How to show light from laser pointer is plane polarised?

AI Thread Summary
To demonstrate that laser pointer light is plane polarized, it is directed through a polarizing filter, which, when rotated, causes a change in intensity. If the light were unpolarized, the transmitted intensity would remain constant at half the incident intensity regardless of the filter's angle. Plane polarized light will show maximum intensity when aligned with the filter and zero intensity when perpendicular. The discussion also clarifies that partially polarized light's transmitted intensity depends on the mix of polarized and unpolarized components. Understanding these principles helps differentiate between fully polarized, partially polarized, and unpolarized light.
songoku
Messages
2,488
Reaction score
393
Homework Statement
A teacher has a laser pointer. The laser pointer emits a narrow beam of plane-polarised light. Describe how the teacher can demonstrate that the light from the laser pointer is plane polarised.
Relevant Equations
Malus Law
The answer key is the light is directed through one polarising filter then filter is rotated and the light changes intensity.

I don't understand how that proves that the light is plane polarised. I think if the light is unpolarised, the intensity will also change when it passes through polarising filter so how can we tell the difference between unpolarised and plane polarised light?

Also, by "plane polarised", does it mean fully polarised or partially polarised or can be both?

Thanks
 
Physics news on Phys.org
songoku said:
The answer key is the light is directed through one polarising filter then filter is rotated and the light changes intensity.

I don't understand how that proves that the light is plane polarised. I think if the light is unpolarised, the intensity will also change when it passes through polarising filter
If unpolarised, the transmitted intensity won't change as the angle changes. The transmitted intensity will remain constant – it will be half of the incident intensity, whatever the angle is.

Suppose the incident intensity is I and you change the angle. What can you say about the maximum and minimum transmitted intensities when:
a) the incident light is fully plane polarised?
b) the incident light is partially plane polarised?
 
Although I have not tried it, I suppose one could bounce the beam off a dielectric surface at Brewster's angle, rotate the laser and watch the reflected intensity. No analyzer needed.
 
Steve4Physics said:
If unpolarised, the transmitted intensity won't change as the angle changes. The transmitted intensity will remain constant – it will be half of the incident intensity, whatever the angle is.
Ah ok, now I understand. I misinterpreted the answer.

Steve4Physics said:
Suppose the incident intensity is I and you change the angle. What can you say about the maximum and minimum transmitted intensities when:
a) the incident light is fully plane polarised?
b) the incident light is partially plane polarised?
a) the maximum will be I and the minimum will be zero

b) sorry I don't know for this one

Thanks
 
songoku said:
a) the maximum will be I and the minimum will be zero.
Yes.

songoku said:
b) sorry I don't know for this one
Sorry, I should have been clearer. It depends on the 'mix'. Suppose the intensities of the polarised and unpolarised incident light are U and P respectively. The total incident intensity is I = U+P.

Half the unpolarised always gets through. Between 0 and P of the polarised light gets through. The min. and max. transmitted intesities are therefore U/2 and P + U/2.
 
Thank you very much for the help and explanation Steve4Physics and kuruman
 
  • Like
Likes berkeman and Steve4Physics
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Thread 'Trying to understand the logic behind adding vectors with an angle between them'
My initial calculation was to subtract V1 from V2 to show that from the perspective of the second aircraft the first one is -300km/h. So i checked with ChatGPT and it said I cant just subtract them because I have an angle between them. So I dont understand the reasoning of it. Like why should a velocity be dependent on an angle? I was thinking about how it would look like if the planes where parallel to each other, and then how it look like if one is turning away and I dont see it. Since...
Thread 'Correct statement about a reservoir with an outlet pipe'
The answer to this question is statements (ii) and (iv) are correct. (i) This is FALSE because the speed of water in the tap is greater than speed at the water surface (ii) I don't even understand this statement. What does the "seal" part have to do with water flowing out? Won't the water still flow out through the tap until the tank is empty whether the reservoir is sealed or not? (iii) In my opinion, this statement would be correct. Increasing the gravitational potential energy of the...
Back
Top