitting the reflector. Area ratio:

\frac{A_{\text{refl}}}{A_{\text{spot}}} = \frac{\pi(0.25)^2}{\pi(249)^2} = \frac{0.0625}{62001} = 1.01\times10^{-6}

Coming back. The reflector is 0.5 m across, so the returning beam diverges by 1.22\lambda/0.5 = 1.30\times10^{-6} rad, giving a spot on Earth of radius:

r' = (1.30\times10^{-6})(3.844\times10^{8}) = 500\ \text{m}

A 1 m receiving telescope catches:

\frac{\pi(0.5)^2}{\pi(500)^2} = \frac{0.25}{250000} = 1.0\times10^{-6}

Total round-trip fraction:

(1.01\times10^{-6})(1.0\times10^{-6}) = 1.0\times10^{-12}

One photon in a trillion.

What to notice. A pulse containing 10^{19} photons — about 4 joules at this wavelength — returns roughly 10^{7} photons before atmospheric absorption, and in practice observatories detect around one photon per pulse after all losses. That is enough, because the timing of that single photon gives the distance: light takes 2.56 s for the round trip, and measuring it to 100 picoseconds gives 1.5 cm. Lunar laser ranging since 1969 has shown the Moon receding at 3.8 cm per year, tested general relativity's predictions to a few parts in 10^{4}, and confirmed that the Moon has a liquid core. All from one photon in a trillion.