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Then we apply Malus' Law and make θ the subject of the equation. θ = cos-1( √0.6) which is about 39 degrees.How about this question?
the first polarising filter removes half of the light though assuming it’s unpolarisedThen we apply Malus' Law and make θ the subject of the equation. θ = cos-1( √0.6) which is about 39 degrees.
this was the actual solution.θ = cos-1( √0.6) which is about 39 degrees.
irst polarising filter removes half of the light though assuming it’s unpolarised
ive found most questions assume unpolarised, these ones seem like outliersthis 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.
Ahh yes, good point. therefore is impossible to transmit 60% of the original intensity.the first polarising filter removes half of the light though assuming it’s unpolarised