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Integrated Concepts(a) On a day when the intensity of sunlight is 1.00 ${kW} / {m}^{2}$ , a circular lens 0.200 ${m}$ in diameter focuses light onto water in a black beaker. Two polarizing sheets of plastic are placed infront of the lens with their axes at an angle of $20.0^{\circ} .$ Assuming the sunlight is unpolarized and the polarizers are 100$\%$ efficient, what is the initial rate of heating of the water in $^{\circ} {C} / {s}$ , assuming it is 80.0$\%$ absorbed? The aluminum beaker has a mass of 30.0 grams and contains 250 grams of water. (b) Do the polarizing filters get hot? Explain.

A. $2.07 \times 10^{-2} \mathrm{C} / \mathrm{s}$B. Yes, the polarizing filters get hot because they absorb some of the lost energy from the sunlight.

Physics 103

Chapter 27

Wave Optics

Cornell University

Rutgers, The State University of New Jersey

University of Washington

Lectures

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emission rate of heating can be found by finding the power observed so the bottle observed can be found. But a by intensity after the second polarizer I took in Britain is the intensity first after the first polarizer times the course square off the angle, which is a 20 degrees. So if I want me here is ah, 500 words for me to square course square over 20. Excuse us I to off 441.5 words per meter square, Reedus square. Then we can write a power that is 44.15 times Bye. Derided by 100. This gives us the power off 13.87 words Warts. So the pollen observed may be observed will be called to 80% or for 13 13.87 13 point Great Soon this will give us a bottle observed off 11.0 mine six wars, then the rate off the heating Initial eating can be found by Delta T divided by time. He's equal Thio power observed be divided by the mass times a specific heat off the harmonium plus the moss off the water times the specific eat off the water. So substituting those values from the given problem, then we get the right off the initial Hedy General, 0.10 34 degrees for a second per second. So this was a party in Bobby? Yes. Yes. The polarizer will observe some off the intensity off a light which is not transmitted, resulting the heating off the polarizer, and it will get hot.

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