Unpolarized light with intensity I0 is incident on two polarizing filters. The axis of the first filter makes an angle of 60.0° with the vertical, and the axis of the second filter is horizontal. What is the intensity of the light after it has passed through the second filter?

Video transcript

Hi everyone. In this practice problem, we are being asked to calculate the transmitted intensity through a polarizing system. We will have a system of two polarizing films where the two films have their polarization axis inclined at 40 degrees to each other. A coated beam of un polarized light with intensity of five milli weber per meter squared is sent into the system. And we're being asked to calculate the transmitted intensity through the polarizing system. The options given are a zero milli Weber per meter squared. B 1.47 m weber per meter squared, C 2.5 milli Weber per meter squared. And lastly D 3.83 milli Weber per meter squared. So the incident light given in the problem statement is going to equals to INS or I inc which is going to be five mili Weber per meter squared. So the incident light here is un polarized. So the intensity of the linearly polarized light transmitted by the first polarizer is going to equals to I one equals to I inc divided by two which will then come out to be five milli weber per meter squared divided by two which is going to be 2.5 milli weber per meter squared. So the second polarizer will then reduces the intensity of 2. mil weber per meter squared by a further factor of I two to be equals to I one multiplied by cosine squared of Phi which in this case, the cosine squared of phi because Phi is going to be degrees, then I one will be multiplied by cosine squared of 40 degrees to then get the I two. Thus, the intensity transmitted through the polarizing system is then going to equal to Iran, which is going to equals to I two, which is going to be I one multiplied by cosine square of degrees, which is 2.5 milli weber per meter squared multiplied by cosine squared of 40 degrees, which is going to come out to be I trends to be equal to 1.47 me Weber or meter square. So the intensity transmitted through the polarizing system after that is going to be I trends which is going to equal to 1.47 milli Weber per meter squared, which is going to correspond to option B and our answer choices. So option B will be the answer to this practice problem and that'll be it for this video. If you guys still have any sort of confusion, please make sure to check out our other lesson videos on similar topic and that'll be it for this one. Thank you.