A beam of unpolarized light of intensity I0 passes through a series of ideal polarizing filters with their polarizing axes turned to various angles as shown in Fig. E33.27. (b) If we remove the middle filter, what will be the light intensity at point C?

Video transcript

Hi, everyone in this practice problem, we're being asked to determine the intensity of a beam. When it emerges through a system of polarizes, we will have a filament lamp slide beam with the intensity of light sent on a series of three polarizer sheets. Each rotated 45 degrees from the one before. As it is shown in the figure, a student rotates the middle polarizes and make the polarization axis of the first and middle polarizes as align, we are being asked to determine the intensity of the beam I when it emerges from the system of polarize. The options given are A I equals zero B I equals I light divided by square root of two C I equals I light divided by two and lastly D I equals I light divided by four. So in order for us to uh determine the intensity of the beam after it emerges through the system of polarize, we have to uh recall that when un polarized light passes through a polarizer, the intensity is going to be reduced by a factor of health and the transmitted light is polarized along the axis of the polarizer. So after passing the first polarizer or after passing first polarizer, I'm just gonna abbreviate that with Paul um the transmitted intensity or I'm gonna write that down as I one is going to be equals to I light divided by two. When the student aligns the axis of polarization of the middle polarizer with the first, the light will pass through the middle polarizer with an intensity of I light divided by two and polarization will be along the vertical axis. This occurs because cosine of cosine squared of phi will equals to be cosine squared of zero because Phi will be zero because the um axis of polarization will align and the cosine squared of zero will equals to one. Hence, because of this I will or I two will also equals to I one or in this in this case equals to I light divided by two. Awesome. So this is after passing the second, after passing second polarizer or po now, um we're looking into the last polaris polarizer. So um the angle between the polarization direction after the middle filter and the axis of the last filter is going to be degrees. The transmitted intensity through the final polaris polarizer is going to be given by after third four I three or essentially I, which is the I that is uh listed in the answer option is going to equals two I max multiplied by cosine squared of. In this case, the Phi is going to be equals to 90 degrees. So I will then equals to I max multiplied by cosine squared of degrees where cosine of 90 degrees is going to be zero. So I will equals to zero. Thus, the intensity of the beam when it emerges from the system of polarizer or after it passes, the last polarizer is going to equals to zero, which is going to correspond to option A in our answer choices and option A will be the answer to this video. So that'll be it for this one. If you guys still have any sort of confusion, please make sure to check out our other lesson videos on similar topics and that'll be it for this video. Thank you.