IEEEhome |  Region 8 |  IPShome |  IEEE Italy Section |  IEEE Xplore

  IEEE Photonics Italy Chapter

Home

About us

Chapter
Board

CEC Documents

Events

News

Awards

Job
Opportunities

Photos

Links

Contact us
    Home
Tuesday, 26 September 2017

Past Problems and Solutions:


PROBLEM No.4
Have the Greek astronomer Eratostene (circa 240 BC) had an (uncalibrated) power meter and have known the radiance invariance law, he could have determined the earth-to-moon distance!  Explain how he could have done.
(Hints: he had just determined the earth diameter. He looked first at the moon cinereous light and made a balance on it)

Winner: A. Bruno, University of Pavia
Winner solution
Award photo

PROBLEM No.3 
Your 60-MHz oscilloscope has a 1-Mohm input resistance and is set at 1mV/div sensitivity. Voltage noise of the resistor is (4kTRB)^0.5 = 1 mV. You should expect the trace on the scope to be larger than 1 division, but it is not at all so. Why?  What is the correct expression of your noise voltage in this case?

winner: D. Sebastio, University of Pavia
Winner solution
Award photo

PROBLEM No.2 
Can it be true the old tale of Archimedes devising a mirror solar-burner of  foe-ships sails? Assuming burning temperature of sail is 300°C,  and mirrors held by hands, with 40 cm diameter and 30% reflectivity (for copper), calculate how many mirrors are required to burn sails at 200 m distance (of course, assume a sunny day, at noon, no clouds).

Solution of problem no.2:
By radiance invariance it is Rsail=Rsun. Writing a power balance of the received and radiated powers at the sail, we have:
 Psail= r Rsun (Amirror / d^2) Asail = (sigma) Tsail^4 2Asail,
where the factor 2 is because both sail surfaces radiate, and r denotes  mirror reflectance.
Since Rsun=(sigma)Tsun^4 / pi  (lambertian emitter) we get:
 Amirror=(2 pi d^2 /r)(Tsail/Tsun)^4
and, inserting numbers,
Amirror= 2 pi 40000 m^2  / 0.3 (600/6000)^4 = 83.8 m^2.
As the mirror area is (pi 0.4^2 m^2) /4 = 0.126 m^2  we require N=665 mirrors.

winner: A. Irace, Università di Napoli

PROBLEM No.1 
Consider an Y branching device on a planar substrate, made of equal lossless waveguides up to the branching region that is smooth and without any geometrical defect. What is the power out of the lower waveguide of the Y when two equal powers P enter the two upper waveguides of the Y? What is the power out of the two upper waveguides when a power P enters the lower waveguide? What is the phaseshift of the fields E1, E2 out of the upper waveguides when a field E enter the lower waveguide (assuming a perfectly symmetrical geometry?)

Winner solution of problem no.1:
By the invariance of radiance, the power out the lower waveguide can’t be larger than P (a 3dB attenuation). At the upper waveguides, input power P is split evenly and, because of symmetry, the relative phaseshift shall be zero. As field vectors E1 and E2 shall add to give E=(P)^0.5, this means that they are both half the amplitude of E, or, in power: P1=P2=P/4.

winner: M. Lenzi, Italtel, Milano
Award photo

Useful Links
IPS Conferences
AEIT events
AICT events
Region8 events

© Copyright 2007 PHOS Italian Chapter – All Rights Reserved.
Use of this website signifies your agreement to the Terms of Use.
For questions or comments, please contact the PHOS Italian Chapter Webmaster