Chapter 25 Lecture Notes - Personal.tcu.edu

Chapter 25 Lecture NotesFor all of these lecture notes files, I’m going to be using theassociated chapter resource files on the 10164 website as ageneral outline. So we will start our coverage of Chapter 25with a basic discussion of light and mirrors. The law ofreflection is pretty simple:The angle of incidence (θi) the angle of reflection (θr).These angles are measured with respect to the NORMAL. For aflat mirror, this is very basic. For a curved surface, it getsa little more complicated. I recommend you take a look at the“Blackboard Optics” Physics Demonstration Video that is linkedat the end of Chapter 25.2 in the WileyPlus eBook. Your studentstudy guide also has some useful images of ray tracing.I’m not going to ask you to do any complicated ray tracing (likefrom a curved surface) in this chapter, but I think it ishelpful to give you a feel for where rays SHOULD converge andwhere you expect to find images for each kind of mirror.We are going to start the worksheets for this chapter with acouple of fairly simple geometry problems applying the law ofreflection. Please try to solve worksheet 25.1. Once you havefinished it or at least seriously attempted it and gotten stuck,please proceed to the next page to see my detailed solution.

If you have any questions about the way I solved worksheet 25.1,please post to the “Chapter 25 worksheets” discussion forum onthe Physics 10164 course shell. I will be checking this forumoften to answer questions.There is similarly a forum for questions about theHomework” for the same purpose, and I plan to havefor each of the six remaining homework assignmentssemester, shared by all three classes, on the same“Chapter 25two forumsthiscourse shell.As you can see, there are a few different geometric rules thatyou are going to need to remember for doing problems like this.First, the three interior angles of a triangle must add up to180 degrees.Second, complementary angles are angles that add up to a rightangle (90 degrees), so if you know one, you need to be able todeduce the other.Also, supplementary angles are angles that form a straight linewhen added together, so they add up to 180 degrees. Again, ifyou know one, you should be able to deduce the other.These rules, combined with the law of reflection and the rulesfor sin, cos and tan for right triangles (on the first page ofyour formula sheet) should be all you need for these kinds ofproblems.Next, you will have a chance to practice these rules again bytracing a simple ray bouncing off a flat surface. Pleaseattempt worksheet 25.2 now. Once you have finished it or atleast seriously attempted it and gotten stuck, please proceed tothe next page to see my detailed solution.

If you have any questions about worksheet 25.2, please post tothe appropriate “Chapter 25 worksheets” forum on d2l.tcu.edu,and I will respond there.—Now we will move on to the last section in Chapter 25, whichdeals with spherical mirrors and images. Please examine the raytracing diagrams in sections 25.4 and 25.5 carefully so youunderstand the difference in the way we treat concave and convexmirrors (the student study guide has some more detailed diagramsas well).The relevant equations for this section are located in themiddle of page 7 of your formula sheet. The sign conventionsfor mirrors, including the rules for focal length, imagedistance and object distance are shown there. I’m going to beusing different notation than the book here.p object distance (instead of do)q image distance (instead of di)Notice some of the important shorthand in that figure on yourformula sheet.When an image distance or object distance is positive, we saythe corresponding image or object is REAL.When an image distance or object distance is negative, we saythe corresponding image or object is VIRTUAL.Watch for that shorthand in problem statements! Knowing thesign of p or q is an important step sometimes in solving theproblem, even if you don’t know p or q at the start.The animated figure 25.18 (located in section 25.5 of youreBook) probably goes into more detail than you will need forthis chapter on how to do ray tracing, but it is definitelyhelpful when you are trying to follow ray tracing diagrams.Continuing our examination of the eBook, we are in chapter 25.5.Please notice the difference between figures 25.18 and 25.19.In figure 25.18, a real image is formed by an object locatedoutside of the focal point of a concave mirror.

From an algebraic standpoint, starting from the equation that wewill apply in pretty much every problem with a curved surface,the OPTICS EQUATION: 1/p 1/q 1/f. We can rearrange theequation to read 1/q 1/f - 1/p.If the object is located OUTSIDE of the focal point of themirror, that means p is bigger than f. That means 1/f - 1/pmust be positive since 1/f is bigger than 1/p. So q ispositive, meaning we get a real image.Read those last three sentences again if you have to.In figure 25.19, a virtual image is formed by an object locatedINSIDE the focal length of the mirror. Can you see why thismust be true algebraically?1/f is now smaller than 1/p, so 1/q ends up negative.negative, that means the image is virtual.When q isYou can conclude these two facts by using algebra OR raytracing. I encourage you to play around with ray tracing inConcept Simulation 25.2, located just below figure 25.20 in youreBook.—Looking at the formulas on page 7 of your formula sheet, I wantto point out two more.First, the definition of radius of curvature. For a sphericalmirror, the radius of curvature is the distance from the surfaceof the mirror to the center of the imaginary sphere from whichthe spherical mirror has been cut. The focal length of themirror has a value equal to R/2.Focal length for a concave mirror is R/2.Focal length for a convex mirror is -R/2.So just because the radius of curvature is always given as apositive number, that doesn’t mean the focal length is alwayspositive! For example, suppose you are told you have a convexmirror with a radius of curvature of 22 cm. What’s the focallength of that mirror?f -R/2 -11cm.Watch out for this!

Next, I want to look at our two equations involvingmagnification. These equations are derived from ray tracing,but we don’t need to recreate those derivations. We will justuse them to algebraically solve some of these problems.First, (height of image) Magnification *(height of object).Notice that Magnification has absolute value brackets around itsince you shouldn’t ever have a negative height when you aretalking about an object or an image. That just wouldn’t makesense.But magnification DOES have a /- sign associated with it.find that with our other formula: M -q/pWeNotice that if p and q have the same sign, then themagnification will be negative.If p and q have opposite signs, the magnification will bepositive.The way we interpret the magnification is also given on theformula sheet. If the magnification is positive, that means thefinal image will be upright. If the magnification is negative,that means the final image will be inverted.I know these are a lot of little details to keep handy, butthat’s why you have a formula sheet. Use it during this chapterto practice putting all of these clues together.For example, suppose you have a virtual image and a positivemagnification. Is the object real or virtual?Think about it If the magnification is positive, that means the object andimage distances have opposite signs, so if the image is virtual,the object must be real.You may think that objects always must be real, but as we willsee in Chapter 26, that’s not always the case.

Before we go on with worksheets, I would like you to takeadvantage of your book’s resources to practice as well. Pleaseread chapter 25.6 (where the magnification equation is coveredin a detail, though they use different notation) and try tosolve examples 4, 5 and 6 in this section of the book.I want to point out that example 7 in the book is just crazy.Don’t bother with it. The book tries to cram all of theequations together into one horrific algebraic mess. I preferto solve these things one step at a time.One last thing before we go to another worksheet.We always “read” optics diagrams from left to right. So if anobject is to the left of a mirror, we say the object is “infront of” the mirror and has a positive object distance.Likewise with images. That’s shown on the formula sheetdiagram, but I wanted to say it explicitly here, too, so I cansay “I told you so” later when you forget!—Let’s try an example with a concave mirror, which is worksheet25.3. Please complete or seriously attempt this worksheet asyou would try to do in lecture for 5-10 minutes, then when youare ready, check out the completed solution on the next page.Refer to your formula sheet for the relevant equations dealingwith magnification, radius of curvature and the optics equation.

If you have any questions about worksheet 25.3, please post tothe appropriate “Chapter 25 worksheets” forum on d2l.tcu.edu,and I will respond there.—Note how I solved for q using fractions and decimals. You canuse whatever method you are comfortable with. Most studentsfind converting all fractions to decimals is faster and lessprone to error, but I kinda like dealing with fractions andcommon denominators. It’s more likely to give me a preciseanswer unaffected by rounding errors.Note also what I did in part b. I found the magnification usingthe formula M -q/p but then I used the absolute value ofmagnification to determine the image height.—Now let’s try worksheet 25.4. This one has a lot of potentialplaces for sign mistakes and mixing up p and q, so proceed withcaution.One hint I will give you before you start is that themagnification in this problem MUST be negative. Can you see whyjust by reading the problem?Please complete or seriously attempt worksheet 25.4 as you wouldtry to do in lecture for 5-10 minutes, then when you are ready,check out the completed solution on the next page.

If you have any questions about worksheet 25.4, please post tothe appropriate “Chapter 25 worksheets” forum on d2l.tcu.edu,and I will respond there.—Notice how I used the image and object heights to find theabsolute value of the magnification and then used the fact thatthe image is inverted to deduce that the magnification must benegative. There is more to this chapter than just blindlyplugging numbers into your calculator. You have to stop and useyour judgement every once in a while. Practice, practice,practice!Our final worksheet for this chapter also takes a little bit ofinterpretation of the problem statement. Note in this nextproblem, we are assuming that the object is real. When there isjust a single mirror in a problem, we will always assume theobject is in front of the mirror and therefore real.Please complete or seriously attempt worksheet 25.5 as you wouldtry to do in lecture for 5-10 minutes, then when you are ready,check out the completed solution on the next page.