Unpolarised light of intensity 4.0Wm^−2 is incident on a polariser placed at 15∘ to the vertical.
State the intensity of the transmitted light.
I've tried 4 x cos(15)^2 = 3.732050808, but it says "Unpolarised light is a random mix of polarisations, so you can't just pick a direction of polarisation for it to have before it hits the polariser."
You are correct in stating that unpolarised light is a random mix of polarisations. This means that the electric field vectors of unpolarised light are oriented in various directions. When unpolarised light passes through a polariser, the polariser selects and transmits light that oscillates in a particular direction while blocking light oscillating perpendicular to that direction.
To determine the intensity of the transmitted light, we need to take into account the nature of unpolarised light. The intensity of unpolarised light is equal in all directions. Therefore, only half of the total intensity reaches the polariser since the remaining half is perpendicular to the polarisation axis.
In this case, the intensity of the transmitted light can be calculated by multiplying the intensity of the unpolarised light by the square of the cosine of the angle between the polariser axis and the vertical direction. However, since the light is unpolarised, we need to consider that only half of the intensity is transmitted.
So, the correct calculation would be:
Transmitted intensity = (1/2) x (4.0 W/m^2) x cos^2(15°)
Using this equation, the transmitted intensity of the light turns out to be approximately 1.835 W/m^2.