(10%) Problem 21: Unpolarized light of intensity Io = 1500 W/m² is incident upon three polarizers. The axis of the first polarizer is vertical. The axis of the second polarizer is rotated at an angle 0=25° from the vertical. The axis of the third polarizer is horizontal. Randomized Variables 10=1500 W/m² 0=25° *000 20% Part (a) What is the intensity of the light in W/m² after it passes through the first polarizer? 20% Part (b) What is the intensity of the light in W/m² after it passes through the second polarizer? 20% Part (c) What is the intensity in W/m² of the light after it passes through the third polarizer? Correct Answer 13=110 Grade Detail Student Answer 13=46.87 Otheexpertta.com Final Grade 0% Correct see above Submission History Student Feedback The answer provided was not correct and no specific feedback is available. 100% Final Answer Credit 0% Answer Hints Feedback 1 13=0 3% 0% 2 13=506.01 3% 0% Note: Feedback not accessed. Note: Feedback not accessed. 13=506 3 13-558.28 3% 0% 3-558.3 4 13-505.97 3% 0% Note: Feedback not accessed. 13-506 5 13=46.87 Totals 0% 12% 0% 0% Totals 0% 3% 0% 3% 0% 3% 0% 3% 0% 0% 0% 12% 20% Part (d) Write an equation for the intensity of light after it has passed through all three polarizers in terms of the intensity of unpolarized light entering the first polarizer Io and the angle of the second polarizer relative to the first, given that the first and third polarizers are crossed (90° between them). Use trigonometric identities to simplify and give the results in terms of a single trigonometric function of = 20. 20% Part (e) What is the smallest angle 02, in degrees, from the vertical that the second polarizer could be rotated such that the intensity of light passing through all three polarizers is 13' = 64 W/m²?