Solutions

A13-1
Official (English)

General Grading Guidelines
When student’s solutions are correct and s/he also show how solutions were obtained, the stduentgets full credit. The scheme oulined below is helpful if the student’s answers are partially correct.
Attention will be paid to the detailed solution so, if the final answer is correct but it is obtained byincorrect method(s) then no credit will be given. Alternative solutions may exist and will be given
due credit.
Partial or full outcomes obtained for later sections in the problem which are incorrect solely becauseof errors being carried forward from previous sections, but are otherwise reasonable, will not be
further penalized. For example a dimensioanlly wrong answer when carried forward will not get
any credit in the subsequent sections. A numerically wrong evaluation when carried forward will get
credit in subsequent sections unless the numerical answer is patently wrong (e.g. the value of g is
981 m/sec2 ! )Incorrect or no labeling of an axis is penalized by -0.1 points
The numerical answer (i) must be correct to +/- 10% AND (ii) must respect significant figures.
It maybe noted that NO micro-marking scheme takes care of all contingencies. A certain
amount of discretion rests with and a certain level of judgement is invested in the academiccommittee.

Maxwell, Rayleigh and Mount Everest: THE SOLUTION
Oscillation of the electron cloud:A.1 (0.5 pt)⃗ is the electric field at the location of the molecule. The equation of motion of the charge in the𝐸(𝑡)⃗ would beabsence of 𝐸(𝑡)𝑦 ̈ = −𝜔02 𝑦,(1)and under forced oscillations𝑦 ̈ = −𝜔02 𝑦 −[a sign mistake or a term missing −0.3]

𝑞𝐸0cos 𝜔𝑡.
𝑚(2)
[0.5]

Solutions

A13-2
Official (English)

A.2 (0.5 pt)
In equation (2) we substitute 𝑦 = 𝑦0 cos 𝜔𝑡 to obtain𝑞𝐸0.
𝑚−𝜔2 𝑦0 = −𝜔02 𝑦0 −

(3)
[0.2]

This implies that the amplitude of oscillation is𝑦0 =𝑞𝐸0 /𝑚.
𝜔2 − 𝜔02

(4)
[0.3]

[a sign mistake or a term missing −0.1]

A.3 (0.5 pt)
Since 𝑦 is the separation between the positive and negative charge clouds, the magnitude 𝑝(𝑡) of thedipole moment is𝑞 2 𝐸0𝑝(𝑡) = 𝑞𝑦(𝑡) ≈cos 𝜔𝑡.
(5)
𝑚𝜔02
[0.5]

[sign mistake −0.1][answer without approximation −0.2]

A.4 (0.5 pt)
We model the atom as a stationary positive point charge 𝑞 surrounded by a spherical negative charge
cloud of total charge −𝑞, radius 𝑟 and mass 𝑚. Now let the charge cloud be displaced by a small
distance 𝑦. The electrostatic force on the electron cloud by the central positive charge is (see Figure)𝐹el⃗ = 𝑚𝑦𝑦̈ ̂ = −𝑞2𝑦𝑦 ̂ = −𝑚𝜔02 𝑦𝑦 ̂
4𝜋𝜖0 𝑟3(6)
[0.4]

−𝑞+𝑞𝑟

𝚥̂𝑦𝚤̂

Figure 1. Model of the atom with a central positive charge and a displaced spherical electron cloudof radius 𝑟Thus, the natural frequency of oscillation is
𝜔0 =𝑞√4𝜋𝜖0 𝑚𝑟3

(7)
[0.1]

Solutions

A13-3
Official (English)

Power radiated:

B.1 (1 pt)
Dimension of power is(8)[𝑠] = kg ⋅ m2 ⋅ s−3[0.1]

Dimension of dipole moment is(9)[𝑝0 ] = C ⋅ m.

[0.1]

We are using SI units. C stands for Coloumb. Dimension of 𝜔 is(10)[𝜔] = s−1 .

[0.1]

Dimension of 𝜖0 is

2[𝜖0 ] = C ⋅ N−12

⋅ m−2 = C ⋅ kg−1(11)⋅ m−3 ⋅ s2 .

[0.1]

Dimension of speed of light 𝑐 is

(12)[𝑐] = m ⋅ s−1 .

[0.1]

Let us take the ansatz 𝑠 = 𝑝0𝛼 𝜔𝛽 𝜖𝛾0 𝑐𝛿 , i.e.,(13)[𝑠] = [𝑝0 ]𝛼 [𝜔]𝛽 [𝜖0 ]𝛾 [𝑐]𝛿 .

[0.1]

We get four equations for the four variables,
𝛼 + 2𝛾 = 0,

𝛾 = −1,(14)− 𝛽 + 2𝛾 − 𝛿 = −3

𝛼 − 3𝛾 + 𝛿 = 2.

(15)This gives
𝛼 = 2,
Implying,

𝛽 = 4,𝛾 = −1,𝑝 2 𝜔4
𝑠=𝑘 0 3𝜖0 𝑐𝛿 = −3.

(16)

(17)
[0.4]

Solutions

A13-4
Official (English)

B.2 (0.2 pt)
We have
𝑠=1 𝑝02 𝜔4
1 𝑞 2 𝑦02 𝜔4
1 𝑞 4 𝐸02 𝜔4==.
12𝜋 𝜖0 𝑐3
12𝜋 𝜖0 𝑐3
12𝜋 𝑚2 𝜖0 𝑐3 𝜔04(18)[0.2]

Attenuation of the Intensity 𝐼(𝑥) :C.1 (1 pt)
Recall that the intensity is the power incident per unit area. Consider a horizontal column of theatmosphere of cross-sectional area 𝐴 and length Δ𝑥. Let the incident intensity be 𝐼(𝑥). Let thetransmitted intensity be 𝐼(𝑥 + Δ𝑥). The drop in the intensity is due to the scattering of light by theair molecules. If the number density of air molecules is 𝑛0 then the total power radiated per unit
volume is 𝑛0 𝑠. Therefore,
𝐼(𝑥)𝐴 − 𝐼(𝑥 + Δ𝑥)𝐴 = 𝑛0 𝑠(𝐴Δ𝑥).

(19)
[0.8]

This gives−𝑑𝐼= 𝑛0 𝑠.
𝑑𝑥

(20)
[0.2]

C.2 (0.5 pt)
Since 𝑠 ∝ 𝐸02 and 𝐼 ∝ 𝐸02 we have
−𝑑𝐼
𝐼
= ,𝑑𝑥
𝐿

(21)
[0.2]

where
𝐿=6𝜋𝜖20 𝑚2 𝑐4 𝜔0 4( ) .
𝑛0 𝑞 4
𝜔The solution to the differential equation as a function of 𝑥 is
𝐼(𝑥) = 𝐼0 𝑒−𝑥/𝐿 ,with 𝐿 given above.

(22)
[0.2]
(23)
[0.1]

C.3 (0.3 pt)
Substituting the numbers we find𝐿=

6𝜋𝜖20 𝑚2 𝑐4 𝜔0 4( ) .
𝑛0 𝑞 4
𝜔

(24)

𝐿 ≈ 130 km.

(25)[points are for numerical calculation 0.3]

Height 𝐻 ′ of the Mountains as seen by an observer :Solutions

A13-5
Official (English)

D.1 (2 pt)
𝐻𝑆𝜃
𝐵𝜃

𝐵′𝑆′𝛼𝑑𝑂𝑃𝑅ℎFigure 2. Great circle on which lie the mountain 𝐵𝑆 at height 𝐻 and the observer 𝑃 at height ℎ.
The figure is not to scale.
[0.7]
In Δ𝑂𝑃 𝐵′ ,𝑂𝐵′ = 𝑂𝑃 sec(𝜃) = (𝑅 + ℎ) sec(𝜃)(26)Now, ∠𝑂𝑆𝑃 = ∠𝐵 𝑆𝑆 and ∠𝑆𝑂𝑃 = ∠𝑆𝐵 𝑆 , hence Δ𝑂𝑆𝑃 and Δ𝐵 𝑆𝑆 are similar. Thus′′′′′′𝐵′ 𝑆 ′𝐵′ 𝑆
𝑂𝑆 − 𝑂𝐵′𝑂𝐵′===1−.
𝑂𝑃
𝑂𝑆𝑂𝑆𝑂𝑆(27)

Noting that 𝐵′ 𝑆 ′ = 𝐻 ′ , 𝑂𝑃 = 𝑅 + ℎ, 𝑂𝑆 = 𝑅 + 𝐻 and using Eq. (26), we obtain𝐻′(𝑅 + ℎ) sec(𝜃)=1−𝑅+ℎ𝑅+𝐻(28)Or𝐻′ = 𝑅 + ℎ −(𝑅 + ℎ)2sec(𝜃).
𝑅+𝐻

[0.8]

Noting that cos(𝜃) ≈ 1 − 𝜃2 /2 and 𝜃 = 𝑑/𝑅 we get,𝐻′ ≃ 𝑅 + ℎ −(29)

𝑑2(𝑅 + ℎ)2(1 +).
𝑅+𝐻2𝑅2(30)

𝑑22𝑅

(31)

Alternative answers such as
𝐻′ = 𝐻 − ℎ −are given credit.
The numerical values are 𝐻 ′ = 6096 m for Mt Kanchenjunga and 𝐻 ′ = 4534 m for Mt Everest.
[0.5]

Solutions

A13-6
Official (English)

E.1 (1 pt)
In Eq.(23) 𝐼0 represents the intensity of the source which would have been perceived by an observerat that location if attenuation effects were absent. If the power of the source is taken to be 𝑃0 , then𝐼0 = 𝑃0 /4𝜋𝑑2 for the location at distance 𝑑.
𝐼=𝑃0𝑑exp [− ]
4𝜋𝑑2
𝐿

(32)

[points only if 1/𝑑2 is recognised 0.5]
The relative intensity of Mt Everest as seen from Darjeeling would be
𝐼Everest
𝐼Kanchenjunga=2𝑑Kanchenjunga
2𝑑Everest𝑑Kanchenjunga𝑑exp [− Everest +]
𝐿𝐿

(33)
(34)

= 0.093[0.3]

Yes, Mt Everest is visible.

[0.2]

Attenuation length 𝐿𝑝 due to aerosol pollution :F.1 (1 pt)
From the information given in the problem we have𝐿𝑝 =18𝑛𝜋𝑟2(35)[ 0.3]

𝜌𝑝𝑚(36)4𝜋 3𝑟 𝜌3(37)𝑟𝜌= 50 km,6𝜌𝑝(38)𝑛=𝑚=This yields
𝐿𝑝 =[ 0.2 (expression)][ 0.5 (evaluation)]where,
𝑟 = 500 × 10−9 m,

𝜌 = 3 × 103 kg/m3 ,𝜌𝑝 = 5 × 10−9 kg/m3 .

Relative intensity and Visibility of Mt. Kanchenjunga and Mt. Everest :(39)Solutions

A13-7
Official (English)

G.1 (1 pt)
The new relation for the intensity attenuation is𝑃0𝑑𝑑
exp [− −].
4𝜋𝑑2
𝐿 𝐿𝑝

(40)

𝐼𝐾𝑑= exp [− 𝐾 ] = 0.22.
𝐼ref
𝐿𝑝(41)

𝐼=For Mt Kanchenjunga[0.3 (expression) + 0.1 (numerical answer)]
The drop in intensity is to 22 % of the reference value. Mt Kanchenjunga will be visible from Darjeeling. For Mt Everest
[0.1]
𝐼𝐸
𝑑
= 0.093 exp [− 𝐸 ] = 0.093 × 0.033 = 0.003.
𝐼ref
𝐿𝑝(42)[0.3 (expression) + 0.1 (numerical answer)]
The drop in intensity is to 0.3 % of the reference value. Mt Everest will not be visible from Darjeeling.
[0.1]