PC1432 Physics IIE: Tutorial 4

Question 1: Toroidal solenoid

A certain toroidal solenoid has a rectangular cross section. It has N uniformly spaced

turns, with air inside. Do not assume the the field to be uniform inside the cross section.

(a) Show that the magnetic flux through the cross section of the toroid is

ΦB =µ0Nih

2πln

(

b

a

)

. (1)

(b) Show that the inductance of the toroidal solenoid is given by

L =µ0N

2h

2πln

(

b

a

)

. (2)

Question 2: Atmospheric Optics

When the sun is either rising or setting and appears to be just on the horizon, it is in fact

below the horizon. The explanation for this apparent paradox is that the light from the

sun bends slightly when entering the Earth’s atmosphere, as shown in the figure below.

Since our perception is based on the idea that light travels in straight lines, we perceive

the light to be coming from an apparent position that is an angle δ above the sun’s true

position.

1

(a) Assuming the atmosphere has uniform density with refractive index n, and extends

to the height h above the Earth’s surface, show that the angle δ is given by

δ = arcsin

(

nR

h+R

)

− arcsin

(

R

h+R

)

. (3)

(b) Calculate δ using n = 1.0003, h = 20 km, and R = 6378 km

Question 3: Polarization

A beam of plane-polarized strikes two polarizing sheets. The polarizing axis of the first

sheet is at an angle of θ with respect to the incident beam, and the polarizing axis of

the second sheet being at 90◦ with respect to the incident beam. Find the angle θ for a

transmitted beam intensity that is 1/10 of the incident beam intensity.

Question 4: Path difference

Consider light generated by a source of frequency f and has wavelength λ0 in vacuum

and corresponding wave number k0 =2π

λ0

.

(a) If the light enters a medium of refractive index n, express its new wavelength λ in

terms of λ0 and new wave number k in terms of k0.

(b) The light propagates between points A and B as shown

x

x = 0 x = L

Direction of propagation

A B

Calculate φ1, the phase change acquired by the light after moving from point A to

B.

(c) Now a piece of plastic with refractive index n is placed between A and B, as shown

x

x = 0 x = L

Direction of propagation

A B

Calculate φ2, the phase change acquired by the light after moving from point A to

B.

2

(d) A piece of plastic is cut shorter into length τ < L.

x

x = 0 x = L

Direction of propagation

A B

x = τ

Calculate φ3, the phase change acquired by the light after moving from point A to

B.

(e) Calculate the phase difference between the lights in Case 2 and 1 (φ2−φ1), and the

phase difference between the lights in Case 3 and 1 (φ3 − φ1).

Question 5: Interference

Consider three very narrow slits with spacings of d and 3d

2as shown in the figure.

The slits are irradiated from the left with a plane wave of monochromatic light with

a wavelength λ = 2d

5. You may assume that the screen is very far away from the slits.

(a) Derive the condition θ must satisfy such that it

is the direction in which waves from all three

slits interfere constructively, thus giving rise

to principal maxima. What are the corre-

sponding values of θ?

(b) Let the bottom slit be covered by a filter which

introduces a half-cycle phase change. How

many principal maxima will now be observed,

and what are their corresponding values of θ?

(c) Now suppose the filter used in part (b) con-

sists of a thin transparent film with an index

of refraction n = 1.33, and it is now placed directly in front of the bottom slit.

Calculate the minimum thickness (expressed in terms of d) of this film that would

be needed to achieve the half-cycle phase change.

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PC1432 Tutorial 4 Basic Concepts

• Self Inductance

– L = ΦB

I [Caution! L depends only on the geometrical shape of the object. Changing ΦB or I will not change L]

– Potential difference across an inductance: V = LdIdt .

– Energy stored: E = 12LI

2.

– General procedure to calculate L:

1. Identify the cross sectional area of the object.

2. Calculate the magnetic field anywhere inside the area.

3. Calculate the flux ΦB =∫~B · d ~A.

4. Divide the result by I, that’s it!

• Maxwell’s Equations and Electromagnetic Wave

– Maxwell’s equations (Integral form):

∗∮~E · d ~A = qenc

ε0(Gauss Law for electric field)

∗∮~B · d ~A = 0 (Gauss Law for magnetic field)

∗∮~E · d~l = −dΦB

dt (Faraday’s Law)

∗∮~B · d~l = µ0Ienc + µ0ε0

dΦE

dt (Generalized Ampere’s Law)

– General solution to sinusoidal wave propagating along positive z direction:

∗ ~E = ~E0 cos(kz − ωt+ φ).

∗ ~B = ~B0 cos(kz − ωt+ φ).

∗ If along negative z direction, replace minus sign above by plus sign. If along x or y direction, replace zby x or y respectively.

∗ Replacing cos by sin in the above solutions is fine, as long as consistent, i.e. replace both expressions by sinnot just one, and don’t forget to include the undetermined phase φ.

∗ ~E ⊥ ~B in such away that the direction of propagation is the same as the direction of ~E× ~B, i.e. can use righthand rule to determine the directions.

– Polarizing Filter, as the name suggests, filter the direction of the electric field. In particular, if a beam of lightpasses through a polarizing filter aligned along x axis, the electric field of the transmitted light will be along xdirection.

• Optics

– Refractive index of a medium: n = cv , where v is the speed of light in the medium.

– In a medium, λ = λ0

n , while f0 = f , where λ0 and f0 are wavelength and frequency in vacuum.

– If a light is traveling in a medium of refractive index n, the optical path distance is xo = nxp, where xp is thephysical (real) distance traveled by the light.

– Snell’s Law: n1 sin(θ1) = n2 sin(θ2), where θ1 and θ2 are measured with respect to normal line.

– Interference:

1. Constructive Interference:

∗ ∆x0 = mλ

∗ ∆φ = 2πm

2. Destructive Interference:

∗ ∆x0 = (m+ 12 )λ

∗ ∆φ = 2π(m+ 12 )

1

where m is a positive integer, xo is the optical path, λ is the wavelength in vacuum, and φ is the phase.

– Thin Film Interference:

∗ Interference between two reflected lights in a thin film.

∗ ∆x0 = 2nt, where n is the refractive index and t is the thickness.

∗ If a light is reflected by a medium with higher refractive index, there is an extra phase difference by π.

∗ If both reflected lights are not affected or both affected by the extra phase difference, the constructiveand destructive interference formula are the same as above.

∗ If only one of the reflected wave is affected by the extra phase difference, the constructive and destructiveinterference formula are swapped.

– Diffraction:

∗ Only care about Fraunhofer diffraction, i.e. when the screen is far away. Fresnel diffraction, i.e. when thescreen is near, is not covered in this module.

∗ Single slit diffraction

· minima condition: a sin(θ) = mλ.

· Intensity profile: II0

= ( sin(x)x )2, where x = πa sin(θ)

λ

∗ Multiple slit diffraction

· Maxima condition: d sin(θ) = mλ

· Minima condition: d sin(θ) = (m− 12 )λ

2

PC1432 Tut 4 Radit’s Solution

1. Concepts used: Inductance, Ampere’s Law

a. Recall the definition of the magnetic flux Φ =∫

B · dA. We therefore need to calculate the magnetic field first.

Notice that the system has a symmetry. If you rotate the toroid with any angle about an axis perpendicular to it,

it will look exactly the same. Therefore, we expect the magnetic field to also not change by any rotation. Similar to

tutorial 1 question 4, this symmetry allows us to extract some information about the behavior of the field. In this

case, the symmetry implies B to be perpendicular to the cross sectional area of the toroid anywhere inside

the toroid. Furthermore, for a given radius from the center of the toroid, the magnitude of the magnetic field

is constant. From these two conditions, if we draw a circular loop with radius r centered around the center of the

toroid, as illustrated by the blue line in the picture below,

Ampere’s Law gives:

∮B · dl = 2πrB

= µ0Ienc ,

namely, we can take out B from the integral.

Since the toroid has N turns, and each turn carries the same current I, the total current enclosed by this loop is

1

NI. Therefore, if we continue the above equation, we should get B = µ0NI2πr . Now, let’s calculate the magnetic flux.

To do so, we take a closer look into a cross-sectional area of the toroid, which according to the question is in a form

of a rectangle of length b− a and width h. Next, we break the toroid into strips, then consider one of the strips as

shown below:

In the above, the gray dots denote the center of the toroid. The magnetic flux passing through the strip shown above

is easy to compute, as it is just dΦ = B(r)hdr = µ0NI2πr hdr. To get the total flux, we have to integrate this expression

to cover the whole rectangle.

Φ =

∫ b

a

µ0NI

2πrhdr

=µ0NIh

2πln

(b

a

)

Summary:

1. Find the magnetic field inside the toroid. You can either copy the result from the textbook or derive it using

Ampere’s law and symmetry. The latter is more recommended.

2. Consider a small portion in the rectangle, and write down the magnetic flux passing through this small portion.

3. Integrate this to get the total magnetic flux. Don’t forget to set the lower and upper limit.

b. Self inductance is defined as L = ΦI . Using this formula and the result from part a, the inductance of one

rectangle in the toroid is L = µ0Nh2π ln

(ba

). Since there are N such rectangles in the toroid, we have to multiply this

result by N to get the total self inductance. Therefore, we obtain L = µ0N2h

2π ln(ba

).

2. Concepts used: Snell’s Law

a. Refer to the figure below:

2

By applying Snell’s law,

nvacuum sin(θ + δ) = nair sin(θ)

sin(θ + δ) = n sin(θ)

θ + δ = arcsin(n sin(θ))

δ = arcsin(n sin(θ)) − θ .

From the trigonometry of the picture above, sin(θ) = RR+h . Therefore, we have

δ = arcsin(nR

R+ h) − arcsin(

R

R+ h) . (1)

b. Just sub in the number, δ = 0.22o.

3. Concepts used: Polarization filter

Tip! When a beam of light pass through a polarizing filter, the direction of the electric field will be

parallel to the polarizing axis of the filter. Since the intensity is proportional to the magnitude of the electric

field squared, the intensity ratio satisfies the formula II0

= cos(θ)2, where θ is the angle between the incident electric

field and the polarization axis.

Now, let’s do this question. When the light passes through the first polarizer, The intensity of the light will be reduced

to I = I0 cos(θ)2. Furthermore, the direction of the electric field is now along the direction of the first polarization

axis. When this light now passes through the second polarizer, since the angle between the polarization axis of the first

and second filter is 90o−θ, the new intensity is now I ′ = I cos(90o−θ)2 = I sin(θ)2 = I0 cos(θ)2 sin(θ)2 = I04 sin(2θ)2,

where I have used the formula sin(2θ) = 2 sin(θ) cos(θ) in writing the last line. Now, we require that II0

= 110 . We

thus have,

3

1

10=

sin(2θ)2

4

sin(2θ) =

√2

5

2θ = 39.23o

θ = 19.62o

Note that this is not the only solution, θ = 90o − 19.62o = 70.38o is also a solution by symmetry.

4. Concepts used: Path difference and Phase of a wave

a. The relationship between the wavelength λn in a medium of refractive index n and λ0, which is the wavelength

in vacuum is λn = λ0

n . Therefore,

kn =2π

λn

= n2π

λ0

= nk0.

b. Recall that the phase of a wave can be generally written as φ = kx− ωt. As the light propagates from A to B,

we thus have

φ1 = φB − φA

= (k0L− ωt) − (0 − ωt)

= k0L .

c. Use part a and part b

φ2 = φB − φA

= (kL− ωt) − (0 − ωt)

= kL

= nk0L .

4

d. In order to find φ2, we have to separate the path taken by the light into 2, the path taken inside the plastic and

the path taken outside the plastic. Inside the plastic, we have shown in part a that the wave number will be related

to that in vacuum according to k = nk0. On the other hand, ω is the same outside and inside the plastic.

• The total distance taken by the light inside the plastic is τ . The phase acquired as a result of traveling inside

the plastic is therefore φplastic = (kτ − ωt) − (0 − ωt) = kτ = nk0τ .

• The total distance taken by the light outside the plastic is L − τ . The phase acquired as a result of traveling

outside the plastic is therefore φvacuum(k0(L− τ) − ωt) − (0 − ωt) = k0(L− τ).

• The total phase acquired by the block, i.e. φ3, is therefore the sum of the two results above,

φ3 = φplastic + φvacuum

= (n− 1)k0τ + k0L .

e. Just use the results from previous parts:

φ2 − φ1 = (n− 1)k0L ,

φ3 − φ1 = (n− 1)k0τ .

5. Concepts used: Diffraction, Interference

a. Recall that the maxima condition (constructive interference condition) for a double slit diffraction is given by

the equation d sin(θ) = mλ, where m is a positive integer. This formula also applies for multiple slit diffraction

(more than two slits), provided all the slits are symmetrical, i.e., separated by the same distance. However, for this

question, the slits are not symmetrical, and we need to consider two slits at a time and ensure that they interfere

constructively. In this case,

• For the top slit to interfere constructively with the center slit, we require d sin(θ) = mλ.

• For the center slit to interfere constructively with the bottom slit, we require 3d2 sin(θ) = nλ, or d sin(θ) = 2n

3 λ.

• For the top slit to interfere constructively with the bottom slit, we require 5d2 sin(θ) = lλ, or d sin(θ) = 2l

5 λ.

In general, we need to find θ that simultaneously satisfies the three equations above. However, notice that if the

first two points are satisfied, the third point is automatically satisfied.1 Tip! For multiple slit diffraction

1To quickly see this, simply add the first two points, d sin(θ) + 3d2

sin(θ) = (m+ n)λ. By relabeling the integer m+ n as l, we get thethird point.

5

with nonsymmetrical slits, you only need to consider the constructive interference between any two

adjacent slits.

One way to solve the above, apart from using brute force method, is to take the ratio between the second and the

first point, which leads to n : m = 3 : 2. That is, one solution corresponds to setting m = 2 and n = 3, in which case

the first and second point above will correspond to the same equation, which can then be solved to find the angle,

sin(θ) =2λ

d

=4

5

θ = 53.1o.

We can of course proceed to set m = 4 and n = 6 (the next set of integers satisfying the ratio above), but you can

easily check that this corresponds to sin(θ) = 85 . Since sin(θ) ≤ 1, there is no solution for θ in this case. 53.1o is

therefore the only solution for θ.

b. If the bottom slit has a half-cycle phase change, it is equivalent to saying that the diffracted light passing through

the bottom slit travel an extra optical path distance λ2 compared to the other slits. As such, the maxima and minima

formulas are swapped. It means that,

• The constructive interference condition between the top and the center slit is not affected, i.e. d sin(θ) = mλ.

• For the constructive interference to occur between the center slit and the bottom slit, we require 3d2 sin(θ) =

(n+ 12 )λ.

• For the constructive interference to occur between the top slit and the bottom slit, we require 5d2 sin(θ) =

(l + 12 )λ.

As before, we need only focus on the first two points above and look for θ which satisfies the first two equations

simultaneously. Again by taking the ratio, m : (n + 12 ) = 2 : 3. We should first try the combination m = 1 and

n = 1. This gives rise to sin(θ) = λd = 2

5 or θ = 23.6o.

Now, let’s try the combination m = 3 and n = 4. This gives rise to sin(θ) = 65 which has no solution. Therefore, the

only maximum is located at angle 23.6o.

c. Just use the result from question 4e,

∆φ = φ3 − φ1

6

= (n− 1)k0τ

= (n− 1)2π

λ0τ.

Half-cycle means ∆φ = π. The above then becomes,

τ =λ0

2(n− 1)

= 0.61d.

Answers to extra practice 4:

1. c = 2.86 × 108 m/s.

2. (a) z-direction.

(b) λ = 200nm.

(c) ~E(x, t) = 106k cos[π × 107x+ 2π × 1015t

].

3. v′ = 2v(n−1n

)when y < h and v′ = v

(n−1n

)when y > h.

7

This is a supplementary containing my remarks about tutorial 4.

Tut 4 FAQ

Q: Why can we write the general sinusoidal wave formula as E = E0 cos(k ·r± ωt+ φ), where ω = |k|c? Where does the sign ± come from?

A: The formal explanation of this can actually be found in many wave theorytextbooks. The following is an intuitive explanation to understand theabove. Suppose the wave is, without loss of generality, travelling towardpositive x-direction. Suppose the source of the plane wave is at x = 0.That is, at x = 0, there is something that generates a changing electricfield E = E0 cos(ωt). At time t = 0, the electric field at x = 0 is E0,while at anywhere else is 0. Let’s focus on this electric field E0. Sincethis EM wave is travelling toward positive x-direction, at time t = T , E0

has reached x = X = cT . In particular, we can write E0 as E0 cos(ω ×0) = E0 cos(ω(T − T )) = E0 cos(ω(T − X

c )) = E0 cos(ωT − kX), wherewe define k = ω

c . From this, we can deduce that in terms of x and t,E(x, t) = E0 cos(ωt − kx). Hopefully now you can see where the minussign comes from.1 Similarly, you can use similar argument to show that fora wave travelling toward negative x-direction, we get something similaras above, but with +kx instead of −kx. Try it! This explanation can begeneralized to get the general formula E(x, t) = E0 sin(ωt± k · x + φ).

Q: In question 3, why does light whose electric field is along another directioncannot pass through the filter, and only those whose electric field is parallelto the axis of the filter can?

A: I’m sure you still remember the properties of conductor from tutorial 1right? There cannot be an electric field inside the conductor. Looselyspeaking, we can regard polarizing filter as a conductor whose axis ofpolarization is some sort of hole. Therefore, any electric field not parallelto the axis polarization cannot go through, it will be absorbed by theconductor (i.e. the free electrons inside the conductor will distribute insuch away to counter this electric field). Since a light is composed ofchanging electric and magnetic field, it cannot go through if the electricfield is absorbed.

Q: Still in question 3, I understand that magnetic field is perpendicular toelectric field and direction of propagation. If the electric field is alongthe direction of the polarization axis, does that mean magnetic field isperpendicular to the polarization axis?

1You may notice that, for example in the lecture note, we have kx − ωt instead of whatI showed. That is actually fine. In general we have the phase constant φ to make the twoconventions equivalent. In wave theory, I prefer to use ωt−kx because it is simpler to explain(Like what I just did). However, in quantum mechanics, I prefer to use kx−ωt instead, sinceit is simpler to explain quantum mechanically.

1

A: First of all, the magnetic field will not be affected by a conductor, unlikeelectric field. Therefore, the polarizer has no direct effect on the magneticfield. However, remember that light is a product of changing electric andmagnetic field perpendicular to each other, traveling in a certain direction.In other words, a polarized light should consist of electric and magneticfield, which are perpendicular to each other. Unpolarized beam of lightconsists of many lights polarized in many directions. After hitting thepolarizing filter, only light with a specific direction of electric field canpass through. This specific direction of electric field is associated witha specific direction of magnetic field as well, which has to be in suchaway that it is perpendicular to both electric field and the direction ofpropagation.

Q: So, if a beam of light hits a polarizing filter, only the component of E fieldalong the axis of polarizing filter can pass through, am I right?

A: That is actually quite tricky. If the incident light is already polarized ina certain direction, then yes. If that is the case, only the component ofE parallel to the axis of polarization filter can pass through. As such, themagnitude of the electric field is decreased by a factor cos(θ), where θ isthe angle between the polarizer and the incident electric field. That is alsothe reason that the intensity ratio is given by cos(θ)2, since the intensityof the light is proportional to the squared amplitude of the electric field.However, if the incident light is unpolarized, we cannot say that only thecomponent of E field along the axis of polarizing filter can pass through.In this case, the beam of light consists of many light components eachwith different direction, and even magnitude, of E. For this situation, wecan only say that the E of the transmitted light is along the direction ofthe axis of polarizer.

Q: What is the phase difference ∆Φ between two lights, how is it related tothe path difference, and what is its significance?

A: Recall that light can be described in terms of an electric field E = E0 cos(kx±ωt + φ). If two lights interfere each other, we need to add the two elec-tric fields corresponding to each light. In general, the electric field ofthe two lights can be described as E1 = E0,1 cos(kx1 ± ωt1 + φ1) andE2 = E0,2 cos(kx2 ± ωt2 + φ2). Assuming E0,1 and E0,2 are point-ing in the same direction, constructive interference occurs if cos(kx1 ±ωt1 + φ1) = cos(kx2 ± ωt2 + φ2), and destructive interference occurs ifcos(kx1 ± ωt1 + φ1) = − cos(kx2 ± ωt2 + φ2). Since we know from highschool math that sine is a periodic function with sin(θ) = − sin(θ ± π) =sin(θ±2π), we see that constructive interference occurs if kx2±ωt2 +φ2 =kx1±ωt1+φ1+2mπ, and destructive interference occurs if kx2±ωt2+φ2 =kx1 ± ωt1 + φ1 + 2(m+ 1

2 )π, where m is an integer. From here, we define∆Φ = kx2±ωt2 +φ2− (kx1±ωt1 +φ1) to be the phase difference, and isnot to be confused with (it has nothing to do with) the electric/magnetic

2

flux. In this module, usually we take φ1 = φ2 (except when one of thelight is reflected by the surface of a medium with a higher refractive in-dex, in which case φ1 = φ2 +π). Furthermore, if the light comes from thesame wavefront, t1 = t2. Therefore, the phase difference between the twowaves normally depends on the path difference between them accordingto ∆Φ = k∆x = 2π∆x

λ .

Q: In question 5, since the slits are not symmetrical, we need to ensure that eachpair of slits interfere constructively. However, why did you only considerthe top-middle and middle-bottom pairs? Should we consider top-bottompair as well?

A: Intuitively, if the top-middle pair interfere constructively, and the middle-bottom pair does as well, it is easy to see by induction that top-bottom pairwill also interfere constructively. Intuitively it goes like this. Rememberthat constructive interference occurs when phase difference is an integermultiple of π. Let’s say the phase of the top slit is φ1. Then constructiveinterference on the top-middle pair means that the phase of the middle slitis φ2 = φ1 +m1π. Meanwhile, the constructive interference on the middle-bottom pair means that the phase of the bottom pair is φ3 = φ2 +m2π =φ1 +(m1 +m2)π, i.e., phase difference between top and bottom slits is alsoan integer multiple of π. Therefore, when dealing with similar questionsinvolving not symmetrical slits, you can simply consider pairs of adjacentslits.

Q: I understand that when light enters a medium, the speed and wavelengthwill be reduced by a factor of 1

n , where n is the refractive index. Howabout the amplitude of the constituent E and B field, namely, E0 andB0? Will they get reduced as well inside the medium?

A: That is a good question! It depends on the medium. The refractive index ofthe medium actually depends on both its permittivity (ε) and permeability(µ). If the medium is a non-magnetic dielectric, n only changes because εchanges, in which case E0 will be affected from electricity theory, but B0

will stay the same. If the medium is magnetic and dielectric, then both εand µ change, and both E0 and B0 will change as well.

Other than that, please make sure you understand the basic concepts for thisweek topics. To check your understanding, make sure you know the answers tothe following basic questions:

1. What is the difference between polarized and unpolarized light?

2. Is it possible to have a uniform magnetic field in 3D space, so that it ispointing along positive z-direction if z > 0, and pointing along negativez-direction if z < 0?

3. What is the difference between Brewster angle and critical angle?

3

4. Write a general formula for the E and B fields of a sinusoidal plane wavetravelling along positive x-direction.

5. What is light interference?

6. What is the condition for a constructive/destructive interference to occur?

7. What is the minima condition for a single slit diffraction?

8. What is the intensity profile of the diffracted light in the single slit diffractioncase?

9. What is the maxima/minima condition for a multiple slit diffraction?

4

PC1432 extra practice 4

Note: The following questions are for extra practices and listed according to difficulties. Please spend some times to do thefollowing problems. Final answers will be provided, but no solutions will be given, unless someone send me his/her correctsolutions.

1. (Easy) One day, your girlfriend (boyfriend) visits your home and randomly gives you a chocolate bar as a present.Since you have just learned the concept of wave and light from PC1432, you want to impress her (him) by claiming thatyou can measure the speed of light using the chocolate. To do so, you put the chocolate inside a microwave oven for about20 s. You then observe that the two points marked by A and B in the chocolate (shown below) are slightly melted, whilethe rest are still perfectly solid. According to the manual, the microwave operates at 2.45 GHz frequency. From all theseinformation, how would you proceed to estimate the speed of light? What is the value of the speed of light you obtainedfrom this experiment?

2. (Medium) A plane of electromagnetic wave is propagating in a medium with index of refraction n = 1.50. It has amagnetic field of the form:

~B(x, t) = (5 × 10−3T )j cos[kx+ (2π × 1015rad/s)t

]where k is the wavenumber of the wave.

1. In which direction is this wave polarized?

2. What is the wavelength of this wave?

3. Write down the corresponding expression for the electric field ~E(x, t). In your expression, all constants should beexpressed in terms of their numerical values.

3. (Challenge Problem)1 In the figure below, a cylindrical tank of height H is initially empty. A light source is placedinside the tank at a height h from the bottom of the tank, and a mirror is placed at the bottom of the tank. At time t = 0, atap water of refractive index n is poured onto the tank so that the water level increases at a constant speed of v. As a result,while the position of the light source is fixed, its reflection from the mirror will appear to move around. At what speed doesthe reflection move around? You need to consider two separate cases when the water level is below and above h.

1Only to those interested. Challenge problems are designed to be brain teasing and may involve deep understanding on the concepts with someclever tricks to solve them.

1