Calculating Intensity of Light from Potential Difference

ausdreamer
Messages
23
Reaction score
0

Homework Statement



It's a problem I'm having with my experiment.

We set up a diode LASER through 2 polarizers, and changed one of the polarizers' angle with respect to the other, to demonstrate Malus's Law. We used some sort of photo detection device (which we were told uses the photoelectric effect to measure intensity of light) which was connected to an Oscilloscope which measured the change in potential difference over the detector. SOMEHOW we are expected to calculate the intensity of the photon beam from the potential difference measurements.

Homework Equations



Well all I know is that the potential difference over the detector is directly proportional to the intensity of the photon beam passing through the 2 polarizers.

The Attempt at a Solution



Read point # 2.
 
Physics news on Phys.org
What does Malus's Law say? How do you need to plot your measured quantities in order to verify it?
 
I spoke to my tutor about my problem and he was confused why I wanted to know the intensity after we'd measured the potential difference. This is because the potential difference is directly proportional to intensity of light (in our experiment) and so we just needed to state the potential difference the LASER made as the 'intensity'.

Just for your own interest, Malus's Law states:

I = Io cos^2 (theta)

Where I is the intensity of light getting through the two polarisers, Io is the maximum intensity if no polarisers are present and theta is the angle the 2nd polariser makes with the 1st polariser. :)
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top