Question

Physics I52 Due: Wednesday, April 24, 2019 Roster No.: Score: Paper Homework #10 For multiple-choice questions, circle the letter of the one best answer (unless more than one answer is asked for). Show your work on free-response questions. Be sure to use proper units and significant figures in your final answers. Organization and neatness count, too. Do NOT use red ink- please reserve that color for grading. 1. You do NOT need to show your work for this question. Be sure to include correct units. ± signs, and sig. figs! a. (4 pts.) You place a 30.0-cm-tall object 90.0 cm away from a converging lens with a 50.0 cm focal length. Using our standard plus/minus sign conventions, find the following values: what is the position of the image? What is the height ofthe image? What is the magnification of the image? Is the image real or virtual? b. (4 pts.) You place the same 30.0-cm-tall object 90.0 cm away from a diverging lens with a 50.0 cm focal length What is the height of the image? What is the position of the image? Is the image real or virtual? What is the magnification of the image? 2. (3 pts.) A spherical, concave mirror has a radius of curvature of 80.0 cm. You have a small, well-lit plastic action figure that is 10.0 cm tall, and you want to create a real image of it that is 2.50 times as large as the original object. (Hint: Watch your signs- the real image will be inverted, so the magnification should be negative.) At what distance from the mirror should you place the object? Show your work completely ppose that a shiny silver spherical Christmas-tree ornament makes a nicely reflective convex mirror "image"). What is the magnification of The ornament has a diameter of 10.0 cm. You position your face (the "object) 750 ha i the magnification of the ornament. When you look at the ornament, you see your reflection (the your reflection? Is it upright or inverted? Show your work completely. 4. (5 pts.) You notice a thin layer of transparent oil floating on the surface of a water puddle. Suppose that sunlight strikes the puddle from nearly directly overhead, and also that you are viewing the puddle from nearly directly above the puddle. You see bright colored light reflected from the oil, and the color gradually varies as the oil's thickness varies. In one particular region, the oil appears to be bright green ( 540. nm). If Mol 1.75 and t how thick (in nanometers) is the oil layer in that region? Since there are an infinite number of solutions, give the three thinnest answers. Show your work clearly, and include a sketch if desired 5. (4 pts.) A collimated beam (parallel rays, non-spreading and non-focusing) of u a flashlight has initial intensity le The beam then passes through three successive plan . a vertical polarizer, then a polarizer oriented at 15.0 to the vertical, then a polarizer oriented at 55.0 to the vertical. Find the final intensity of the beam of light that emerges from the third polarizer, in terms of le. Express your final answer as a decimal fraction of I, to 3 significant figures Show your work Be sure to use the correct angle θ in each application of Malus. Law!) (Hint: 6. When you look at the horizontal surface of a puddle of water or a pond (n-1.33) outdoors, you see a (dim) reflection of the surrounding buildings, trees, sky, etc. (2 pts.) At what angle of reflection is the reflected light completely (100%) horizontally polarized? Show your work BONUS: (+1 should wear sunglasses that are... A. horizontally polarized C. direction of polarization doesn't matter B. vertically polarized

Answer 1

1.

The position of the image is given by 1 1 1 + - f dd 1 1 1 df d fd d-f (50 cm)(90 cm) 90 сm - 50 cm =1 12.5 cm Therefore, the position of the image is 112.5 cm The height of the image is given by, d de 112.5 cm 3D- (30 сm 90 cm =-37.5 cm Therefore, the height of the image is -37.5 cm. The magnification of the image is given by =_ 112.5 cm 90 cm =-1.25 cm Therefore, the magnification of the image is -1.25 cm The position of the image is 112.5 cm, which is greater than zero. Hence, the image is real

The position of the image is given by, 1 1 1 f dd 1 1 1 d fd di d-f (-50 cm)(90 cm 90 cm-(-50 cm) C =-32.14 cm Therefore, the position of the image is -32.14 cm The height of the image is given by, d h h -32.14 cm =(30 cm 90 cm 10.7 cm n Therefore, the height of the image is 10.7 cm The magnification of the image is given by, d. -32.14 cm 90 cm 0.35 cm C Therefore, the magnification of the image is 0.35 cm Hence, the image is virtual The position of the image is -32.14 cm, which is less than zero.