2007 Bulgarian IPhO Team Selection Test – Solutions

Short Exam 2
Problem. A monochromatic (λ = 500 nm) point source S illuminates a rectangular mirror O
which rotates at a frequency of ν = 16 Hz. The distance between the source and the mirror
is L = 100 m. The reflected light is sent to a detector D of negligible size located close to the
source. The mirror rotates about an axis that lies in the plane of the mirror and is perpendicular
to the plane in which the reflected ray moves. You can assume that the source, the detector,
and the mirror are collinear. Accounting for diffraction, find the duration ∆t of the light pulse
registered by the detector. What is the width a of the mirror that minimises ∆t? What is the
value of ∆t then? The height of the mirror is much larger than a.

Figure 1

Solution. There are two factors to consider here. The first is related to ray optics. The mirror
will get to rotate by some small angle θg starting from the position in the figure, after which
no points on the mirror will reflect anything towards the detector1 .

Referring to the figure, this happens when the cone of the rays emanating from the image S ′
no longer includes the source S. This is the same as requiring ∠OSS ′ ≥ ∠OSM . Given that
SS ′ ⊥ OM , we have ∠OSS ′ = θ, and since the mirror’s tilt is small, we can write ∠OSM = a/2
L
.
a
Thus θg = 2L .
We’ll also need to account for diffraction. The reflection from the mirror isn’t perfect, and it
will essentially act as an obstacle of size a which creates a diffraction pattern centered in the
direction of the reflected ray n̂. The angular width of the central maximum of the pattern as
seen from the mirror is 2λa
, with λa on each side. The light within the maximum will land on
1
In the original problem statement, the detector is said to be near the mirror rather than the source. But
in that case you would definitely need the exact distance between the mirror and the √ detector, which isn’t
specified. I also have it on good authority that the answer for the optimal a is indeed λL, which can’t be
obtained unless the detector is actually near the source.

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the detector even after the detector goes out of range for the geometrically reflected rays. This
will cease after an additional rotation of θd , such that ∠SON = 2θd = λa . In total, the mirror
a λ
rotates by θg + θd = 2L + 2a until the end of the pulse. The total length of the pulse corresponds
to twice that value as the detector is accessible from before the mirror is vertical. Thus,

2(θg + θd ) (a/L) + (λ/a)

∆t = = .
ω 2πν
To find the minimum of ∆t, we set the derivative with respect to a to zero, whereby
1 λ √
− 2 =0 ⇒ a= λL = 7.1 mm.
L a
Returning to the expression for ∆t, we have
r
1 λ
∆t = = 1.4 × 10−6 s.
πν L

Theoretical Exam
Problem 4. Figure 2 shows the I-V curve of a bulb. This bulb is connected in parallel with
a resistance R = 2.00 Ω to a source of EMF E = 15 V and internal resistance r = 3.00 Ω. Using
the I-V curve, find the power P dissipated in the bulb. Estimate the accuracy ∆P of your
result.

Figure 2

Solution. Using Kirchhoff’s rules, we will find a constraint on the state of the lightbulb (U, I).
Firstly, note that the voltage across the resistance R is also U , and hence the current through
it is U/R. Then, the total current through r is I + UR , and so the loop rule gives us
 
U E R+r
E− I+ r−U =0 ⇒ I = − U ⇒ I = 5 A − (0.833 Ω−1 ) U.
R r Rr
The voltage and the current of the lightbulb correspond to the point on the I-V curve which
follows the constraint. On the graph, this constraint looks like a straight line. After we plot it

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and intersect it with the I-V curve, we estimate U = 4.8 V and I = 0.98 A. Our results are very
sensitive to the slope of the line, so we should have generous error estimates, e.g. ∆U = 0.1 V
and ∆I = 0.02 A. The dissipated power is P = U I = 4.704 W, while the error in P is
∆P = ∆(U I) = ∆U I + U ∆I = 0.194 W.
To present our answer properly, we need to round the error to one significant digit, and work
with the same precision for P . Thus, P = (4.7 ± 0.2) W .

Problem 5. A coil of inductance L = 2.0 µH and internal resistance r = 1.0 Ω is connected

in parallel to a resistance R = 2.0 Ω. These have been connected to a constant voltage source
E = 3.0 V for a long time. At time t = 0 the source is removed from the rest of the circuit.
Find the time dependence of the current through the coil I(t). Find the total heat Q dissipated
in the coil until the current ceases to flow.
Solution. The coil in this problem can be modelled as an ideal inductance L connected in series
with a resistance r. Before the source is removed, the circuit is in a steady state – the currents
through the components are constant, and there’s no back EMF from the inductance. More
specifically, the currents through the inductor and the resistor are I1 = E/r and I2 = E/R,
respectively. When the source is disconnected, the inductor doesn’t allow any sudden change
in the current through it. Since the back EMF is given by UL = −L dI dt
, any significant jump in
I within the instant dt when we’re disconnecting the source will result in UL diverging, which
is unphysical. It follows that the initial current (with no source) is I1 = E/r along the whole
loop of L, r, and R.
Now, the loop equation is  
dI
−L − Ir − IR = 0,
dt
which has the solution
 
− R+r E E R+r
I = Ae L
t
, I(0) = I1 = ⇒ I(t) = e− L t .
r r

The heat dissipated at r (i.e. at the coil) is then

Z ∞
E 2L
Q= I 2 r dt = = 3 µJ.
0 2(R + r)r

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Problem 6. A parallel beam of monochromatic light with wavelength λ = 589.0 nm is normally
incident on a grating of period d = 2.5 µm and N = 10000 slits. Find the angular width of the
diffraction maximum of order m = 2. Derive the relevant formula and calculate the numerical
value in arcminutes.
Solution. This is essentially a problem about N -slit interference, where will we have to derive
an expression for the intensity pattern. We will use complex numbers to avoid dealing with
cumbersome trigonometric identities. For diffraction at an angle θ from the normal, the waves
due to any two adjacent slits will have a phase difference of ∆φ = 2π λ
d sin θ. After we assign
an amplitude A to each slit, the superposition of all the waves can be found as the sum of a
geometric series:
1 − eiN ∆φ
A + Aei∆φ + Ae2i∆φ + · · · + Ae(N −1)i∆φ e−iωt = Ae−iωt ·


.
1 − ei∆φ
This corresponds to a wave of amplitude
1 − eiN ∆φ 1 − eiN ∆φ |(1 − cos N ∆φ) − i(sin N ∆φ)|
A′ = A i∆φ
= A i∆φ
=A .
1−e |1 − e | |(1 − cos ∆φ) − i(sin ∆φ)|
We know from the theory of waves that the intensity is proportional to the square of the
amplitude. Then, the intensity I(θ) from the grating relates to the intensity I0 from a single
slit like
 ′ 2
sin2 N ∆φ


A (1 − cos N ∆φ)2 + (sin N ∆φ)2 2 − 2 cos N ∆φ 2
I(θ) = I0 = I0 = I0 = I0
.
A (1 − cos ∆φ)2 + (sin ∆φ)2 2 − 2 cos ∆φ sin2 ∆φ
2

Now to interpret this expression. Both the numerator and the denominator cycle between 0
and 1 as the angle θ increases, but the numerator does so much more rapidly. It seems that
the maxima will be observed when the denominator is small. Indeed, in the limiting case when
both the sines are small, we have
N ∆φ 2


I(θ) = I0 2
= N 2 I0 ,
∆φ 2


2

Thinking again in terms of wave superposition, this corresponds to the case where all the waves
are in phase, so that A′ = N A. This requires ∆φ = 2πm, m ∈ Z. It it those m that we use to
indicate the order of the maxima. We should mention that there are also additional subsidiary
maxima which occur every time the numerator in the general expression for I(θ) reaches unity,
no matter the value of the denominator. These, however, are negligible in the case of an ideal
grating (N → ∞), so they’re not part of the convention for labelling maxima. When we say
m = 2, we mean the second-order primary maximum where ∆φ = 4π and hence sin θ = 2λ d
. To
illustrate all this, here’s the intensity pattern for N = 5:

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Let’s proceed with finding the width of the m = 2 maximum. You’ll need to know that ‘width’
means the distance between the two minima on either side of the maximum. These minima
correspond to the instances of sin2 N ∆φ


2
= 0 which lie nearest to ∆φ = 4π. In that case
2π 2λ λ
∆φ = 4π ± N , or sin θ = d ± N d . These values for θ, which we’ll denote by θ+ and θ− , are
very close. We seek their difference (θ+ − θ− ) ≡ ∆θ. Subtracting the sines, we find
   
2λ ∆θ θ+ + θ−
sin θ+ − sin θ− = = 2 sin cos = ∆θ cos θ.
Nd 2 2
p q
2
2
Here cos θ = 1 − sin θ = 1 − mλ d
, so

2λ
∆θ = p = 0.184′ .
2
N d − (mλ) 2

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Experimental Exam
Problem 1. The speed of sound in air.
Equipment:
Tone generator, two loudspeakers, microphone with a jack (female XLR connector), rectifier,
resistance, two-channel oscilloscope, wire connectors, wires, screwdriver, self-retracting tape
measure, tape, graph paper.
(a) Connect the setup to the microphone jack. The jack has three pins called base (−),
middle (+), and top (signal). The middle and top pins are shorted. Carefully examine
the jack. Using the wire connectors and the wires, connect the jack and the resistance in
series with the rectifier. Also connect the oscilloscope to the microphone so as to measure
the voltage across it. Sketch the circuit. Supply a voltage of 3 V to the microphone and
test it. (1.5 pt)
Note: To avoid damaging the microphone when turning it on, follow these instructions:
- Before connecting the microphone to the rectifier, set the voltage to zero using the
potentiometers.
- First, turn on the rectifier.
- Then connect the wires to the rectifier’s terminals, being mindful of the polarity.
- Slowly increase the voltage to 3 V.

Note: To avoid damaging the microphone when turning it off, do not turn off the rectifier
while the microphone is connected. First turn down the voltage to zero, detach the wires and
only then turn off the rectifier.
(b) Connect the two loudspeakers to the tone generator. Put the microphone very close
to one of the loudspeakers. Vary the frequency of the tone generator in the range
[2 kHz, 20 kHz]. Write down the frequency at which the voltage across the microphone
is largest. Work with this frequency from now on. (1 pt)
Design a setup for observing two-source interference. The distance d between the loudspeakers
should be around 30 cm to 50 cm. The distance L between the line through the loudspeakers and
the line along which you will move the microphone should be around 1 m. Put the loudspeakers
and the microphone on separate tables. Remove all objects that could reflect the sound waves.
Be mindful of where you stand during the measurements.
(c) Move the microphone around and measure the coordinates xk of at least 6 consecutive
minima around the central maximum. Use a voltage low enough so that the loudspeakers
emit monochromatic waves (without any harmonics). Write down the value of this
voltage. Write down the values of xk in a table. Use those to find an accurate value for
the coordinate of the central maximum x0 . (3 pt)
(d) Derive an exact formula for the optical path difference ∆ of the interfering sound waves.
Express ∆ in terms of L, d, x, and x0 . (1 pt)
(e) Plot a graph of the optical path difference ∆ for the measured minima against some
number that corresponds to the physical condition for observing such minima. (1 pt)
(f) Find the wavelength of the sound waves λ. (1 pt)
(g) Calculate the speed of sound in air c for your experimental setup. (0.5 pt)
(h) Estimate the error in your value for c. (1 pt)

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