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Calculating Intensity of Light from Potential Difference

  • Thread starter ausdreamer
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  • #1
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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.
 

Answers and Replies

  • #2
kuruman
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What does Malus's Law say? How do you need to plot your measured quantities in order to verify it?
 
  • #3
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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. :)
 

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