# Homework Help: Polarisation of Light

1. Jun 28, 2007

### ku1005

1. The problem statement, all variables and given/known data
Unpolarised light is shone onto a set of three transmission Polaroid sheets. The intensity of light after the first Polaroid sheet is measured to be I = E2. The transmission axis of the second and third Polaroid sheets are at angles of alpha and beta with respect to the transmission axis of the first Polaroid sheet. The amplitude of light emerging from the set of Polaroid sheets is:

2. Relevant equations

Intensity = I (MAX) * (cos(theta))^2

3. The attempt at a solution
My workings follow :

http://img407.imageshack.us/img407/8982/polaroidphysicsqbq3.png [Broken]

Why, is the answer then :
http://img224.imageshack.us/img224/334/answertoplroidti7.png [Broken]

which part of my workings are incorrect/show lack of understanding??

(see above)
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

Last edited by a moderator: May 2, 2017
2. Jun 28, 2007

### ku1005

just thinking now, it because the amplitude of light is unaffected by the intensity of the light, and therefore, throughout the 3 polaroids, the amplitude of the light will simply change via:

E-> first

Ecos(alpha) through second

Ecos(alpha)cos(beta-alpha) through third

We only need the component of Amplitude transmitted, not the intensity, the intensity bit is a distraction/trap

is this correct thinking?

3. Jun 28, 2007

### Dick

It is correct thinking.

4. Jul 9, 2007

### PhillipKP

You had it right the first time.

You forgot that the question is asking for the final amplitude. This can be deduced from the final intensity because:

$$I=E^2$$

So all you needed after your third equation was to take the square root.

Amplitude is a vector and is can be thought of as a wave; thus it requires an imaginary phase.

Intensity is the square of amplitude and can be measured and thus has no imaginary parts. It is not a wave.

Last edited: Jul 9, 2007
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