13.07.2015 Views

Solutions to Homework Questions 8

Solutions to Homework Questions 8

Solutions to Homework Questions 8

SHOW MORE
SHOW LESS

You also want an ePaper? Increase the reach of your titles

YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.

Physics 112 <strong>Homework</strong> 8 (solutions) (2004 Fall)<strong>Solutions</strong> <strong>to</strong> <strong>Homework</strong> <strong>Questions</strong> 8Chapt22, Problem-6: The two mirrors in Figure P22.6 meetat a right angle. The beam of light in the vertical plane P strikes mirror 1 asshown. (a) Determine the distance the reflected light beam travels beforestriking mirror 2. (b) In what direction does the light beam travel after beingreflected from mirror 2?Solution:The first point <strong>to</strong> note here is that everything happens in a thedashed Plane P (only specular reflection need be considered). Thusit is possible (and useful) <strong>to</strong> redraw the geometry in only 2D(ie the situation as ‘seen’ in plane P).Once we have done that, the problem becomes one of ‘simple’trigonometry plus the law of reflection (applied twice – once ateach mirror).Mirror 2The 2D-view is shown right. So ….(a) Where we are asked for the distance d.Well we apply the law of reflection for mirror1, then given thedistance of that that reflection from mirror2, we can use simpletrig <strong>to</strong> calculate the distance requested1.25 m = d sin 40.0° , so d = 1.94 m.(b) Where we are asked for the direction following the 2 nd reflectionHere (again) using simple trig, we can calculate all the anglesassociated with the “angle of incidence” of the 2 nd reflection..Armed with this we simply apply the Law of Reflection again <strong>to</strong> obtain50.0° above horizontal , or parallel <strong>to</strong> the incident ray.40°50°i 2 = 50°d1.25 m40°i 1 = 40°50° 50°Mirror 1Chapt22, Problem-7: An underwater scuba diver sees the Sun at an apparent angle of 45.0° from the vertical.What is the actual direction of the Sun?Solution:Here one simply has <strong>to</strong> use Snell’s Law – obviously a sketch helps…n 1 sin ! 1 = n 2 sin ! 2sin ! 1 = 1.333 sin 45.0°sin ! 1 = (1.333)(0.707) = 0.943so ! 1 = 70.5° " 19.5° above the horizontaln 2 = 1.333! 1n 1 = 1.00! 21


Physics 112 <strong>Homework</strong> 8 (solutions) (2004 Fall)Chapt22, Problem-13: A ray of light is incident on the surface of a block of clear ice at an angle of 40.0°with the normal. Part of the light is reflected and part is refracted. Find the angle between the reflected and refracted light.Solution:Again, a straight application of Snell’s law:! 2 = sin "1 # n 1 sin! 1&% ( = sin "1 #( 1.00)sin 40.0° &%( = 29.4° ,$ n 2 ' $ 1.309 'and the law of reflection gives ! = " 1 = 40.0° .Hence, the angle between the reflected and refracted rays is! = 180° "# 2 "$ = 180° " 29.4° " 40.0°= 111°Chapt22, Problem-22: A submarine is 300 m horizontally out from the shore and 100 m beneath the surfaceof the water. A laser beam is sent from the sub so that it strikes the surface of the water at a point 210 m from the shore. Ifthe beam just strikes the <strong>to</strong>p of a building standing directly at the water’s edge, find the height of the building.Solution:The angle of incidence at the watersurface is! 1 = tan "1 # 90.0 m &% ( = 42.0°$ 100 m 'Then, Snell’s law gives the angle ofrefraction as! 2 = sin "1 # n water sin! 1&% ( = sin "1 # ( 1.333)sin 42.0° &%( = 63.1° ,$ n air ' $ 1.00 'so the height of the building is h = 210 m = 210 mtan! 2tan63.1° = 107 mChapt22, Problem-24: A narrow beam of ultrasonic wavesreflects off the liver tumor in Figure P22.24. If the speed of the wave is10.0% less in the liver than in the surrounding medium, determinethe depth of the tumor.Solution:"From Snell’s law, sin! = $#n mediumn liver%' sin 50.0° .&But, n medium= cv medium= v liver= 0.900 ,n livercv liverv mediumso, ! = sin "1 [( 0.900)sin 50.0° ]= 43.6° .From the law of reflection,12.0 cmd = = 6.00 cm , and h = d2tan! =ice"! 2h! 2210 mwater! 1n = 1.33350.0#dh!6.00 cmtan 43.6°airairn = 1.0090.0 m! 140.0# $Submarine12.0 cm!Tumor100 mn mediumn liver2


Physics 112 <strong>Homework</strong> 8 (solutions) (2004 Fall)Chapt22, Problem-40: A jewel thief hides a diamond byplacing it on the bot<strong>to</strong>m of a public swimming pool. He places a circularraft on the surface of the water directly above and centered on the diamond,as shown in Figure P22.40. If the surface of the water is calm and the poolis 2.00 m deep, find the minimum diameter of the raft that would prevent thediamond from being seen.Solution:The circular raft must cover the area of the surface throughwhich light from the diamond could emerge. Thus, it must form the base of a cone (with apex at thediamond) whose half angle is q, where q is greater thanor equal <strong>to</strong> the critical angle.The critical angle at the water-air boundary is! c = sin "1 # n air&% ( = sin "1 # 1.00 &% ( = 48.6°$ ' $ 1.333'n waterThus, the minimum diameter of the raft is2 r min = 2 h tan! min = 2 h tan! c = 22.00 m ( )tan 48.6°= 4.54 m!! h r min! !Chapt22, Problem-54: When you look through a window, by how much time is the light you see delayedby having <strong>to</strong> go through glass instead of air? Make an order-of-magnitude estimate on the basis of data you specify. By howmany wavelengths is it delayed?Solution:Consider glass d = 4.0 mm thick, and with index of refraction n = 1.5 . The speed of light in the glass isv = c 3.0 ! 108 ms=n 1.5The extra travel time is= 2.0! 10 8 ms!t = d v " d c = 4.0 #10 "3 m2.0 #10 8 ms " 4.0 #10 "3 m3.0 #10 8 ms = 6.7 # 10"12 s ~10 !11 sFor light of wavelength ! 0 = 600 nm in vacuum, the wavelength in the glass is! = ! 0600 nm= = 400 nmn 1.5and the extra path length as a number of wavelengths is!d = d " # d = 4.0 $ 10#3 m" 0400 $10 #9 m # 4.0 $10 #3 m600 $ 10 #9 m = 3.3 $ 103 ~10 3Chapt22, Conceptual-1: Under certain circumstances, sound can be heard from extremely far away. Thisfrequently happens over a body of water, where the air near the water surface is cooler than the air at higher altitudes. Explainhow the refraction of sound waves could increase the distance over which sound can be heard.Solution:Sound radiated upward at an acute angle <strong>to</strong> the horizontal is bent back <strong>to</strong>wards the Earth by refraction.This means that the sound can reach the listener by this path as well as by a direct path. Thus, the soundis louder, allowing it <strong>to</strong> be heard from a greater distance.4


Physics 112 <strong>Homework</strong> 8 (solutions) (2004 Fall)Chapt22, Conceptual-2: What are some of the reasons that most ceilings are made of white, texturedmaterial?Solution:Ceilings are generally painted white (or another light color) so they will reflect more light, making the roombrighter. Textured materials are often used on the ceiling <strong>to</strong> diffuse the reflected light and reduce ‘glare’(specular reflections).Chapt22, Conceptual-3: The color of an object is said <strong>to</strong> depend on wavelength. So, if you view coloredobjects under water, in which the wavelength of light will be different, does the color change?Solution:The color will not change, for two reasons. First, despite the popular statement that the color depends onwavelength, it actually depends on the frequency of the light, which does not change under water. Second,when the light enters the eye, it travels through thhe fluid within the eye. Thus, even if color did dependon wavelength, the important wavelength is that of the light in the ocular fluid, which does of course doesnot change whether your head is under water or not.Chapt22, Conceptual-5: Pass white light through a prism <strong>to</strong> produce a multicolored spectrum. Let thatspectrum fall on a piece of cardboard that has a slit in it such that only green light passes through. Nw let this green lightfall on a second prism. Describe the light that comes out of the second prism.Solution:A prism creates a multicolored spectrum when white light passes through it because white light containsall visible wavelengths. Because the index of refraction varies wavelength, the different wavelengthsemerge from the prism traveling in slightly different directions. This produces the observed rainbow ofcolors. If only one wavelength of light (or a very small range of wavelengths such as the small range knownas ‘green’) is allowed <strong>to</strong> enter the prism, as in the case of the second prism in the question, a singlewavelength (or small range - ‘green’) will emerge from it.Chapt22, Conceptual-17: Why do astronomers looking at distant galaxies talk about looking back intime?Solution:Light travels through the vacuum of space at a speed of 3x10 8 m s -1 . Thus an image we see from a distantstar or galaxy must have been generated some time ago. For example, the star Altair is 16 lightyears away;if we look at an image of Altair taken <strong>to</strong>day, we are actually seeing what Altair looked like 16 years ago.Hence we are effectively ‘looking back in time’5

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!