Past exam of the physics 2 paper of the University of Oxford

Solution

The non-inverting input is grounded, so the finite voltage gain gives

V_{OUT} = - A V^{-}

, or

V^{-} = - V_{OUT} / A

. Infinite input impedance makes Kirchhoff's current law at the inverting input

\frac{V _{IN} - V ^{-}}{R _{1}} + \frac{V _{OUT} - V ^{-}}{R _{2}} = 0.

Substitute

V^{-} = - V_{OUT} / A

and multiply by

R_{1} R_{2}

R_{2} V_{IN} + (R_{1} + \frac{R _{1} + R _{2}}{A}) V_{OUT} = 0.

Hence the inverting operational amplifier has

\frac{V _{OUT}}{V _{IN}} = - \frac{R _{2}}{R _{1} + ( R _{1} + R _{2} ) / A}, B = \frac{R _{1} + R _{2}}{A} .

A \to \infty

, the voltage gain tends to

- R_{2} / R_{1}

. Setting

V^{-} = 0

before taking that limit would discard precisely the correction being calculated.

3

i

Solution

Assume the electrical conductor is nonmagnetic, with magnetic permeability

μ_{0}

, and neglect end effects. Rotational symmetry and translational symmetry make the magnetic flux density azimuthal and constant along a coaxial circle. The magnetostatic Ampère law therefore gives

2 π r B (r) = μ_{0} I_{enc} (r) .

Uniform current density is

j = I / (π a^{2})

, so

I_{enc} (r) = jπ r^{2} = I r^{2} / a^{2}

. Thus

B (r) = \frac{μ _{0} I r}{2 π a ^{2}} (0 \leq r \leq a) .

The direction follows the right-hand rule around the electric current. For a uniform linear magnetic material, replace

μ_{0}

by its magnetic permeability.

ii

Solution

Write the axial current density as

j (r) = Kr

. Normalize it using the total electric current:

I = \int_{0}^{a} Kr 2 π r d r = \frac{2 π K a ^{3}}{3}, K = \frac{3 I}{2 π a ^{3}} .

The enclosed electric current at radius

r

I_{enc} (r) = 2 π K r^{3} /3 = I (r / a)^{3}

. The magnetostatic Ampère law now yields

B (r) = \frac{μ _{0} I r ^{2}}{2 π a ^{3}} (0 \leq r \leq a) .

Both profiles join the exterior magnetic flux density

μ_{0} I / (2 πa)

at the surface, but the second vanishes quadratically at the axis because less electric current is concentrated near it.

4

Solution

Choose zero electrostatic potential at infinity. By Coulomb's law and the superposition principle, the two point charges give

Φ (r, θ) = \frac{1}{4 π ϵ _{0}} [\frac{q _{1}}{r ^{2} + r _{1}^{2} - 2 r r _{1} cos θ} + \frac{q _{2}}{r ^{2} + r _{2}^{2} - 2 r r _{2} cos θ}] .

Here

ϵ_{0}

is the vacuum permittivity. For a nondegenerate zero-equipotential curve, the electric charges must have opposite signs. Let

d_{1}, d_{2}

be the two distances. The equation

Φ = 0

becomes

q_{1} d_{2} = - q_{2} d_{1}

, and hence

q_{1}^{2} (r^{2} + r_{2}^{2} - 2 r r_{2} cos θ) = q_{2}^{2} (r^{2} + r_{1}^{2} - 2 r r_{1} cos θ) .

For a circle

r = ρ > 0

centred at the origin, its

cos θ

coefficient must vanish:

q_{1}^{2} r_{2} = q_{2}^{2} r_{1}

. If

r_{1}, r_{2} > 0

and

r_{1} \neq = r_{2}

, this gives

q_{2} = - q_{1} r_{2} / r_{1} .

Substitute

q_{2}^{2} / q_{1}^{2} = r_{2} / r_{1}

in the constant term:

r_{1} ρ^{2} + r_{1} r_{2}^{2} = r_{2} ρ^{2} + r_{2} r_{1}^{2}, (r_{1} - r_{2}) (ρ^{2} - r_{1} r_{2}) = 0.

(4)

Therefore

ρ = r_{1} r_{2} .

(5)

This is an Apollonius circle: the distance ratio

d_{2} / d_{1}

is constant. The distinct-position assumption matters. For example,

r_{1} = r_{2} = 1

q_{2} = - q_{1}

gives zero electrostatic potential everywhere away from the coincident point charges, so there is no distinguished zero circle. If only one position is at the origin and the other is not, no nontrivial positive-radius circle centred at the origin is possible.

5

i

Solution

In charge-free, current-free vacuum, Maxwell's equations are

\nabla \cdot E = 0, \nabla \cdot B = 0, \nabla \times E = - \frac{\partial B}{\partial t}, \nabla \times B = μ_{0} ϵ_{0} \frac{\partial E}{\partial t} .

Taking the curl of the third equation and using the fourth, the curl of the curl identity, and the zero divergence of the electric field gives

\nabla^{2} E = μ_{0} ϵ_{0} \frac{\partial ^{2} E}{\partial t ^{2}} .

The magnetic flux density obeys the same wave equation. For the plane wave factor

e^{i (ω t - k x)}

, the spatial Laplacian contributes

- k^{2}

and the second time derivative contributes

- ω^{2}

. Thus

k^{2} = μ_{0} ϵ_{0} ω^{2}, c = \frac{ω}{k} = \frac{1}{μ _{0} ϵ _{0}}

for propagation in the positive

x

direction. The zero-divergence conditions require

E_{0 x} = B_{0 x} = 0

, while the curl equations require

B_{0} = (k / ω) \hat{x} \times E_{0}

. These conditions show that the proposed plane waves satisfy all four Maxwell's equations, rather than only their derived wave equations. Physical electric fields and magnetic fields are the real parts of the complex expressions.

ii

Solution

Using

\nabla \times E = - \partial_{t} B

directly with the specified plane wave gives

- ik \hat{x} \times E_{0} = - iω B_{0}, B_{0} = \frac{E _{0}}{c} \hat{z} .

The electric field, magnetic flux density and propagation direction form a right-handed triad, and for a nonzero wave amplitude

\frac{∣ E _{0} ∣}{∣ B _{0} ∣} = c .

The electric field and magnetic field have the same wave phase in vacuum.

6

i

Solution

The distance from the origin to the incident straight light ray is

d_{a}

(taking

d_{a} \geq 0

). At its intersection with the sphere, let

i

and

t

be the angle of incidence and angle of refraction, measured from the radial normal vector. Elementary right triangle geometry gives

d_{a} = r_{b} sin i

. By Snell's law,

n_{a} sin i = n_{b} sin t .

The refracted straight light ray inside the sphere has distance

d_{b} = r_{b} sin t

from the origin. Therefore

d_{b} = \frac{n _{a}}{n _{b}} d_{a}, r cos (θ - θ_{b}) = \frac{n _{a}}{n _{b}} d_{a} .

The angle

θ_{b}

describes the normal vector of the new line; it need not equal the propagation angle. The existence of a crossing requires

∣ d_{a} ∣ < r_{b}

. A signed convention for

d_{a}

gives the same line equation after choosing the orientation of its normal vector consistently.

ii

Solution

Approximate the atmosphere by concentric spherical layers of refractive index. At every boundary, Snell's law conserves

n r sin ψ

, where

ψ

is the angle between the light ray and the radius. Straight propagation between boundaries conserves the same quantity because

r sin ψ

is the distance of that line from the origin. In the smooth limit this is the spherical ray invariant

p = n (r) r sin ψ .

At the surface, a light ray apparently on the horizon is tangent, so

p = n_{s} R

. In vacuum far from the Earth,

n = 1

, and the straight incident line has distance

p

from the centre. Removing the atmosphere while retaining that incident line gives minimum altitude

H = p - R = (n_{s} - 1) R = 0.0003 (6400 km) = 1.92 km .

The detailed variation of refractive index with altitude changes the local curvature of the light ray, but not the conserved

p

. This conclusion assumes a spherically stratified atmosphere and a light ray which actually connects the surface to space; nonspherical structure or a trapping layer would require separate analysis.

7

i

Solution

Take coordinate

y

across the slit and observation angle

θ

from its normal vector. Under Fraunhofer diffraction, the relative optical path difference from

y

y sin θ

. Uniform coherent illumination makes the complex amplitude proportional to the aperture integral:

E (θ) = K \int_{- w /2}^{w /2} e^{- ik y s i n θ} d y = K w \frac{s i n β}{β}, β = \frac{π w s i n θ}{λ}, k = \frac{2 π}{λ} .

The light intensity is proportional to

∣ E ∣^{2}

, so

I (θ) = I (0) [\frac{sin ( π w sin θ / λ )}{π w sin θ / λ}]^{2} .

The central value is the continuous limit of the sinc function. The first minima satisfy

sin θ = \pm λ / w

, if

λ \leq w

. Infinite slit height removes spreading in the other transverse direction; this expression describes the slit-width direction.

Strictly, distance

D ≫ w

alone does not ensure Fraunhofer diffraction: one also needs

w^{2} / (λ D) ≪ 1

, so that quadratic aperture-dependent wave phase is negligible, or observation in a suitable focal plane. For example,

w = 1 mm

λ = 500 nm

and

D = 1 m

have

D / w = 1000

but

w^{2} / (λ D) = 2

; the Fresnel diffraction correction is then substantial. The displayed result uses the intended far-field regime of the question and the usual scalar aperture approximation.

ii

Solution

For viewing near the nadir, the required angular resolution is

δ θ ≃ \frac{s}{H} = \frac{100}{840000} = 1.19048 \times 1 0^{- 4} rad .

Using the Rayleigh criterion with the first minimum of the single-slit diffraction pattern gives

δ θ ≃ λ / w

. Thus the minimum collecting width in that model is

w_{m i n} = \frac{λ H}{s} = \frac{12 \times 1 0 ^{- 6} \times 840000}{100} m = 0.1008 m ≃ 10.1 cm .

The numerical coefficient depends on the aperture shape and the chosen resolution criterion. If the mirror has an unobstructed circular aperture, the Airy disk version of the Rayleigh criterion is

δ θ = 1.22 λ / D

, giving

D_{m i n} = 0.12298 m ≃ 12.3 cm

. The question does not specify mirror shape;

10.1 cm

follows the slit-width model of part (i).

8

a

Solution

Use the

e^{iω t}

phasor convention and interpret impedance matching as conjugate matching for the maximum power transfer theorem. The load electrical impedance of the series circuit is

Z_{RC} = R - \frac{i}{ω C} .

For

Z = A + iω B

, the condition

Z_{RC} = Z^{*}

equates the real parts and imaginary parts:

R = A, ω^{2} BC = 1, ω = \frac{1}{BC} .

These formulas require

B > 0

C > 0

and a passive realization with

A \geq 0

, evaluated at the matching angular frequency; if

A, B

depend on

ω

, the equation for

ω

is implicit.

There is a convention ambiguity in the word “matched”. Literal equality

Z_{RC} = Z

, as required for equal bridge arms in part (b), instead gives

R = A

and

ω^{2} BC = - 1

, requiring

B < 0

. The inductive cases below use conjugate matching; literal equality is impossible for either passive inductive configuration at a finite positive angular frequency because its electrical reactance is positive while that of the series RC impedance is negative.

i

Solution

Let the unknown series circuit contain electrical resistance

R_{s} > 0

and inductance

L > 0

. Its electrical impedance is

Z = R_{s} + iω L

. Thus

A = R_{s}

B = L

, and conjugate impedance matching is possible with

R = R_{s}, ω = \frac{1}{L C} .

At this angular frequency, the inductor and load capacitor have opposite electrical reactances of equal magnitude. Literal equality of their separate electrical impedances is impossible; for example,

R_{s} = L = C = 1

gives

1 + iω

versus

R - i / ω

for every

ω > 0

ii

Solution

For electrical resistance

R_{p} > 0

in a parallel circuit with inductance

L > 0

, add electrical admittances and invert:

Z = (\frac{1}{R _{p}} + \frac{1}{iω L})^{- 1} = \frac{R _{p} ω ^{2} L ^{2}}{R _{p}^{2} + ω ^{2} L ^{2}} + iω \frac{R _{p}^{2} L}{R _{p}^{2} + ω ^{2} L ^{2}} .

Conjugate impedance matching to the series RC impedance requires

\frac{1}{ω C} = \frac{ω R _{p}^{2} L}{R _{p}^{2} + ω ^{2} L ^{2}}, R_{p}^{2} = ω^{2} L (C R_{p}^{2} - L) .

It is possible at a finite positive angular frequency exactly when

C R_{p}^{2} > L

, in which case

ω = \frac{R _{p}}{L ( C R _{p}^{2} - L )}, R = \frac{L}{C R _{p}} .

The expression for

R

follows by substituting this

ω^{2}

into the real part of

Z

. If

C R_{p}^{2} \leq L

, no such finite matching angular frequency exists. Under literal equality, both passive inductive configurations fail by the sign of their imaginary parts.

b

Solution

Let the source's top terminal have voltage phasor

V_{0}

and its bottom terminal zero. Denote the left and right junctions by

L

and

R

. The two left resistors are equal, so at balance

V_{L} = V_{0} /2

. Zero electric current through the ammeter requires

V_{R} = V_{L}

, and therefore equal upper and lower right-arm electrical impedances:

Z = \frac{R _{x}}{1 + iω R _{x} C _{x}} = W, W = R_{v} + \frac{1}{iω C} = R_{v} - \frac{i}{ω C} .

Write

x = ω R_{x} C_{x}

. Separating real parts and imaginary parts gives

R_{v} = \frac{R _{x}}{1 + x ^{2}}, \frac{1}{ω C} = \frac{ω R _{x}^{2} C _{x}}{1 + x ^{2}} .

The second equation implies

1 + x^{2} = ω^{2} R_{x}^{2} C C_{x}

. Subtract

x^{2} = ω^{2} R_{x}^{2} C_{x}^{2}

to obtain

ω_{0} = \frac{1}{R _{x} C _{x} ( C - C _{x} )}, \frac{C _{x}}{C} = 1 - \frac{R _{v}}{R _{x}} .

A finite balance with positive components requires

0 < C_{x} < C

, and then

0 < R_{v} < R_{x}

. This equal-ratio RC bridge uses equality of arm electrical impedances, independent of the conjugate-matching convention in part (a).

Detector convention. The stated negligible-electrical impedance ammeter is an ideal short circuit, whose actual terminal voltage is zero at every angular frequency. The nontrivial bridge output usually intended is the open-circuit junction difference, obtained with the detector removed or replaced by a high-input impedance voltmeter. Define its polarity by

V_{oc} = V_{R} - V_{L} = V_{0} (\frac{W}{Z + W} - \frac{1}{2}) = \frac{V _{0} ( W - Z )}{2 ( W + Z )} .

(4)

For a detector of finite electrical impedance

Z_{A}

, its terminal voltage is instead

V_{A} = V_{oc} \frac{Z _{A}}{Z _{A} + Z _{th}}, Z_{th} = \frac{R}{2} + \frac{Z W}{Z + W} .

(5)

Here

Z_{th}

is the Thévenin equivalent electrical impedance, found by shorting the ideal source. This exhibits why the ideal-ammeter limit

Z_{A} \to 0

has zero terminal voltage even when the bridge is unbalanced.

i

Solution

At zero angular frequency, the lower right capacitor is an open circuit, while the upper right electrical impedance tends to

R_{x}

. With the detector removed, no electric current flows down that branch, so

V_{R} = V_{0}

and

V_{L} = V_{0} /2

. Therefore

V_{oc} (0) = \frac{V _{0}}{2}, V_{A} (0) = 0 for an ideal ammeter .

For the connected ideal ammeter, the resulting electric current from right to left is

V_{0} / (R + 2 R_{x})

, despite its zero terminal voltage. This follows from

I_{A} = V_{oc} / Z_{th}

with

Z_{th} = R /2 + R_{x}

ii

Solution

At the balance angular frequency,

Z = W

and the two voltage dividers have the same ratio. Thus

V_{oc} (ω_{0}) = 0, V_{A} (ω_{0}) = 0, I_{A} (ω_{0}) = 0.

This zero detector electric current is the operational condition used to find

R_{v}

and

ω_{0}

iii

Solution

ω \to \infty

, both capacitors approach short circuits. Thus

Z \to 0

W \to R_{v} > 0

, and the unloaded right junction tends to the source's top-terminal voltage. Hence

ω \to \infty lim V_{oc} = \frac{V _{0}}{2}, V_{A} = 0 for an ideal ammeter .

The connected ideal ammeter carries limiting electric current

V_{0} / R

from right to left because

Z_{th} \to R /2

. Reversing the output polarity reverses both nonzero open-circuit answers, without changing their magnitudes.

9

a

Solution

Set the grounded plane at

z = 0

and the real point charge at

(0, 0, h)

. The image charge

- q

(0, 0, - h)

makes the electrostatic potential vanish throughout the plane. The uniqueness theorem for the electrostatic potential then identifies the physical solution in

z > 0

Φ (a, z) = \frac{q}{4 π ϵ _{0}} [\frac{1}{a ^{2} + ( z - h ) ^{2}} - \frac{1}{a ^{2} + ( z + h ) ^{2}}] .

The normal electric field is

E_{z} = - \partial_{z} Φ

. Differentiation gives

E_{z} (a, z) = \frac{q}{4 π ϵ _{0}} [\frac{z - h}{[ a ^{2} + ( z - h ) ^{2} ] ^{3/2}} - \frac{z + h}{[ a ^{2} + ( z + h ) ^{2} ] ^{3/2}}],

and therefore

E_{z} (a, 0^{+}) = - \frac{q}{2 π ϵ _{0}} \frac{h}{( a ^{2} + h ^{2} ) ^{3/2}} .

For

q > 0

the electric field points into the electrical conductor, consistently with negative induced surface charge density.

b

Solution

Apply the superposition principle to the result of part (a). If

F (a, h) = - q h / [2 π ϵ_{0} (a^{2} + h^{2})^{3/2}]

, then the exact normal electric field from the two real point charges and their image charges is

E_{z} = F (a, h) - F (a, h + d) = \frac{q}{2 π ϵ _{0}} [\frac{h + d}{[ a ^{2} + ( h + d ) ^{2} ] ^{3/2}} - \frac{h}{( a ^{2} + h ^{2} ) ^{3/2}}] .

For

d ≪ h

, Taylor's theorem gives

E_{z} = - d \partial_{h} F + O (d^{2})

. Since

\frac{d}{d h} \frac{h}{( a ^{2} + h ^{2} ) ^{3/2}} = \frac{a ^{2} - 2 h ^{2}}{( a ^{2} + h ^{2} ) ^{5/2}},

the leading electric dipole result is

E_{z} (a, 0^{+}) = \frac{q d}{2 π ϵ _{0}} \frac{a ^{2} - 2 h ^{2}}{( a ^{2} + h ^{2} ) ^{5/2}} + O (\frac{q d ^{2}}{ϵ _{0} h ^{4}}) .

The electric dipole moment is

p = - q d \hat{z}

. The circle

a = 2 h

is the zero of the leading normal electric field and hence of the induced surface charge density. Inside it, induced surface charge density is negative; outside it, positive. The tangential electric field is zero everywhere on the grounded electrical conductor, so the leading full electric field vanishes on that circle. The corresponding dipole-induced charge on a grounded plane integrates to zero over the whole plane: the negative central region is balanced by the positive exterior.

For finite

d

a = 2 h

is only the leading-order zero. With

u = h^{2/3}

and

v = (h + d)^{2/3}

, the exact zero satisfies

a^{2} = uv (u + v)

, giving

a = 2 (h + d /2) + O (d^{2} / h)

. For example, inserting any

d > 0

in the exact expression at

a = 2 h

does not give zero. This is a controlled electric dipole approximation, not an exact finite-separation identity.

Figure 1.
Sign reversal and enclosed induced electric charge for a normal electric dipole above a grounded plane
. **Details**.

c

Solution

The electric field within a conductor in electrostatic equilibrium vanishes. The normal electric field boundary condition, with outward normal vector

+ \hat{z}

, therefore gives

σ = ϵ_{0} E_{z} (0^{+})

. Directly below the monopole,

a = 0

σ_{mono} (0) = - \frac{q}{2 π h ^{2}} .

For the electric dipole,

σ_{dip} (0) = - \frac{q d}{π h ^{3}} + O (q d^{2} / h^{4}) .

The exact finite-separation answer is

σ (0) = \frac{q}{2 π} [\frac{1}{( h + d ) ^{2}} - \frac{1}{h ^{2}}],

whose Taylor series agrees with the boxed leading term. Although the real point charges have zero total electric charge, their unequal distances leave negative induced surface charge density immediately beneath them.

10

a

Solution

Use the source orientation, anticlockwise viewed from

+ z

. The position and oriented line element are

r = a (cos θ, sin θ, 0), d l = a (- sin θ, cos θ, 0) d θ .

The Lorentz force on the electric current element is

d F = I d l \times B

. Evaluating the cross product gives

d F = I a (B_{z} cos θ, B_{z} sin θ, - B_{y} sin θ - B_{x} cos θ) d θ .

Each component integrates to zero around

0 \leq θ < 2 π

, so

F_{net} = 0

. Equivalently, uniform magnetic flux density permits

I \oint d l \times B = I (\oint d l) \times B = 0

. Uniformity is essential; a nonuniform magnetic field can exert a net force on a magnetic dipole.

b

Solution

The elemental torque about the origin is

d τ = r \times d F

. Substitution of part (a) gives

d τ = I a^{2} (- B_{y} sin^{2} θ - B_{x} sin θ cos θ, B_{y} sin θ cos θ + B_{x} cos^{2} θ, 0) d θ .

Only its

x

component drives the allowed rotation about the

x

axis. Using

\int_{0}^{2 π} sin^{2} θ d θ = π

and

\int_{0}^{2 π} sin θ cos θ d θ = 0

, the projected torque is

τ_{about x} = - I π a^{2} B_{y} \hat{x} .

The magnetic dipole moment is

m = I π a^{2} \hat{z}

, so the full electromagnetic torque is

m \times B = I π a^{2} (- B_{y} \hat{x} + B_{x} \hat{y}) .

Its

x

component agrees exactly. The

y

component is a real electromagnetic torque, but the bearing constraining motion to the

x

axis supplies a cancelling reaction torque. For an allowed infinitesimal rotation

d α = \hat{x} d α

, only

τ \cdot d α = τ_{x} d α

does mechanical work.

c

Solution

Define positive

α

by the right-hand rule about

+ x

, starting from the loop in the

x

y

plane at

α = 0

. Its magnetic dipole moment becomes

m (α) = m (0, - sin α, cos α), m = I π a^{2} .

The corresponding

x

component of magnetic dipole torque is

τ_{x} = - m (B_{z} sin α + B_{y} cos α)

. Integrating

τ_{x} = - d E / d α

gives the mechanical magnetic dipole potential energy

E = - m \cdot B = m (B_{y} sin α - B_{z} cos α),

where the arbitrary additive constant is chosen as zero. Set

B_{⊥} = B_{y}^{2} + B_{z}^{2}

. For

B_{⊥} > 0

, a positive-amplitude form is

E = E_{0} cos (α - α_{0}), E_{0} = m B_{⊥}, cos α_{0} = - \frac{B _{z}}{B _{⊥}}, sin α_{0} = \frac{B _{y}}{B _{⊥}} .

Equivalently,

α_{0} = atan2 (B_{y}, - B_{z})

modulo

2 π

. With this positive choice, minimum potential energy occurs at

α = α_{0} + π

, where the rotating magnetic dipole moment aligns with the projection of

B

into the

y

z

plane. If

B_{y} = B_{z} = 0

, the allowed torque vanishes,

E_{0} = 0

and

α_{0}

is arbitrary. Holding electric current constant includes exchange with its supply; differentiating only the isolated loop's stored magnetic energy would not give the appropriate mechanical potential energy.

11

a

Solution

Write

T

for transit time and

ℓ (y) = t_{0} - α y^{2}

for lens thickness to avoid using

t

for both. In the thin lens approximation, the two geometrical distances to the lens plane are

a^{2} + d^{2}

and

b^{2} + d^{2}

. Their sum counts the lens segment as vacuum. Replacing that segment by material of refractive index

n

adds

(n - 1) ℓ (d)

to the optical path length. Thus, to the retained thin-element accuracy,

c T = a^{2} + d^{2} + b^{2} + d^{2} + (n - 1) (t_{0} - α d^{2}) .

Using the Taylor expansion

1 + u = 1 + u /2 + O (u^{2})

for

d ≪ a, b

gives

T = \frac{1}{c} [a (1 + \frac{d ^{2}}{2 a ^{2}}) + b (1 + \frac{d ^{2}}{2 b ^{2}}) + (n - 1) (t_{0} - α d^{2})] .

The omitted geometrical terms include

O (d^{4} / a^{3} + d^{4} / b^{3})

in optical path length, together with finite-thickness corrections discarded by the thin lens and paraxial approximations. The addition is

n - 1

, because the underlying distance already includes propagation across the thickness once in vacuum.

b

Solution

Fermat's principle makes the optical path length, or transit time in a stationary medium, stationary under small variations of the light ray with fixed endpoints. It need not be a global minimum. In the result of part (a), the coefficient of

d^{2}

\frac{1}{2 a} + \frac{1}{2 b} - α (n - 1) .

For a family of paraxial light rays from the object to its image, stationarity with respect to

d

requires this coefficient to vanish. Therefore the thin lens equation is

\frac{1}{a} + \frac{1}{b} = 2 α (n - 1) = \frac{1}{f} .

For a distant object,

a \to \infty

, and the image position

b = f

defines the focal length. Hence

f = \frac{1}{2 α ( n - 1 )} .

The axial member

d = 0

is stationary by symmetry for arbitrary

b

; requiring the neighbouring non-axial light rays to meet the same image is what determines

b

c

Solution

In surrounding refractive index

n^{'}

, the geometrical segments contribute

n^{'} [a^{2} + d^{2} + b^{2} + d^{2}]

to the optical path length, while replacement by the lens adds

(n - n^{'}) ℓ (d)

. Fermat's principle therefore gives

n^{'} (\frac{1}{a} + \frac{1}{b}) = 2 α (n - n^{'}) .

The physical focal length measured in the surrounding medium is

f (n^{'}) = \frac{n ^{'}}{2 α ( n - n ^{'} )} = \frac{1}{2 α ( n / n ^{'} - 1 )} .

For

α > 0

and

n > 1

, increasing

n^{'}

from

1

towards

n

makes the positive focal length grow monotonically from its vacuum value to infinity:

df / d n^{'} = n / [2 α (n - n^{'})^{2}] > 0

. At

n^{'} = n

, there is no refraction at the boundary, so the element has zero optical power and no finite optical focus. If

n^{'} > n

, its optical power changes sign and the same shape becomes diverging.

d

Solution

For the convex vacuum void, use internal refractive index

1

, external refractive index

n

, and the original

α > 0

in the immersed thin lens formula:

f_{void} = \frac{n}{2 α ( 1 - n )} = - \frac{n}{2 α ( n - 1 )} .

A concave glass thin lens of the same curvature magnitude in vacuum has

α_{concave} = - α

and therefore

f_{concave} = - \frac{1}{2 α ( n - 1 )}, f_{void} = n f_{concave} .

Both elements diverge, but their physical focal lengths differ by a factor

n

for

n > 1

. For example,

n = 1.5

makes the void's focal length magnitude

1.5

times that of the concave glass element. Their reduced optical powers

n_{outside} / f

are equal; physical focal length also depends on the surrounding refractive index.

2021 resit (A10288W1)

Source paper: www.whatdotheyknow.com/request/physics_preliminary_past_papers/response/2680788/attach/3/2018%202021%20Physics%20prelims.zip?cookie_passthrough=1 (archive member 2020/L/a10288w1.pdf; the cover dates this sitting to 7 September 2021).

The lettered subdivisions added to otherwise unlabelled paragraphs below follow their order in the source. Both sections and every optional question are solved.

1

a

Solution

In vacuum, Gauss's law states that the outward electric flux through a closed surface is the enclosed electric charge divided by the vacuum permittivity:

\oint_{S} E \cdot d A = \frac{Q _{enc}}{ϵ _{0}} .

Take opposite electric charges

\pm Q

on the two electrical conductors, with surface charge densities

\pm σ = \pm Q / A

. A Gaussian pillbox crossing one isolated large sheet gives

2 ES = σ S / ϵ_{0}

, so each sheet contributes

σ / (2 ϵ_{0})

on either side. By the superposition principle, the electric fields add between the sheets and cancel outside. Ignoring the fringing field of a capacitor, the gap electric field is

E = σ / ϵ_{0}

. The electric potential difference and capacitance of this parallel-plate capacitor are consequently

V = \int_{0}^{d} E d x = \frac{Q d}{ϵ _{0} A}, C = \frac{Q}{V} = \frac{ϵ _{0} A}{d} .

The approximation requires

d

much smaller than both transverse sheet dimensions. A uniform filling with permittivity

ϵ

replaces

ϵ_{0}

ϵ

b

Solution

Put electric charge

+ Q

on the inner electrical conductor and

- Q

on the outer one. In the gap, spherical symmetry makes the electric field radial and constant on a spherical Gaussian surface. Gauss's law gives

4 π r^{2} E (r) = \frac{Q}{ϵ _{0}}, a < r < b .

The inner-to-outer electric potential difference is therefore

V = \int_{a}^{b} \frac{Q}{4 π ϵ _{0} r ^{2}} d r = \frac{Q}{4 π ϵ _{0}} (\frac{1}{a} - \frac{1}{b}) .

Thus the capacitance of the spherical capacitor is

C = 4 π ϵ_{0} \frac{ab}{b - a} .

b \to \infty

this gives the capacitance of an isolated conducting sphere,

4 π ϵ_{0} a

; for

b - a ≪ a

it approaches the parallel-plate capacitor result with area

4 π a^{2}

c

Solution

Use opposite electric charges

\pm Q

over a common length

L

, and write the inner linear charge density as

λ = Q / L

. Away from the ends, axisymmetry and Gauss's law on a coaxial cylindrical Gaussian surface of length

ℓ

give

E (r) 2 π r ℓ = \frac{λ ℓ}{ϵ _{0}}, E (r) = \frac{λ}{2 π ϵ _{0} r}, a < r < b .

The electric potential difference and capacitance of the coaxial cylindrical capacitor are

V = \int_{a}^{b} E (r) d r = \frac{Q}{2 π ϵ _{0} L} lo g \frac{b}{a}, C = \frac{2 π ϵ _{0} L}{lo g ( b / a )} .

This neglects the fringing field of a capacitor at the ends, as justified when

L ≫ b

2

a

Solution

The real image is upside down. A converging optical lens produces negative transverse magnification on the screen: a point above the optical axis sends a central light ray through the optical lens centre to a point below the optical axis. The same happens in the horizontal direction. Viewed directly from behind, the screen therefore shows a picture rotated through

18 0^{\circ}

relative to the scene.

b

Solution

The image is upright vertically when viewed down from behind, but left and right are interchanged. To track the image formation, let

+ x

point forwards and

+ z

upwards. Incoming light travels along

- x

. The

4 5^{\circ}

flip plane mirror sends it along

+ z

. Before this reflection, the optical lens has inverted the scene's top into the

- z

direction; the law of reflection maps that direction into

+ x

on the horizontal screen. The screen's far edge is the top of the picture for someone looking down from behind. The horizontal inversion from the optical lens remains, since this plane mirror does not change the lateral coordinate.

c

Solution

The minimum is two additional plane mirrors, making three including the flip plane mirror. Their reflecting planes can meet at

4 5^{\circ}

in the vertical section, as in a pentaprism. For example, in the

x z

section take the first added reflecting plane at

67. 5^{\circ}

+ x

and the second at

112. 5^{\circ}

. The upward beam first travels diagonally forwards and upwards, then horizontally backwards towards an eye looking forwards. The two reflections together rotate directions through

9 0^{\circ}

: the propagation direction

+ z

becomes

- x

, while the screen's top direction

+ x

becomes

+ z

at the eye. Thus the picture remains upright vertically.

One additional plane mirror would also send

+ z

into

- x

, but it would send the top direction

+ x

into

- z

, making the image upside down. This proves minimality for ordinary plane mirrors. The three-reflection arrangement still reverses left and right; correcting that additionally requires a roof arrangement or another suitable image-erecting system. Vertical erection alone is what is asked here.

Figure 1.
An original vertical-section ray construction with the minimum two added plane mirrors
. **Details**.

d

Solution

Use a converging thin lens as a magnifying glass, with the screen at or just inside its front focal plane. It presents an upright virtual image at infinity or at the eye's near point, respectively. Let the near point distance be

D

; the conventional numerical assumption is

D = 250 mm

, which is not specified in the paper.

Under the usual approximation that the eye is close to the eyepiece, interpreting

50 mm

as the focal length

f

gives the familiar angular magnifications

M_{\infty} = \frac{D}{f} = 5, M_{near} = 1 + \frac{D}{f} = 6 (f = 50 mm, D = 250 mm) .

The first is for a relaxed eye with the screen at the front focal plane. For the second, the virtual image is at signed distance

v = - D

and the thin lens equation gives screen distance

u = f D / (D + f) = 41.7 mm

. Thus the often intended maximum

6

requires a screen-to-eye distance of approximately

41.7 mm

, rather than exactly

50 mm

50 mm

instead fixes the physical screen-to-eye separation

s

, an eye adjacent to the optical lens sees the angular magnification

M = \frac{D}{s} = 5 (s = 50 mm, D = 250 mm) .

This is achieved by

f = s

for relaxed viewing, or by

f = sD / (D - s) = 62.5 mm

for a virtual image at the near point. The latter follows from

1/ f = 1/ s - 1/ D

More generally, the source does not specify the eyepiece-to-eye distance either. If that distance is

e

, then

u = s - e

and a virtual image at the near point has transverse magnification

(D - e) / u

. Its angular magnification relative to unaided viewing at

D

M = \frac{D - e}{s - e} .

For example,

s = 50 mm

and

e = 25 mm

give

M = 9

, with

u = 25 mm

and

f = 28.125 mm

. This is a concrete counterexample to a universal maximum of

5

at fixed screen-to-eye separation. In ideal geometrical optics,

e \to s^{-}

allows arbitrarily large

M

with an arbitrarily short focal length; practical aperture and optical aberrations constrain this. A unique numerical maximum therefore needs the conventional near-eye eyepiece geometry and near point assumption. The conventional

f = 50 mm

answer is

6

, with those assumptions stated explicitly.

U > 0

be the drop in electric potential along the positive electric charge's path. By conservation of energy,

q U = m v^{2} /2

, so the nonrelativistic exit speed is

v = 2 q U / m

. Apply a uniform magnetic field perpendicular to the incident velocity and to the desired orbital plane. The magnetic Lorentz force is perpendicular to the velocity, does no work, and supplies the centripetal force:

q v B = \frac{m v ^{2}}{R}, R = \frac{m v}{qB} = \frac{1}{B} \frac{2 U}{q / m} .

For example,

v = v x

and

B = B z

give initial deflection along

- y

for positive

q

. Reversing the magnetic field reverses the bending. A uniform electric field alone would give constant acceleration and a parabolic path, not uniform circular motion. The displayed result assumes

q U ≪ m c^{2}

; the exact relativistic momentum relation would give

R = q^{2} U^{2} / c^{2} + 2 m q U / (qB)

b

Solution

Add a uniform electric field satisfying

E = - v_{0} \times B

. Then the Lorentz force

q (E + v_{0} \times B)

vanishes for the desired velocity. With the axes used above,

E = E y

, perpendicular to both

v_{0}

and

B

, with

E = B v_{0} = B 2 U (q / m)_{0}, (q / m)_{0} = \frac{E ^{2}}{2 U B ^{2}} .

The crossed fields form a velocity selector. All particles must first have traversed the same electric potential difference from rest: that is what converts selection of speed into selection of charge-to-mass ratio. Equal charge-to-mass ratios yield equal speeds even if the separate masses and charges differ. For other ratios, the initial transverse force is

q (E - v B)

and is nonzero.

v (t) = Re (V e^{jω t})

i (t) = Re (I e^{jω t})

, and

j^{2} = - 1

. The complex impedance is

Z = V / I = X + jY

: its modulus is the voltage-to-current amplitude ratio, and its complex argument is the voltage-minus-current phase difference. Here

X

is the electrical resistance part and

Y

the signed electrical reactance; the source's

X, Y

notation differs from the common

R, X

notation.

The component laws

v_{R} = R i

v_{L} = L d i / d t

and

i_{C} = C d v_{C} / d t

give the complex impedances of an ideal resistor, inductor and capacitor:

Z_{R} = R, Z_{L} = jω L, Z_{C} = \frac{1}{jω C} = - \frac{j}{ω C} .

In a series circuit the same electric current flows through both components, so Kirchhoff's voltage law adds their voltage phasors. The resulting series RC impedance is

Z = R - \frac{j}{ω C}, ∣ Z ∣ = R^{2} + \frac{1}{ω ^{2} C ^{2}}, ar g Z = - arctan \frac{1}{ω CR} .

For positive

R, C, ω

, the negative phase difference means that electric current leads the applied electric potential difference.

b

Solution

In the parallel LC circuit both components have the same voltage phasor. Adding their current phasors by Kirchhoff's current law gives the total electrical admittance:

\frac{1}{Z} = \frac{1}{jω L} + jω C = j (ω C - \frac{1}{ω L}) .

Thus

Z = \frac{jω L}{1 - ω ^{2} L C}, ∣ Z ∣ = \frac{ω L}{∣1 - ω ^{2} L C ∣}, ar g Z = {+ π /2, - π /2, ω < 1/ L C, ω > 1/ L C .