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程序代写案例-PHYS4031

时间：2021-04-07

SHOWING SOLUTIONS

Tuesday, 30 April 2019

09:30 – 11:00

EXAMINATION FOR THE DEGREES OF B.SC. (DESIGNATED AND

HONOURS), M.SCI. (HONOURS) AND M.SC.

[ PHYS4031 ]

Waves & Diffraction

Candidates should answer Question 1 (16 marks)

and either Question 2A or Question 2B (24 marks each)

Answer each question in a separate booklet

Candidates are reminded that devices able to store or display text or images

may not be used in examinations without prior arrangement.

Approximate marks are indicated in brackets as a guide for candidates.

PHYS4031 Waves & Diffraction

Fundamental constants

name symbol value

speed of light c 2.998× 108 m s−1

permeability of free space µ0 4pi × 10−7 H m−1

permittivity of free space 0 8.854× 10−12 F m−1

electronic charge e 1.602× 10−19 C

Avogadro’s number N0 6.022× 1023 mol−1

electron rest mass me 9.110× 10−31 kg

proton rest mass mp 1.673× 10−27 kg

neutron rest mass mn 1.675× 10−27 kg

Faraday’s constant F 9.649× 10−4 C mol−1

Planck’s constant h 6.626× 10−34 J s

fine structure constant α 7.297× 10−3

electron charge to mass ratio e/me 1.759× 1011 C kg−1

quantum/charge ratio h/e 4.136× 10−15 J s C−1

electron Compton wavelength λe 2.426× 10−12 m

proton Compton wavelength λp 1.321× 10−15 m

Rydberg constant R 1.097× 107 m−1

Bohr radius a0 5.292× 10−11 m

Bohr magneton µB 9.274× 10−24 J T−1

nuclear magneton µN 5.051× 10−27 J T−1

proton magnetic moment µp 1.411× 10−26 J T−1

universal gas constant R 8.314 J K−1 mol−1

normal volume of ideal gas – 2.241× 10−2 m3 mol−1

Boltzmann constant kB 1.381× 10−23 J K−1

First radiation constant 2pihc2 c1 3.742× 10−16 W m2

Second Radiation constant hc/kB c2 1.439× 10−2 m K

Wien displacement constant b 2.898× 10−3 m K

Stefan-Boltzmann constant σ 5.670× 10−8 W m−2 K−4

gravitational constant G 6.673× 10−11 m3 kg−1 s−2

impedance of free space Z0 3.767× 102 Ω

Derived units

quantity dimensions∗ derived unit

energy ML2T−2 J

force MLT−2 N

frequency T−1 Hz

gravitational field strength LT−2 N kg−1

gravitational potential L2T−2 J kg−1

power ML2T−3 W

entropy ML2T−2 J K−1

heat ML2T−2 J

capacitance M−1L−2T 4I2 F

charge IT C

current I A

electric dipole moment LTI C m

electric displacement L−2TI C m−2

electric polarisation L−2TI C m−2

electric field strength MLT−3I−1 V m−1

electric (displacement) flux TI C

electric potential ML2T−3I−1 V

inductance ML2T−2I−2 H

magnetic dipole moment L2I A m2

magnetic field strength L−1I A m−1

magnetic flux ML2T−2I−1 Wb

magnetic induction MT−2I−1 T

magnetisation L−1I A m−1

permeability MLT−2I−2 H m−1

permittivity M−1L−3T 4I2 F m−1

resistance ML2T−3I−2 Ω

resistivity ML3T−3I−2 Ω m

∗ M = mass, L = length, T = time, I = current

SHOWING SOLUTIONS

1 (a) A string of length 2 m and total mass 0.1 kg is placed under a tension of

30 N.

i. Calculate the propagation speed of transverse waves on the string. [2]

Solution: Unseen problem requiring recall of standard relations and

deduction of appropriate units.

The linear density ρ = 0.12 = 0.05 kg m

−1 .

c =

√

T

ρ

=

√

30

0.05

=

√

600 = 10

√

6 = 24.5 m s−1 .

[. . . 2]

ii. Define the impedance of a stretched string. Calculate the impedance of

the string with parameters given above. Ensure you quote appropriate

units for your result. [3]

Solution: The impedance of a stretched string is the ratio of transverse

force to transverse velocity. The transverse force is −T ∂y∂x . For

non-dispersive propagation ∂y∂x = −1c ∂y∂t , so the transverse force is Tc ∂y∂t . Thus

Z = Tc =

T√

T/ρ

=

√

Tρ . [. . . 2]

So for the problem given, Z =

√

30× 0.05 = √1.5 = 1.22 N s m−1 . [. . . 1]

iii. The ratio of reflected amplitude to incident amplitude for waves inci-

dent from a medium of impedance Z1 on a boundary with a medium of

impedance Z2 is

Z1 − Z2

Z1 + Z2

.

What is the ratio of reflected energy to incident energy? Hence deduce an

expression for the ratio of transmitted energy to incident energy. Finally,

deduce an expression for the ratio of the reflected wave amplitude to the

incident amplitude. [4]

Solution: Bookwork, with a small unseen element.

The ratio of reflected energy to incident energy = Z1Z1

[

Z1−Z2

Z1+Z2

]2

where

the first term is included here – though the numerator and denominator

cancel – as a reminder that the impedance of the string in which a wave

propagates must be taken into account when moving from wave amplitude to

energy. This will not be required in a student’s solution. [. . . 1]

Conservation of energy then dictates that the ratio of transmitted

energy to incident energy must be

1− Z1

Z1

[

Z1 − Z2

Z1 + Z2

]2

PHYS4031 Waves & Diffraction 3/16 Q 1 continued over. . .

Q 1 continued

SHOWING SOLUTIONS

=

(Z1 + Z2)

2 − (Z1 − Z2)2

(Z1 + Z2)2

=

4Z1Z2

(Z1 + Z2)2

.

[. . . 1]

The ratio of transmitted energy to incident energy is Z2Z1

C2

A2

, where we

denote the incident wave amplitude by A and the transmitted wave

amplitude by C . [. . . 1]

The ratio of reflected amplitude to incident amplitude is

C

A

=

√

C2

A2

=

√

Z1

Z2

[

4Z1Z2

(Z1 + Z2)2

]

=

√

4Z1

2

(Z1 + Z2)2

=

2Z1

Z1 + Z2

.

[. . . 1]

(b) Waves propagating in a particular medium follow a dispersion relation

given by

ω = ak3 − bk2 ,

where a and b are constants and k is the wavevector. Obtain expressions for

the phase and group velocities of the propagating waves and show that they

are equal for k = b/(2a) . [3]

Solution: Standard knowledge plus application to an unseen problem.

Phase velocity = ωk = ak

2 − bk . [. . . 1]

Group velocity = ∂ω∂k = 3ak

2 − 2bk . [. . . 1]

Apart from the trivial solution when k = 0 these velocities are equal when

ak2 − bk = 3k2 − 4bk ⇒ ak − b = 3ak − 2b⇒ k = b2a . [. . . 1]

Sketch the 2-dimensional Fraunhofer diffraction pattern that would arise

from plane wave illumination of a circular aperture. Referring to your diagram,

give an expression for the angle subtended at the aperture by the first signif-

icant feature in the diffraction pattern. Without detailed calculation, sketch

the intensity profile along a straight line through the midpoint of the pattern. [4]

Solution:

Bookwork

2 marks for circular ring pattern plus formula, and 2 marks for a rough sketch

of the radial intensity distribution.

PHYS4031 Waves & Diffraction 4/16 Q 1 continued over. . .

Q 1 continued

SHOWING SOLUTIONS

[. . . 4]

PHYS4031 Waves & Diffraction 5/16 Paper continued over. . .

SHOWING SOLUTIONS

2A A string of length L and linear density ρ lies along the x-axis and is stretched

under tension T . The ends of the string are fixed at x = 0 and x = L .

(a) Sketch the first four normal modes of vibration of the string. [2]

Solution: Standard knowledge

Mode shapes for n = 1...4 are plotted below. Half a mark will be gained for

each of the four plots.

(In the plot, the horizontal axis extends from x = 0 to x = 1, corresponding to

the 0 to L points along the string length. The vertical axis is the transverse

displacement. All modes are illustrated with the same amplitude.) [. . . 2]

(b) The string is initially undisplaced and at time t = 0 is given a velocity

profile

v(x) =

4v0

L2

x(L− x)

for a positive constant v0. Sketch this velocity profile. Without calculation,

discuss the expected excitation of normal modes of vibration of the string that

would result from this initial perturbation. [4]

Solution: Unseen problem; related problems have featured in examples sheet

and in past papers.

The initial velocity profile is a quadratic in x. The velocity is 0 at x = 0 and

x = L and has a maximum value of v0 at x = L/2.˙

PHYS4031 Waves & Diffraction 6/16 Q 2A continued over. . .

Q 2A continued

SHOWING SOLUTIONS

(In the plot, the horizontal axis extends from x = 0 to x = 1, corresponding to

the 0 to L points along the string length. The vertical v(x) axis is scaled so that 1

corresponds to the maximum initial transverse velocity v0 .) [. . . 2]

The shape of the velocity profile is similar to that of the n = 1 mode. It is not

precisely the same, however, so the expectation would be that the fundamental

mode would be the dominant one excited, with smaller amplitudes of higher order

modes. The initial profile has a large value at the midpoint of the string, so even

order modes – which would have a node at the midpoint – cannot be excited. [. . . 2]

(c) The motion of such a string can, in general, be described by a Fourier

decomposition of the form

y(x, t) =

∞∑

n=1

sin

(npix

L

)[

An cos

(npict

L

)

+Bn sin

(npict

L

)]

.

Explaining carefully each step of your calculation, show that for the given

initial condition the subsequent string motion can be expressed as

y(x, t) =

∞∑

n=1,3,5,...

32v0L

n4pi4c

sin

(npix

L

)

sin

(npict

L

)

,

where c is the propagation speed of transverse waves on the string. [10]

PHYS4031 Waves & Diffraction 7/16 Q 2A continued over. . .

Q 2A continued

SHOWING SOLUTIONS

Solution: Standard type of problem, but not seen before in this form.

Note that for educational purposes the solution given below includes even

relatively minor steps in the argument. A typical student-presented solution could

be expected to be significantly more concise.

y(x, t) =

∞∑

n=1

sin

(npix

L

)[

An cos

(

npict

L

)

+Bn sin

(

npict

L

)]

.

The string is initially undisplaced, so y(x, 0) = 0 for all x along the string.

y(x, 0) =

∞∑

n=1

sin

(npix

L

)

[An] = 0 ⇒ An = 0

for all n . [. . . 2]

y(x, t) =

∞∑

n=1

Bn sin

(npix

L

)

sin

(

npict

L

)

.

We are given an initial velocity profile so we need

y˙(x, t) =

∞∑

n=1

Bn

(npic

L

)

sin

(npix

L

)

cos

(

npict

L

)

⇒ y˙(x, 0) =

∞∑

n=1

Bn

(npic

L

)

sin

(npix

L

)

.

Thus,

4v0

L2

x(L− x) =

∞∑

n=1

Bn

(npic

L

)

sin

(npix

L

)

.

[. . . 2]

We solve this by multiplying both sides by sin mpixL and then integrating both

sides from 0 to L , making use of the orthogonality of the sine function to evaluate

the right hand side. Here are the steps:

4v0

L2

x(L− x) sin

(mpix

L

)

=

∞∑

n=1

Bn

(npic

L

)

sin

(npix

L

)

sin

(mpix

L

)

∫ L

0

4v0

L2

x(L− x) sin

(mpix

L

)

dx =

∫ L

0

∞∑

n=1

Bn

(npic

L

)

sin

(npix

L

)

sin

(mpix

L

)

dx

∫ L

0

4v0

L2

x(L− x) sin

(mpix

L

)

dx =

∞∑

n=1

∫ L

0

Bn

(npic

L

)

sin

(npix

L

)

sin

(mpix

L

)

dx

4v0

L2

∫ L

0

x(L− x) sin

(mpix

L

)

dx =

∞∑

n=1

Bn

(npic

L

)∫ L

0

sin

(npix

L

)

sin

(mpix

L

)

dx

PHYS4031 Waves & Diffraction 8/16 Q 2A continued over. . .

Q 2A continued

SHOWING SOLUTIONS

The set of integrals under the sum each evaluate to 0 for m 6= n ; for m = n the

integral evaluates to L/2 . [. . . 1]

To find the Bm coefficients we then have to solve

Bm

(mpic

L

) L

2

L2

4v0

= Bm

(

mpicL2

8v0

)

=

∫ L

0

x(L− x) sin

(mpix

L

)

dx.

[. . . 1]

Using the given integrals (which are readily obtained by integration by parts)

the integral on the right hand side becomes

L

∫ L

0

x sin

mpi

L

xdx−

∫ L

0

x2 sin

mpi

L

xdx

= L

[

− Lx

mpi

cos

(mpix

L

)

+

L2

m2pi2

sin

(mpix

L

)]L

0

−

[

2L3

m3pi3

cos

(mpix

L

)

− Lx

2

mpi

cos

(mpix

L

)

+

2L2x

m2pi2

sin

(mpix

L

)]L

0

= L

[

− L

2

mpi

cosmpi + 0− (−0)− 0

]

−

[

2L3

m3pi3

cosmpi − L

3

mpi

cosmpi + 0− 2L

3

m3pi3

+ 0− 0

]

= − L

3

mpi

cosmpi − 2L

3

m3pi3

cosmpi +

L3

mpi

cosmpi +

2L3

m3pi3

=

2L3

m3pi3

[1− cosmpi] .

[. . . 2]

This equals 0 for m even - as expected. For m odd, we then have the result

Bm

(

mpicL2

8v0

)

=

2L3

m3pi3

[1− (−1)]

⇒ Bm =

(

8v0

mpicL2

)

2L3

m3pi3

[1− (−1)]

⇒ Bm = 32v0L

m4pi4c

.

[. . . 2]

(d) The energy associated with the nth normal mode of vibration of the string

can be expressed as

ρL

4

ωn

2

(

An

2 +Bn

2

)

,

where ωn is the angular frequency of the mode.

Sum the energies of all the excited normal modes to show that the total

energy of vibration is 415Mv0

2 , where M is the total mass of the string. [4]

PHYS4031 Waves & Diffraction 9/16 Q 2A continued over. . .

Q 2A continued

SHOWING SOLUTIONS

Solution: Standard type of calculation, but a new example.

Since, by earlier analysis, or by inspection of the result to be proved in part

(c), An = 0 for all n , [. . . 1]

then En =

ρL

4 ωn

2Bn

2 . Substituting

ωn =

npic

L

,

[. . . 1]

we then have for n odd (Bn = 0 for n even)

En =

ρL

4

(npic

L

)2( 32v0L

m4pi4c

)2

= 256ρLv0

2 1

pi6

1

n6

.

Since M = ρL, we have, for odd n ,

En = Mv0

2 256

pi6

1

n6

and the total energy, summed over all modes, is

Etotal = Mv0

2 256

pi6

∑

n odd

1

n6

.

[. . . 1]

The summation required is the one given, so the final result is

Etotal = Mv0

2 256

pi6

pi6

960

= Mv0

2 4

15

.

[. . . 1]

(e) Calculate the total kinetic energy given to the string by the initial velocity

profile and show that this agrees with the sum of normal mode energies. [4]

Solution: The initial energy is all kinetic. We start with an expression for

the KE in a segment dx of the string:

Eelement =

1

2

(ρdx) v2 =

1

2

(ρdx)

(

4v0

L2

x(L− x)

)2

[. . . 1]

=

8ρv0

2

L4

[x(L− x)]2

and the total KE is then obtained by evaluating the integral

Etotal =

8ρv0

2

L4

∫ L

0

x2(L− x)2dx

PHYS4031 Waves & Diffraction 10/16 Q 2A continued over. . .

Q 2A continued

SHOWING SOLUTIONS

[. . . 1]

=

8ρv0

2

L4

∫ L

0

(L2x2 − 2Lx3 + x4) dx = 8ρv0

2

L4

[

L2x3

3

− 2Lx

4

4

+

x5

5

]L

0

=

8ρv0

2

L4

∫ L

0

(L2x2 − 2Lx3 + x4) dx = 8ρv0

2

L4

[

L5

3

− L

5

2

+

L5

5

]

= 8ρLv0

2

[

1

3

− 1

2

+

1

5

]

= 8ρLv0

2

[

10− 15 + 6

30

]

= 8ρLv0

2

[

1

30

]

=

4

15

Mv0

2,

as expected. [. . . 2]

You may find the following relations useful:∫

x sin ax dx = −x

a

cos ax+

1

a2

sin ax

∫

x2 sin ax dx =

2

a3

cos ax− x

2

a

cos ax+

2x

a2

sin ax

∞∑

n=1,3,5,...

1

n6

=

pi6

960

PHYS4031 Waves & Diffraction 11/16 Paper continued over. . .

SHOWING SOLUTIONS

2B The Fresnel-Kirchhoff diffraction formula for the field at a point P due to

diffraction at an aperture Q can be expressed as

ψP = − i

λ

∫

Q

eik r

r

ψQ

(

n + n′

2

)

· dΣ ,

where ψQ is the field at the aperture and other symbols have their usual

meanings.

(a) With the aid of a diagram, explain the meaning of the symbols that appear

in the Fresnel-Kirchhoff diffraction formula. [4]

Solution: General understanding

Figure: From Applications of Classical Physics, by

Roger D. Blandford and Kip S. Thorne, available at

http://www.pmaweb.caltech.edu/Courses/ph136/yr2012/1208.1.K.pdf.

Formally the integral to calculate the field at an interior point P starts as

being evaluated over a closed surface around P that includes the aperture Q .

Taking most of the surface to be far away from the aperture (plus some other -

irrelevant here - mathematical tricks near the aperture) makes only the

contribution to the integral coming from evaluation over the aperture significant. [. . . 1]

The i term gives a phase offset and e

ikr

r gives the amplitude of a

spherically-expanding wavelet of wavevector k at a distance r. [. . . 1]

The vector area element dΣ, pointing inwards, is the element of the surface of

integration. It’s shown here for a generic location on the closed surface, however,

PHYS4031 Waves & Diffraction 12/16 Q 2B continued over. . .

Q 2B continued

SHOWING SOLUTIONS

for the reason given above, a student may well jump to showing it only as an

element of the surface within the aperture.

Regardless, r is the distance from the area element to the point of interest, P . [. . . 1]

The unit vector n′ is in the direction of propagation of the illuminating plane

waves incident on the aperture, and the unit vector n is in the direction from the

aperture to the point P . [. . . 1]

When evaluating the Fraunhofer case, integrating over the aperture the

obliquity factor

(

n+n′

2

)

· dΣ becomes dΣ . This factor suppresses

backward-propagating rays. [. . . 1]

NB. Full marks will be awarded for solutions containing a majority of the

above information.

(b) Discuss briefly the ways in which this result was an improvement over the

previous Huygens-wavelet approach to modeling diffraction. [3]

Solution: General understanding

The Fresnel-Kirchhoff treatment correctly includes the wavelength dependence

of the diffraction process. It also provides the correct prediction of phase. The

inclusion of the obliquity factor removes what would otherwise be a backward

propagating wave from the spherical Huygens wavelets. [. . . 3]

(c) When applied to the diffraction of light, the Fresnel-Kirchhoff result is

not complete in its description of the diffraction process. Why is this? [2]

Solution: General understanding

The Fresnel-Kirchhoff treatment calculates diffraction of a scalar field. Light,

being an electromagnetic wave, is a vector quantity. The principal omission is that

the scalar treatment cannot include consideration of polarisation effects. [. . . 2]

(d) A diffraction apparatus is set up with parallel illumination incident on

a rectangular aperture – of width a in the x-direction and height 3a in the

y-direction – followed by a lens of focal length f that focuses the resulting

diffraction pattern onto a screen.

Explain why this setup can produce a Fraunhofer diffraction pattern on

the screen.

By taking the Fourier Transform of the diffracting aperture, show that

the intensity in the Fraunhofer diffraction pattern as a function of spatial

frequencies kx and ky is given by

I(kx, ky) = I0 sinc

2

[

3aky

2

]

sinc2

[

akx

2

]

,

PHYS4031 Waves & Diffraction 13/16 Q 2B continued over. . .

Q 2B continued

SHOWING SOLUTIONS

where I0 is the intensity on the optic axis. [5]

Solution: General understanding applied to unseen but standard type of

problem.

Fraunhofer diffraction is observed in the far-field where the diffracted waves

are essentially parallel plane waves. The focusing lens converges these to form a

more local image in the focal plane of the lens – which is where the screen should

be placed. [. . . 2]

With the aperture centred about the origin, it is described by an aperture

function A(x, y) that is unity for −a/2 ≤ x ≤ a/2 and −3a/2 ≤ y ≤ 3a/2 and zero

elsewhere. Then the transmitted amplitude becomes

A′ (kx, ky) =

3a

2∫

− 3a

2

a

2∫

−a

2

e−i

[

x kx+ y ky

]

dxdy

=

3a

2∫

− 3a

2

e−iy ky dy ×

a

2∫

−a

2

e−ix kx dx

[. . . 1]

=

[

− 1

iky

] [

e−iy ky

]+ 3a

2

− 3a

2

×

[

− 1

ikx

] [

e−ix kx

]+a

2

−a

2

=

[

− 1

iky

] [

−2i sin 3aky

2

]

×

[

− 1

ikx

] [

−2i sin akx

2

]

[. . . 1]

=

[

3a

sin

3aky

2

3aky

2

]

×

[

a

sin akx2

akx

2

]

= 3a2 sinc

[

3aky

2

]

sinc

[

akx

2

]

⇒ I(kx, ky) = I0 sinc2

[

3aky

2

]

sinc2

[

akx

2

]

[. . . 1]

(e) Sketch this diffraction pattern intensity, making clear in your sketch the

relative orientations of the aperture and the diffraction pattern and the relative

spatial frequency values of intensity nulls. [5]

Solution: General understanding applied to unseen but standard type of

problem.

A plot along the lines of the picture below is expected. There should be an

indication of the 3:1 (kx : ky) difference in null spacing along the two axes, the

rectangular shape of the maxima should be oriented orthogonally to the the axes

of the rectangular slit, and there should be an indication that the intensity in the

secondary maxima diminishes as spatial frequency increases (e.g. by sketching of

rough contours). [. . . 5]

PHYS4031 Waves & Diffraction 14/16 Q 2B continued over. . .

Q 2B continued

SHOWING SOLUTIONS

(f) If the illuminating light has a wavelength of 500 nm, the width of the

aperture in the x-direction is 30µm and the focal length of the lens is 0.6 m, find

the position coordinates on the screen of the first two nulls in each dimension

of the diffraction pattern. [5]

Solution: Unseen numerical problem

The nulls are when the sinc functions first become zero. These will occur at

3aky

2 = ±pi and akx2 = ±pi , ⇒ ky = ±2pi3a and kx = ±2pia . [. . . 2]

The coordinates on the screen are determined by the angle corresponding to

these spatial frequencies and the focal length f of the image-forming lens. The lens

converges parallel rays originating at an angle θ, to a point that is displaced from

the (0, 0) point on the screen by a distance fθ, provided θ is small. [. . . 1]

So in this case, the first two nulls in the x-direction are located at

X = ±fθX = ±f

kx

k

= ±f

[

2pi

a

] [

1

k

]

= ±f λ

a

= ±0.6×

[

500× 10−9

30× 10−6

]

= ±10 mm .

Similarly, for the y-direction, the first two nulls are located at

Y = ±fθY = ±f

ky

k

= ±f

[

2pi

3a

] [

1

k

]

= ±f λ

3a

= ±0.6×

[

500× 10−9

90× 10−6

]

= ±3.3 mm .

[. . . 2]

PHYS4031 Waves & Diffraction 15/16 Paper continued over. . .

SHOWING SOLUTIONS

End of Paper

NOTE: SHOWING SOLUTIONS

PHYS4031 Waves & Diffraction 16/16 END

学霸联盟

Tuesday, 30 April 2019

09:30 – 11:00

EXAMINATION FOR THE DEGREES OF B.SC. (DESIGNATED AND

HONOURS), M.SCI. (HONOURS) AND M.SC.

[ PHYS4031 ]

Waves & Diffraction

Candidates should answer Question 1 (16 marks)

and either Question 2A or Question 2B (24 marks each)

Answer each question in a separate booklet

Candidates are reminded that devices able to store or display text or images

may not be used in examinations without prior arrangement.

Approximate marks are indicated in brackets as a guide for candidates.

PHYS4031 Waves & Diffraction

Fundamental constants

name symbol value

speed of light c 2.998× 108 m s−1

permeability of free space µ0 4pi × 10−7 H m−1

permittivity of free space 0 8.854× 10−12 F m−1

electronic charge e 1.602× 10−19 C

Avogadro’s number N0 6.022× 1023 mol−1

electron rest mass me 9.110× 10−31 kg

proton rest mass mp 1.673× 10−27 kg

neutron rest mass mn 1.675× 10−27 kg

Faraday’s constant F 9.649× 10−4 C mol−1

Planck’s constant h 6.626× 10−34 J s

fine structure constant α 7.297× 10−3

electron charge to mass ratio e/me 1.759× 1011 C kg−1

quantum/charge ratio h/e 4.136× 10−15 J s C−1

electron Compton wavelength λe 2.426× 10−12 m

proton Compton wavelength λp 1.321× 10−15 m

Rydberg constant R 1.097× 107 m−1

Bohr radius a0 5.292× 10−11 m

Bohr magneton µB 9.274× 10−24 J T−1

nuclear magneton µN 5.051× 10−27 J T−1

proton magnetic moment µp 1.411× 10−26 J T−1

universal gas constant R 8.314 J K−1 mol−1

normal volume of ideal gas – 2.241× 10−2 m3 mol−1

Boltzmann constant kB 1.381× 10−23 J K−1

First radiation constant 2pihc2 c1 3.742× 10−16 W m2

Second Radiation constant hc/kB c2 1.439× 10−2 m K

Wien displacement constant b 2.898× 10−3 m K

Stefan-Boltzmann constant σ 5.670× 10−8 W m−2 K−4

gravitational constant G 6.673× 10−11 m3 kg−1 s−2

impedance of free space Z0 3.767× 102 Ω

Derived units

quantity dimensions∗ derived unit

energy ML2T−2 J

force MLT−2 N

frequency T−1 Hz

gravitational field strength LT−2 N kg−1

gravitational potential L2T−2 J kg−1

power ML2T−3 W

entropy ML2T−2 J K−1

heat ML2T−2 J

capacitance M−1L−2T 4I2 F

charge IT C

current I A

electric dipole moment LTI C m

electric displacement L−2TI C m−2

electric polarisation L−2TI C m−2

electric field strength MLT−3I−1 V m−1

electric (displacement) flux TI C

electric potential ML2T−3I−1 V

inductance ML2T−2I−2 H

magnetic dipole moment L2I A m2

magnetic field strength L−1I A m−1

magnetic flux ML2T−2I−1 Wb

magnetic induction MT−2I−1 T

magnetisation L−1I A m−1

permeability MLT−2I−2 H m−1

permittivity M−1L−3T 4I2 F m−1

resistance ML2T−3I−2 Ω

resistivity ML3T−3I−2 Ω m

∗ M = mass, L = length, T = time, I = current

SHOWING SOLUTIONS

1 (a) A string of length 2 m and total mass 0.1 kg is placed under a tension of

30 N.

i. Calculate the propagation speed of transverse waves on the string. [2]

Solution: Unseen problem requiring recall of standard relations and

deduction of appropriate units.

The linear density ρ = 0.12 = 0.05 kg m

−1 .

c =

√

T

ρ

=

√

30

0.05

=

√

600 = 10

√

6 = 24.5 m s−1 .

[. . . 2]

ii. Define the impedance of a stretched string. Calculate the impedance of

the string with parameters given above. Ensure you quote appropriate

units for your result. [3]

Solution: The impedance of a stretched string is the ratio of transverse

force to transverse velocity. The transverse force is −T ∂y∂x . For

non-dispersive propagation ∂y∂x = −1c ∂y∂t , so the transverse force is Tc ∂y∂t . Thus

Z = Tc =

T√

T/ρ

=

√

Tρ . [. . . 2]

So for the problem given, Z =

√

30× 0.05 = √1.5 = 1.22 N s m−1 . [. . . 1]

iii. The ratio of reflected amplitude to incident amplitude for waves inci-

dent from a medium of impedance Z1 on a boundary with a medium of

impedance Z2 is

Z1 − Z2

Z1 + Z2

.

What is the ratio of reflected energy to incident energy? Hence deduce an

expression for the ratio of transmitted energy to incident energy. Finally,

deduce an expression for the ratio of the reflected wave amplitude to the

incident amplitude. [4]

Solution: Bookwork, with a small unseen element.

The ratio of reflected energy to incident energy = Z1Z1

[

Z1−Z2

Z1+Z2

]2

where

the first term is included here – though the numerator and denominator

cancel – as a reminder that the impedance of the string in which a wave

propagates must be taken into account when moving from wave amplitude to

energy. This will not be required in a student’s solution. [. . . 1]

Conservation of energy then dictates that the ratio of transmitted

energy to incident energy must be

1− Z1

Z1

[

Z1 − Z2

Z1 + Z2

]2

PHYS4031 Waves & Diffraction 3/16 Q 1 continued over. . .

Q 1 continued

SHOWING SOLUTIONS

=

(Z1 + Z2)

2 − (Z1 − Z2)2

(Z1 + Z2)2

=

4Z1Z2

(Z1 + Z2)2

.

[. . . 1]

The ratio of transmitted energy to incident energy is Z2Z1

C2

A2

, where we

denote the incident wave amplitude by A and the transmitted wave

amplitude by C . [. . . 1]

The ratio of reflected amplitude to incident amplitude is

C

A

=

√

C2

A2

=

√

Z1

Z2

[

4Z1Z2

(Z1 + Z2)2

]

=

√

4Z1

2

(Z1 + Z2)2

=

2Z1

Z1 + Z2

.

[. . . 1]

(b) Waves propagating in a particular medium follow a dispersion relation

given by

ω = ak3 − bk2 ,

where a and b are constants and k is the wavevector. Obtain expressions for

the phase and group velocities of the propagating waves and show that they

are equal for k = b/(2a) . [3]

Solution: Standard knowledge plus application to an unseen problem.

Phase velocity = ωk = ak

2 − bk . [. . . 1]

Group velocity = ∂ω∂k = 3ak

2 − 2bk . [. . . 1]

Apart from the trivial solution when k = 0 these velocities are equal when

ak2 − bk = 3k2 − 4bk ⇒ ak − b = 3ak − 2b⇒ k = b2a . [. . . 1]

Sketch the 2-dimensional Fraunhofer diffraction pattern that would arise

from plane wave illumination of a circular aperture. Referring to your diagram,

give an expression for the angle subtended at the aperture by the first signif-

icant feature in the diffraction pattern. Without detailed calculation, sketch

the intensity profile along a straight line through the midpoint of the pattern. [4]

Solution:

Bookwork

2 marks for circular ring pattern plus formula, and 2 marks for a rough sketch

of the radial intensity distribution.

PHYS4031 Waves & Diffraction 4/16 Q 1 continued over. . .

Q 1 continued

SHOWING SOLUTIONS

[. . . 4]

PHYS4031 Waves & Diffraction 5/16 Paper continued over. . .

SHOWING SOLUTIONS

2A A string of length L and linear density ρ lies along the x-axis and is stretched

under tension T . The ends of the string are fixed at x = 0 and x = L .

(a) Sketch the first four normal modes of vibration of the string. [2]

Solution: Standard knowledge

Mode shapes for n = 1...4 are plotted below. Half a mark will be gained for

each of the four plots.

(In the plot, the horizontal axis extends from x = 0 to x = 1, corresponding to

the 0 to L points along the string length. The vertical axis is the transverse

displacement. All modes are illustrated with the same amplitude.) [. . . 2]

(b) The string is initially undisplaced and at time t = 0 is given a velocity

profile

v(x) =

4v0

L2

x(L− x)

for a positive constant v0. Sketch this velocity profile. Without calculation,

discuss the expected excitation of normal modes of vibration of the string that

would result from this initial perturbation. [4]

Solution: Unseen problem; related problems have featured in examples sheet

and in past papers.

The initial velocity profile is a quadratic in x. The velocity is 0 at x = 0 and

x = L and has a maximum value of v0 at x = L/2.˙

PHYS4031 Waves & Diffraction 6/16 Q 2A continued over. . .

Q 2A continued

SHOWING SOLUTIONS

(In the plot, the horizontal axis extends from x = 0 to x = 1, corresponding to

the 0 to L points along the string length. The vertical v(x) axis is scaled so that 1

corresponds to the maximum initial transverse velocity v0 .) [. . . 2]

The shape of the velocity profile is similar to that of the n = 1 mode. It is not

precisely the same, however, so the expectation would be that the fundamental

mode would be the dominant one excited, with smaller amplitudes of higher order

modes. The initial profile has a large value at the midpoint of the string, so even

order modes – which would have a node at the midpoint – cannot be excited. [. . . 2]

(c) The motion of such a string can, in general, be described by a Fourier

decomposition of the form

y(x, t) =

∞∑

n=1

sin

(npix

L

)[

An cos

(npict

L

)

+Bn sin

(npict

L

)]

.

Explaining carefully each step of your calculation, show that for the given

initial condition the subsequent string motion can be expressed as

y(x, t) =

∞∑

n=1,3,5,...

32v0L

n4pi4c

sin

(npix

L

)

sin

(npict

L

)

,

where c is the propagation speed of transverse waves on the string. [10]

PHYS4031 Waves & Diffraction 7/16 Q 2A continued over. . .

Q 2A continued

SHOWING SOLUTIONS

Solution: Standard type of problem, but not seen before in this form.

Note that for educational purposes the solution given below includes even

relatively minor steps in the argument. A typical student-presented solution could

be expected to be significantly more concise.

y(x, t) =

∞∑

n=1

sin

(npix

L

)[

An cos

(

npict

L

)

+Bn sin

(

npict

L

)]

.

The string is initially undisplaced, so y(x, 0) = 0 for all x along the string.

y(x, 0) =

∞∑

n=1

sin

(npix

L

)

[An] = 0 ⇒ An = 0

for all n . [. . . 2]

y(x, t) =

∞∑

n=1

Bn sin

(npix

L

)

sin

(

npict

L

)

.

We are given an initial velocity profile so we need

y˙(x, t) =

∞∑

n=1

Bn

(npic

L

)

sin

(npix

L

)

cos

(

npict

L

)

⇒ y˙(x, 0) =

∞∑

n=1

Bn

(npic

L

)

sin

(npix

L

)

.

Thus,

4v0

L2

x(L− x) =

∞∑

n=1

Bn

(npic

L

)

sin

(npix

L

)

.

[. . . 2]

We solve this by multiplying both sides by sin mpixL and then integrating both

sides from 0 to L , making use of the orthogonality of the sine function to evaluate

the right hand side. Here are the steps:

4v0

L2

x(L− x) sin

(mpix

L

)

=

∞∑

n=1

Bn

(npic

L

)

sin

(npix

L

)

sin

(mpix

L

)

∫ L

0

4v0

L2

x(L− x) sin

(mpix

L

)

dx =

∫ L

0

∞∑

n=1

Bn

(npic

L

)

sin

(npix

L

)

sin

(mpix

L

)

dx

∫ L

0

4v0

L2

x(L− x) sin

(mpix

L

)

dx =

∞∑

n=1

∫ L

0

Bn

(npic

L

)

sin

(npix

L

)

sin

(mpix

L

)

dx

4v0

L2

∫ L

0

x(L− x) sin

(mpix

L

)

dx =

∞∑

n=1

Bn

(npic

L

)∫ L

0

sin

(npix

L

)

sin

(mpix

L

)

dx

PHYS4031 Waves & Diffraction 8/16 Q 2A continued over. . .

Q 2A continued

SHOWING SOLUTIONS

The set of integrals under the sum each evaluate to 0 for m 6= n ; for m = n the

integral evaluates to L/2 . [. . . 1]

To find the Bm coefficients we then have to solve

Bm

(mpic

L

) L

2

L2

4v0

= Bm

(

mpicL2

8v0

)

=

∫ L

0

x(L− x) sin

(mpix

L

)

dx.

[. . . 1]

Using the given integrals (which are readily obtained by integration by parts)

the integral on the right hand side becomes

L

∫ L

0

x sin

mpi

L

xdx−

∫ L

0

x2 sin

mpi

L

xdx

= L

[

− Lx

mpi

cos

(mpix

L

)

+

L2

m2pi2

sin

(mpix

L

)]L

0

−

[

2L3

m3pi3

cos

(mpix

L

)

− Lx

2

mpi

cos

(mpix

L

)

+

2L2x

m2pi2

sin

(mpix

L

)]L

0

= L

[

− L

2

mpi

cosmpi + 0− (−0)− 0

]

−

[

2L3

m3pi3

cosmpi − L

3

mpi

cosmpi + 0− 2L

3

m3pi3

+ 0− 0

]

= − L

3

mpi

cosmpi − 2L

3

m3pi3

cosmpi +

L3

mpi

cosmpi +

2L3

m3pi3

=

2L3

m3pi3

[1− cosmpi] .

[. . . 2]

This equals 0 for m even - as expected. For m odd, we then have the result

Bm

(

mpicL2

8v0

)

=

2L3

m3pi3

[1− (−1)]

⇒ Bm =

(

8v0

mpicL2

)

2L3

m3pi3

[1− (−1)]

⇒ Bm = 32v0L

m4pi4c

.

[. . . 2]

(d) The energy associated with the nth normal mode of vibration of the string

can be expressed as

ρL

4

ωn

2

(

An

2 +Bn

2

)

,

where ωn is the angular frequency of the mode.

Sum the energies of all the excited normal modes to show that the total

energy of vibration is 415Mv0

2 , where M is the total mass of the string. [4]

PHYS4031 Waves & Diffraction 9/16 Q 2A continued over. . .

Q 2A continued

SHOWING SOLUTIONS

Solution: Standard type of calculation, but a new example.

Since, by earlier analysis, or by inspection of the result to be proved in part

(c), An = 0 for all n , [. . . 1]

then En =

ρL

4 ωn

2Bn

2 . Substituting

ωn =

npic

L

,

[. . . 1]

we then have for n odd (Bn = 0 for n even)

En =

ρL

4

(npic

L

)2( 32v0L

m4pi4c

)2

= 256ρLv0

2 1

pi6

1

n6

.

Since M = ρL, we have, for odd n ,

En = Mv0

2 256

pi6

1

n6

and the total energy, summed over all modes, is

Etotal = Mv0

2 256

pi6

∑

n odd

1

n6

.

[. . . 1]

The summation required is the one given, so the final result is

Etotal = Mv0

2 256

pi6

pi6

960

= Mv0

2 4

15

.

[. . . 1]

(e) Calculate the total kinetic energy given to the string by the initial velocity

profile and show that this agrees with the sum of normal mode energies. [4]

Solution: The initial energy is all kinetic. We start with an expression for

the KE in a segment dx of the string:

Eelement =

1

2

(ρdx) v2 =

1

2

(ρdx)

(

4v0

L2

x(L− x)

)2

[. . . 1]

=

8ρv0

2

L4

[x(L− x)]2

and the total KE is then obtained by evaluating the integral

Etotal =

8ρv0

2

L4

∫ L

0

x2(L− x)2dx

PHYS4031 Waves & Diffraction 10/16 Q 2A continued over. . .

Q 2A continued

SHOWING SOLUTIONS

[. . . 1]

=

8ρv0

2

L4

∫ L

0

(L2x2 − 2Lx3 + x4) dx = 8ρv0

2

L4

[

L2x3

3

− 2Lx

4

4

+

x5

5

]L

0

=

8ρv0

2

L4

∫ L

0

(L2x2 − 2Lx3 + x4) dx = 8ρv0

2

L4

[

L5

3

− L

5

2

+

L5

5

]

= 8ρLv0

2

[

1

3

− 1

2

+

1

5

]

= 8ρLv0

2

[

10− 15 + 6

30

]

= 8ρLv0

2

[

1

30

]

=

4

15

Mv0

2,

as expected. [. . . 2]

You may find the following relations useful:∫

x sin ax dx = −x

a

cos ax+

1

a2

sin ax

∫

x2 sin ax dx =

2

a3

cos ax− x

2

a

cos ax+

2x

a2

sin ax

∞∑

n=1,3,5,...

1

n6

=

pi6

960

PHYS4031 Waves & Diffraction 11/16 Paper continued over. . .

SHOWING SOLUTIONS

2B The Fresnel-Kirchhoff diffraction formula for the field at a point P due to

diffraction at an aperture Q can be expressed as

ψP = − i

λ

∫

Q

eik r

r

ψQ

(

n + n′

2

)

· dΣ ,

where ψQ is the field at the aperture and other symbols have their usual

meanings.

(a) With the aid of a diagram, explain the meaning of the symbols that appear

in the Fresnel-Kirchhoff diffraction formula. [4]

Solution: General understanding

Figure: From Applications of Classical Physics, by

Roger D. Blandford and Kip S. Thorne, available at

http://www.pmaweb.caltech.edu/Courses/ph136/yr2012/1208.1.K.pdf.

Formally the integral to calculate the field at an interior point P starts as

being evaluated over a closed surface around P that includes the aperture Q .

Taking most of the surface to be far away from the aperture (plus some other -

irrelevant here - mathematical tricks near the aperture) makes only the

contribution to the integral coming from evaluation over the aperture significant. [. . . 1]

The i term gives a phase offset and e

ikr

r gives the amplitude of a

spherically-expanding wavelet of wavevector k at a distance r. [. . . 1]

The vector area element dΣ, pointing inwards, is the element of the surface of

integration. It’s shown here for a generic location on the closed surface, however,

PHYS4031 Waves & Diffraction 12/16 Q 2B continued over. . .

Q 2B continued

SHOWING SOLUTIONS

for the reason given above, a student may well jump to showing it only as an

element of the surface within the aperture.

Regardless, r is the distance from the area element to the point of interest, P . [. . . 1]

The unit vector n′ is in the direction of propagation of the illuminating plane

waves incident on the aperture, and the unit vector n is in the direction from the

aperture to the point P . [. . . 1]

When evaluating the Fraunhofer case, integrating over the aperture the

obliquity factor

(

n+n′

2

)

· dΣ becomes dΣ . This factor suppresses

backward-propagating rays. [. . . 1]

NB. Full marks will be awarded for solutions containing a majority of the

above information.

(b) Discuss briefly the ways in which this result was an improvement over the

previous Huygens-wavelet approach to modeling diffraction. [3]

Solution: General understanding

The Fresnel-Kirchhoff treatment correctly includes the wavelength dependence

of the diffraction process. It also provides the correct prediction of phase. The

inclusion of the obliquity factor removes what would otherwise be a backward

propagating wave from the spherical Huygens wavelets. [. . . 3]

(c) When applied to the diffraction of light, the Fresnel-Kirchhoff result is

not complete in its description of the diffraction process. Why is this? [2]

Solution: General understanding

The Fresnel-Kirchhoff treatment calculates diffraction of a scalar field. Light,

being an electromagnetic wave, is a vector quantity. The principal omission is that

the scalar treatment cannot include consideration of polarisation effects. [. . . 2]

(d) A diffraction apparatus is set up with parallel illumination incident on

a rectangular aperture – of width a in the x-direction and height 3a in the

y-direction – followed by a lens of focal length f that focuses the resulting

diffraction pattern onto a screen.

Explain why this setup can produce a Fraunhofer diffraction pattern on

the screen.

By taking the Fourier Transform of the diffracting aperture, show that

the intensity in the Fraunhofer diffraction pattern as a function of spatial

frequencies kx and ky is given by

I(kx, ky) = I0 sinc

2

[

3aky

2

]

sinc2

[

akx

2

]

,

PHYS4031 Waves & Diffraction 13/16 Q 2B continued over. . .

Q 2B continued

SHOWING SOLUTIONS

where I0 is the intensity on the optic axis. [5]

Solution: General understanding applied to unseen but standard type of

problem.

Fraunhofer diffraction is observed in the far-field where the diffracted waves

are essentially parallel plane waves. The focusing lens converges these to form a

more local image in the focal plane of the lens – which is where the screen should

be placed. [. . . 2]

With the aperture centred about the origin, it is described by an aperture

function A(x, y) that is unity for −a/2 ≤ x ≤ a/2 and −3a/2 ≤ y ≤ 3a/2 and zero

elsewhere. Then the transmitted amplitude becomes

A′ (kx, ky) =

3a

2∫

− 3a

2

a

2∫

−a

2

e−i

[

x kx+ y ky

]

dxdy

=

3a

2∫

− 3a

2

e−iy ky dy ×

a

2∫

−a

2

e−ix kx dx

[. . . 1]

=

[

− 1

iky

] [

e−iy ky

]+ 3a

2

− 3a

2

×

[

− 1

ikx

] [

e−ix kx

]+a

2

−a

2

=

[

− 1

iky

] [

−2i sin 3aky

2

]

×

[

− 1

ikx

] [

−2i sin akx

2

]

[. . . 1]

=

[

3a

sin

3aky

2

3aky

2

]

×

[

a

sin akx2

akx

2

]

= 3a2 sinc

[

3aky

2

]

sinc

[

akx

2

]

⇒ I(kx, ky) = I0 sinc2

[

3aky

2

]

sinc2

[

akx

2

]

[. . . 1]

(e) Sketch this diffraction pattern intensity, making clear in your sketch the

relative orientations of the aperture and the diffraction pattern and the relative

spatial frequency values of intensity nulls. [5]

Solution: General understanding applied to unseen but standard type of

problem.

A plot along the lines of the picture below is expected. There should be an

indication of the 3:1 (kx : ky) difference in null spacing along the two axes, the

rectangular shape of the maxima should be oriented orthogonally to the the axes

of the rectangular slit, and there should be an indication that the intensity in the

secondary maxima diminishes as spatial frequency increases (e.g. by sketching of

rough contours). [. . . 5]

PHYS4031 Waves & Diffraction 14/16 Q 2B continued over. . .

Q 2B continued

SHOWING SOLUTIONS

(f) If the illuminating light has a wavelength of 500 nm, the width of the

aperture in the x-direction is 30µm and the focal length of the lens is 0.6 m, find

the position coordinates on the screen of the first two nulls in each dimension

of the diffraction pattern. [5]

Solution: Unseen numerical problem

The nulls are when the sinc functions first become zero. These will occur at

3aky

2 = ±pi and akx2 = ±pi , ⇒ ky = ±2pi3a and kx = ±2pia . [. . . 2]

The coordinates on the screen are determined by the angle corresponding to

these spatial frequencies and the focal length f of the image-forming lens. The lens

converges parallel rays originating at an angle θ, to a point that is displaced from

the (0, 0) point on the screen by a distance fθ, provided θ is small. [. . . 1]

So in this case, the first two nulls in the x-direction are located at

X = ±fθX = ±f

kx

k

= ±f

[

2pi

a

] [

1

k

]

= ±f λ

a

= ±0.6×

[

500× 10−9

30× 10−6

]

= ±10 mm .

Similarly, for the y-direction, the first two nulls are located at

Y = ±fθY = ±f

ky

k

= ±f

[

2pi

3a

] [

1

k

]

= ±f λ

3a

= ±0.6×

[

500× 10−9

90× 10−6

]

= ±3.3 mm .

[. . . 2]

PHYS4031 Waves & Diffraction 15/16 Paper continued over. . .

SHOWING SOLUTIONS

End of Paper

NOTE: SHOWING SOLUTIONS

PHYS4031 Waves & Diffraction 16/16 END

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