Photonics Essentials: Chapter 10 Problems

Source

Thomas P. Pearsall, Photonics Essentials: An Introduction with Experiments (McGraw-Hill, 2003), Chapter 10, Measurements in Photonics, Problems 10.1–10.3, printed pages 242–243.

Problem 10.1: Lock-in amplifier experiment

This is an experimental protocol, not a problem with one numerical answer. Do not invent the phase or noise readings.

Setup

1. Mount the silicon photodiode rigidly and connect it to the lock-in input using short, shielded leads.

2. Place the chopping wheel between a stable visible source and detector.

3. Connect the chopper reference output to the lock-in reference input.

4. Start with a long time constant, a sensitivity range that cannot overload, and the manufacturer’s recommended input configuration.

5. Adjust reference phase for the maximum in-phase signal.

Measurements

Chop frequency	Time constant	Source position	Phase at maximum	Lock-in amplitude	Oscilloscope amplitude/noise

Predicted behavior

The lock-in multiplies the detector signal by a phase-coherent reference and low-pass filters the result. A desired sinusoidal component \(V_s\cos(\omega t+\phi)\) produces a DC term proportional to \(V_s\cos\phi\); unrelated light and electrical noise average toward zero.

Moving the source can change optical path, detector capacitance coupling, and signal-to-background ratio, but it should not create a large propagation phase change at laboratory distances. Raising chopping frequency eventually reduces response when the photodiode, amplifier, or selected time constant cannot follow it. Room lighting is strongly rejected unless it contains a component near the reference frequency.

On the oscilloscope, the same chopped signal is visible but rides on broadband noise and ambient-light offsets. This direct comparison demonstrates why phase-sensitive detection can recover a small periodic signal.

Problem 10.2: F-number and aperture angle

For focal length \(f\), clear diameter \(D\), and f-number \(N=f/D\), the marginal-ray half-angle is

\[\theta=\tan^{-1}\left(\frac{D/2}{f}\right) =\tan^{-1}\left(\frac1{2N}\right).\]

Lens	Half-angle \(\theta\)	Full cone angle \(2\theta\)
f/2	14.04 degrees	28.07 degrees
f/5.6	5.10 degrees	10.20 degrees
f/8	3.58 degrees	7.15 degrees

The requested plotting function is

\[\boxed{2\theta(N)=2\tan^{-1}\left(\frac1{2N}\right)}.\]

Full cone angle as a function of f-number.

It decreases monotonically and is approximately \(1/N\) radians for large f-number.

Problem 10.3: Filling a screen through a slit

The book’s figure places a \(2\ \mathrm{cm}\) screen \(L=6\ \mathrm{cm}\) behind the slit. Place the lens so that the focused parallel beam forms its waist at the slit. Beyond the waist the cone expands with

\[\tan\theta=\frac1{2N}.\]

The illuminated height at the screen is

\[H=2L\tan\theta=\frac{L}{N}.\]

Therefore:

Lens	Illuminated height at screen	Fraction of 2 cm screen
f/2	3.0 cm	100% illuminated; 1.0 cm spills outside
f/4	1.5 cm	75%
f/8	0.75 cm	37.5%

To fill the screen exactly,

\[N=\frac{L}{H_{\mathrm{screen}}} =\frac{6}{2} =\boxed{3}.\]

Thus an f/3 cone exactly fills the one-dimensional screen under the problem’s thin-lens and point-slit assumptions.