Your Comments. Also, that 30/60/90 triangle prism question on the last homework...holy geometry batman!
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1 Your Comments Also, that 30/60/90 triangle prism question on the last homework...holy geometry batman! o thats why I'm always up side down when I look at the inside o my cereal spoon, regardless o how I turn my head! Is there any way we can estimate our grades or the semester to know what we are aiming or in the class like the grade estimator rom last semester? Are lenses on the exam? Mats told us that the exam only cover lectures 19-25, yet in the exam 3 activities there were lenses...am I missing something? are you going to upload a video o you taking exam 3? That exam 3 activity that was sent to us was incredibly helpul! I don't understand how it's possible to have a "virtual" image. wait a second... you cant look at a mirror rom the other side... This was a very simple prelecture. No problems! Can you go over the sign conventions or dierent types o lenses? This prelecture was very clear about sign conventions, but the lens one was not. Electricity & Magnetism Lecture 27, lide 1
2 Physics 212 Lecture 27 Today s Concept: Mirrors Exam Tomorrow Night at 7:00 Covers material in Lectures (no lenses) Bring your ID: Rooms determined by discussion section (see link) Electricity & Magnetism Lecture 27, lide 2
3 Relection Angle o incidence = Angle o relection q i = q r q i q r That s all o the physics everything else is just geometry! Electricity & Magnetism Lecture 27, lide 3
4 Flat Mirror All you see is what reaches your eyes You think object s location is where rays appear to come rom. q r q i Flat Mirror Object All rays originating rom peak will appear to come rom same point behind mirror! Image Electricity & Magnetism Lecture 27, lide 4
5 Flat Mirror 1) Draw irst ray perpendicular to mirror 0 = q i = q r 2) Draw second ray at angle. q i = q r 3) Lines appear to intersect a distance d behind mirror. This is the image location. q i q r Virtual Image: No light actually gets here d d Electricity & Magnetism Lecture 27, lide 5
6 Clicker Question A woman is looking at her relection in a lat vertical mirror. The lowest part o her body she can see is her knee. I she stands closer to the mirror, what will be the lowest part o her relection she can see in the mirror. A) Above her knee B) Her knee C) Below her knee Electricity & Magnetism Lecture 27, lide 6
7 Clicker Question A woman is looking at her relection in a lat vertical mirror. The lowest part o her body she can see is her knee. I she stands closer to the mirror, what will be the lowest part o her relection she can see in the mirror. A) Above her knee B) Her knee C) Below her knee I the light doesn t get to your eye then you can t see it Electricity & Magnetism Lecture 27, lide 7
8 You will also get Images rom Curved Mirrors: Electricity & Magnetism Lecture 27, lide 8
9 Concave: Consider the case where the shape o the mirror is such that light rays parallel to the axis o the mirror are all ocused to a common spot a distance in ront o the mirror: Note: analogous to converging lens Real object can produce real image These mirrors are oten sections o spheres (assumed in this class). For such spherical mirrors, we assume all angles are small even though we draw them big to make it easy to see Electricity & Magnetism Lecture 27, lide 9
10 Aside: For a spherical mirror, R = 2 R 2 center o sphere sometimes labeled C Electricity & Magnetism Lecture 27, lide 10
11 Recipe or Finding Image: 1) Draw ray parallel to axis relection goes through ocus 2) Draw ray through ocus relection is parallel to axis object 2 image You now know the position o the same point on the image Note: any other ray rom tip o arrow will be relected according to q i = q r and will intersect the two rays shown at the image point. Electricity & Magnetism Lecture 27, lide 11
12 > 2 image is: real inverted smaller > 0 s > 0 s > = M = object 2 image Electricity & Magnetism Lecture 27, lide 12
13 CheckPoint 1a The diagram above shows three light rays relected o a concave mirror. Which ray is NOT correct? A B C C is not correct as it does not go through the ocal point. Electricity & Magnetism Lecture 27, lide 13
14 CheckPoints 1b The diagram above shows two light rays relected o a concave mirror. The image is A. Upright and reduced B. Upright and enlarged C. Inverted and reduced D. Inverted and enlarged Electricity & Magnetism Lecture 27, lide 14
15 > > 0 rays no longer meet in ront o the mirror but they do meet behind the mirror object image (virtual) Electricity & Magnetism Lecture 27, lide 17
16 > > 0 image is: virtual upright bigger object image (virtual) > 0 s > 0 s > = M = < 0 Electricity & Magnetism Lecture 27, lide 18
17 Convex Mirror Convex: Consider the case where the shape o the mirror is such that light rays parallel to the axis o the mirror are all ocused to a common spot a distance behind the mirror: Note: analogous to diverging lens Real object will produce virtual image Electricity & Magnetism Lecture 27, lide 19
18 > 0 image is: virtual upright smaller > 0 s > 0 s > = M = object image (virtual) < 0 > 0 < 0 Electricity & Magnetism Lecture 27, lide 20
19 Executive ummary Mirrors & Lenses: > 2 real inverted smaller > 0 2 > > real inverted bigger Concave (Converging) Converging > > 0 virtual upright bigger 1 1 = 1 M = < 0 > 0 virtual upright smaller Convex (diverging) Diverging Electricity & Magnetism Lecture 27, lide 21
20 It s Always the ame: 1 1 = 1 M = You just have to keep the signs straight: s is positive or a real image is positive when it can produce a real image Lens sign conventions : positive i object is upstream o lens : positive i image is downstream o lens : positive i converging lens Mirrors sign conventions : positive i object is upstream o mirror : positive i image is upstream o mirror : positive i converging mirror (concave) Electricity & Magnetism Lecture 27, lide 22
21 CheckPoint 2a The image produced by a concave mirror o a real object is A. Always upright B. Always inverted C. ometimes upright & sometimes inverted 2 > > image is: real inverted bigger = M = > > 0 rays no longer meet in ront o the mirror but they do meet behind the mirror 2 object object image (virtual) image I the object is behind the ocal length it will relect an inverted image. I the object is in ront o the ocal length it will produce a virtual upright image. Electricity & Magnetism Lecture 27, lide 23
22 CheckPoint 2b The image produced by a convex mirror o a real object is A. Always upright B. Always inverted C. ometimes upright & sometimes inverted > 0 image is: virtual upright smaller object image (virtual) < 0 > 0 < 0 Electricity & Magnetism Lecture 27, lide 24
23 Calculation An arrow is located in ront o a convex spherical mirror o radius R = 50cm. The tip o the arrow is located at ( 20cm, 15cm). (-20,-15) y x Where is the tip o the arrow s image? Conceptual Analysis Mirror Equation: 1/s 1/s = 1/ Magniication: M = s /s trategic Analysis Use mirror equation to igure out the x coordinate o the image Use the magniication equation to igure out the y coordinate o the tip o the image Electricity & Magnetism Lecture 27, lide 25
24 Calculation An arrow is located in ront o a convex spherical mirror o radius R = 50cm. The tip o the arrow is located at ( 20cm, 15cm). (-20,-15) y x What is the ocal length o the mirror? A) = 50cm B) = 25cm C) = 50cm D) = 25cm For a spherical mirror = R/2 = 25cm. Rule or sign: Positive on side o mirror where light goes ater hitting mirror y R = 25 cm < 0 Electricity & Magnetism Lecture 27, lide 26
25 Calculation An arrow is located in ront o a convex spherical mirror o radius R = 50cm. The tip o the arrow is located at ( 20cm, 15cm). y = 25 cm x What is the x coordinate o the image? (-20,-15) A) 11.1 cm B) 22.5 cm C) 11.1 cm D) 22.5cm Mirror equation 1 = = 1 1 ( 25)(20) = s = 20 cm = 25 cm = 11.1 cm ince s < 0 the image is virtual (on the other side o the mirror) Electricity & Magnetism Lecture 27, lide 27
26 Calculation An arrow is located in ront o a convex spherical mirror o radius R = 50cm. The tip o the arrow is located at ( 20cm, 15cm). y x = 11.1cm = 25 cm x What is the y coordinate o the tip o the image? (-20,-15) A) 11.1 cm B) 10.7 cm C) 9.1 cm D) 8.3cm Magniication equation M = s = 20 cm s = 11.1 cm M = y image = 0.55 y object = 0.556*( 15 cm) = 8.34 cm Electricity & Magnetism Lecture 27, lide 28
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