sun focus image of sun

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1 These are slightly modified book questions. They do NOT represent the entire subject material of the test. Moreover, I have not fully vetted these questions or the supplied answers. They are supplied as is and they may have typos or other difficulties. Let me know if you find any issues. Answers (hopefully correct J ) supplied on last page 1. Focused spot of light A camper is trying to start a fire by focusing sunlight onto a piece of paper. The diameter of the sun is 1.39 x10 9 m, and its mean distance from the earth is 1.50x10 11 m. The camper is using a converging lens whose focal length is 10.4 cm. sun focus image of sun (a) What is the area of the sun's image on the paper? Hint: consider the figure below. A ray of light passing through the very center of the lens continues in a straight line. Hence, the two triangles shown are similar. (b) If W of sunlight passes through the lens, what is the intensity of the sunlight at the paper?

2 2. Angular Magnification You have already measured in lab your eye's angular resolution. You can use a magnifying glass to improve your angular resolution. A jeweler whose near point N is 74 cm from his eye (yours is about 25 cm) uses a magnifying glass to examine a watch. The watch is held 4.0 cm from the magnifying glass. (a) In the figure below, what is the minimum value of d i? Explain why in a sentence. (b) Find the angular magnification of the magnifying glass relative to uncorrected near point viewing. Use the figure below to guide your reasoning.

3 3. Radiated Power A. What is the ratio of the emitted radiative power from a 310K surface and the same surface at 300K? Assume a black body radiator. B. What is the ratio of the emitted radiative power from a 300K surface and the same temperature object with ½ the surface area? Assume a black body radiator. C. Which surface gives off the most number of red photons at 600nm. Briefly say why. 1500K object of area A or 800K object of area A D. Which surface in part C looks redder to your eye? Briefly say why. 4. More Radiated Power The power radiated by an object of temperature T is balanced against the radiative power absorbed from the surrounding. In equilibrium these are equal. The net power emitted by an object of temperature T in a room of temperature To is then Pnet = Aσ( T 4 To 4 ), that is, just the difference between radiative power emitted and radiative power absorbed. Here we assume a perfect black body. A. A human body is about 2 m 2. Let s assume it is a black body, too (not too bad of an approximation). Bare skin temperature is about 306 K but a room is about 293 K. From this determine if net power is absorbed or emitted and how much. B. A friend stands 2 m away from you. Make some reasonable assumptions and determine how much energy you deliver to your friend in the form of radiated heat in 1 minute. C. How many photons of 540nm light is this equivalent to?

4 5. Refractive power At age forty, a certain woman requires contact lenses (f = 63.9 cm) to read a book held 25.0 cm from her eyes (and she can't move it any closer w/o blur). At age forty-five, while wearing these contacts the closest she can hold a book is 27.7 cm from her eyes. (a) What is her near point at age forty? (b) What is her near point at age forty-five? (c) What diopter lenses does she require at age forty-five to read a book at 25.0 cm? 6. Refraction (Etkina 21.45) When reaching a boundary between two media, an incident ray is partially reflected and partially refracted. A. At what angle of incidence is the reflected ray perpendicular to the incident ray? Assume the indexes of refraction for the two media are known. B. Why can t this condition of perpendicular rays happen when the angle of incidence is 45 degrees?

5 Answers (subject to corrections!) 1. (a) 7.3 x 10-7 m 2 (b) 7.5 x 10 5 W/m 2 2. (a) di >= 74 cm Must be farther away than n.p. (b) (a) 1.14 (b) 2 (c) 1500 K Higher temp means more photons at all wavelengths (d) 800 K Skew of spectrum in visible is towards red as seem in PHeT demo in class 4. (a) 159 W (b) 190 J Assume your are naked. Only half of friend s surface receives light so area is 1m 2 (c) about 5 x photons at 540 nm 5. (a) 41.1 cm (b) 48.9 cm (c) Diopters 6. (a) tan(theta_i) = n2/n1 (b) If theta_i = 45 deg then n1=n2 and there would be no reflection!