Lasing experiment and the measurement of intensity

[AlexCdeP]

I have just completed an experiment on principles of lasing, constructing a laser and measuring the beams intensity distribution. This distribution of the beam was measured using a detector mounted on a traveling microscope. The detector had a pinhole cover such that only a small fraction of the beam entered the detector at once, and from this I could measure the current from the detector on an ammeter against the horizontal displacement. Now I must be going crazy because I spoke to a demonstrator in the lab and she said I can just plot this current because current and intensity are proportional. Indeed I proceeded to plot the data and the measurements fitted the predicted values. However I checked and current is not proportional to intensity, it is proportional to the square root of the intensity which mucks everything up. Can anyone think what I have missed.

Homework Equations

P=I^{2}R

and I=\frac{P}{A}

so I^{2}\proptoIntensity

The Attempt at a Solution

The plot comes out as a Gaussian distribution so I=I_{0}e^{\frac{x}{w_{0}}}

w is the beam width, I is intensity and, x is the horizontal position.

So if the current were proportional to the square root of intensity we wouldn't get this distribution.

Thanks in advance to anyone that can help.

 

[Basic_Physics]

The current through the dectector is generated by the absorbed photons. That is most likely why it is directly proportional to its intensity.

 

[collinsmark]

The key is to better model the detector.

Your equation of P = I²R is correct, but don't forget that in a typical photo-detector, the resistance R of the photo-detector is not constant, and changes with the light's intensity. The fact that R is not constant is typical of most photo-detectors. Many photo-detectors work at a constant voltage (roughly speaking) rather than have a constant resistance.

Let's step back a moment and consider an ideal photo-detector. For example, let's consider an ideal solar cell that converts light power to electrical power with a constant efficiency. Also, let's assume that the potential difference (aka. voltage) between the terminals of this ideal solar cell is constant. (In practical solar cells this constant voltage [Edit: I should probably reword that as constant "emf"] idea is a fairly good approximation of their behavior.) Now what is the relationship between the light intensity and the current, given that?

Everything I've said in this post depends upon the characteristics of the detector though. If you do some investigation and find out that the detector's voltage doesn't change much over varying light intensity*, then I think you've found your answer.

*(At least in the range of intensities that you were measuring in the experiment.)

 

[AlexCdeP]

Thanks so much for your replies! Collinsmark, I think you are most probably right. I will do some further investigation into the detector and I'll probably find that the voltage doesn't change significantly. If the the resistance of the detector were to decrease at the same rate as the intensity were to increase(not by the square), this would cancel the effect of the square term on the current I think(not including some constant term)? Thus leaving P proportional to I, which is what I need for my equation to work, so this is what I expect to see when I check tomorrow. Thanks again, I'll look into it and point out any error in my thinking here if you see it :)

 

[ehild]

I

Homework Equations

P=I^{2}R

and I=\frac{P}{A}

so I^{2}\proptoIntensity

The Attempt at a Solution

The plot comes out as a Gaussian distribution so I=I_{0}e^{\frac{x}{w_{0}}}

w is the beam width, I is intensity and, x is the horizontal position.

So if the current were proportional to the square root of intensity we wouldn't get this distribution.

You mix two different things, both denoted by I, and also two different powers, denoted by P.

In P=I²R, I is the electric current flowing through a resistor R, and P is the power dissipated in the resistor.

The intensity of light is also denoted by I, and it is the power of the laser light (P) falling onto an area A , divided by the area. The light intensity is proportional to the number of photons, perpendicularly crossing an unit area in unit time.

The photodetector usually works on the principle of photoelectric effect, and measures the photocurrent, the number of electrons released by the light from the photocathode in unit time.
The number of electrons released is roughly proportional to the number of incident photons.

ehild