Polariser (1 Viewer)

[Bedmajo21]

Could someone explain why the initial intensity is not halved? I got A but it was C. There was a similar q. in my trial and idk if I'm overcomplicating this.

 

[carrotsss]

It should be A

 

[wizzkids]

The question is supposed to be examining Malus' Law. If you just apply the equation I/I_o=cos²θ = 0.5
However there is a trap buried inside this question that probably makes it invalid. Unpolarised light, presumably from the Sun, has reflected off the ocean and during this reflection it becomes partially polarised light. This effect is strongest along the vertical plane joining the Sun and the observer.
Then this partially polarised light passes through the hotel window. We don't know the angle between the optical axis of the window polariser and the polarisation of the partially polarised light coming off the ocean. Therefore we don't know the amount of attenuation that occurs when passing through the first window polariser.
[By the way, this is why polarizing sunglasses are so popular amongst fishermen and boaties, because the partially polarized glare off the ocean can be effectively blocked by polarizing sunglasses, while letting the non-polarized light pass through.]

 

[Bedmajo21]

How about this question?

 

[wizzkids]

How about this question?

Then we apply Malus' Law and make θ the subject of the equation. θ = cos^-1( √0.6) which is about 39 degrees.

 

[Bedmajo21]

Yeah thats fine then.

 

[carrotsss]

Then we apply Malus' Law and make θ the subject of the equation. θ = cos^-1( √0.6) which is about 39 degrees.

the first polarising filter removes half of the light though assuming it’s unpolarised

 

[Bedmajo21]

θ = cos-1( √0.6) which is about 39 degrees.

this was the actual solution.

When i did it though my working got a bit crazy bc i assumed unpolarised.

irst polarising filter removes half of the light though assuming it’s unpolarised

So idk where to go with these questions, bc they seem to just use the formula but at the same time, don't seem to make sense when you factor in unpolarised light.

 

[carrotsss]

this was the actual solution.

When i did it though my working got a bit crazy bc i assumed unpolarised.




So idk where to go with these questions, bc they seem to just use the formula but at the same time, don't seem to make sense when you factor in unpolarised light.

ive found most questions assume unpolarised, these ones seem like outliers

 

[Bedmajo21]

yeah ikr

 

[wizzkids]

the first polarising filter removes half of the light though assuming it’s unpolarised

Ahh yes, good point. therefore is impossible to transmit 60% of the original intensity.