polarisation (1 Viewer)

[=)(=]

Why is it b? Like if Polariser 2 is 45 degrees then 1/2 goes to polariser 3 at which intensity halves again giving a

 

[wizzkids]

Yes, the intensity of the light will halve after passing through the first filter, but it will not halve after passing through the second filter. The law about halving the intensity only applies to unpolarised light that has passed through an ideal polarising filter.
Once the light has become polarised, then Malus' Law applies. When polarised light of intensity I_o passes through a second ideal polarising filter, the maximum intensity after passing through the filter is I_o.

 

[=)(=]

ohh okie but how would you work out the amplitude?

 

[scaryshark09]

drop the subject if you still can

 

[carrotsss]

Yes, the intensity of the light will halve after passing through the first filter, but it will not halve after passing through the second filter. The law about halving the intensity only applies to unpolarised light that has passed through an ideal polarising filter.
Once the light has become polarised, then Malus' Law applies. When polarised light of intensity I_o passes through a second ideal polarising filter, the maximum intensity after passing through the filter is I_o.

Right but there is not situation where polariser 1,2,3 are all in the same direction (1 & 3 are perpendicular from the start and never turn), so by Malus Law it can’t ever be I_o/2

Im pretty confident it’s A

 

[wizzkids]

Right but there is not situation where polariser 1,2,3 are all in the same direction (1 & 3 are perpendicular from the start and never turn), so by Malus Law it can’t ever be Io/2.
I'm pretty confident it’s A

This is a common misconception, but the influence of the first polarizing filter over the third polarizing filter (or the second analyser if you want to refer to it that) is irrelevant once you introduce the middle polarising filter. I have done this experiment many times in class with three polarizing filters, trust me, it doesn't work the way that you think.
For example, take two polarizing filters (#1 and #3) and set them as show in the original post. Start with polarizing filters #1 and #3 crossed. No light is passed (see below).
Then introduce the polarising filter #2 at 45 degrees to both filter #1 and filter #3.
More light is now allowed to pass through the three filters. How do you explain that?

 

[synthesisFR]

drop the subject if you still can

this is the only thing we will ever agree on

 

[carrotsss]

This is a common misconception, but the influence of the first polarizing filter over the third polarizing filter (or the second analyser if you want to refer to it that) is irrelevant once you introduce the middle polarising filter. I have done this experiment many times in class with three polarizing filters, trust me, it doesn't work the way that you think.
For example, take two polarizing filters (#1 and #3) and set them as show in the original post. Start with polarizing filters #1 and #3 crossed. No light is passed (see below).
Then introduce the polarising filter #2 at 45 degrees to both filter #1 and filter #3.
More light is now allowed to pass through the three filters. How do you explain that?
View attachment 40126

By Malus law, correct me if I’m wrong but shouldn’t the first polarising filter halve the unpolarised light, the second polarising filter halve the light intensity (I=I_ocos²45) and then the third polarising filter is a further 45 degrees from the new axis of polarisation so it should be halved again by the same formula - so it should be I_o/8 if anything right? Like purely removing the existence of the first filter and that unpolarised light and treating the second filter as a polarised light source of intensity I_o/4, the third filter rotated by 45 degrees physically cannot increase the intensity of the light by the Law of Conservation of Energy, and should reduce it by a half by Malus Law. I feel like I’m missing something.

 

[wizzkids]

In the original post, it says I_o is the intensity after leaving polarising filter #1. Therefore Malus' Law only applies to the interaction between polarizing filter #2 and filter #3. Therefore the intensity after leaving filter #3 varies from zero to 0.5 I_o.

 

[carrotsss]

In the original post, it says I_o is the intensity after leaving polarising filter #1. Therefore Malus' Law only applies to the interaction between polarizing filter #2 and filter #3. Therefore the intensity after leaving filter #3 varies from zero to 0.5 I_o.

im an idiot

 

[wizzkids]

im an idiot

No you're not. You're doing just fine. Just make sure you read all parts of the question, and underline the important facts.

 

[carrotsss]

No you're not. You're doing just fine. Just make sure you read all parts of the question, and underline the important facts.

yeah I think this is something I need to work on, I usually try to consciously underline every word in exams bc for some reason my brain just can’t focus on reading big blocks of working or questions like that