Lect27 - Physics 212 Lecture 27: Mirrors Lecture Mirrors...

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Unformatted text preview: Physics 212 Lecture 27: Mirrors Lecture Mirrors Physics 212 Lecture 27, Slide 1 Physics Music Who is the Artist? A) B) C) D) E) Marvin Gaye Aaron Neville Michael Bolton Al Green Ray Charles 1. 2. 2. 3. 3. 4. 5. 5. 6. 7. 7. 8. 8. 9. 9. 10. 10. Pretty Woman Unchained Melody Funny How Time Slips Away Funny Ain't No Mountain High Enough How Can you Mend A Broken Heart Take Me To The River I’m So Lonesome I Could Cry I Can't Get Next To You My Girl Light My Fire Something upbeat Something for the day before the exam before Different.. Different.. Mostly covers.. Very good… Very Roy Orbison Roy Righteous Brothers Righteous Willie Nelson Willie Marvin Gaye Marvin Bee Gees Al Green Al Hank Williams Hank Temptations Temptations Temptations Temptations Doors Physics 212 Lecture 27, Slide 2 Physics Your Comments “can we really take either of the 2 finals even can if we don't have a conflict?!?” if YES, BUT YOU MUST SIGN UP TODAY “I'm still a little confused about the sign I'm convention as it applies to the formulas relating image length, object length, and focal length” image We will We Do Examples Do Clarify Sign Conventions “The specific similarities between convex The mirrors and converging lens' and concave mirrors and diverging lens‘” mirrors Note How Similar This Note Lecture is to the Lenses Lecture !! “Just examples to help stuff sink in, please.” “Only one more prelecture?! No! Don't leeeeave us!” 30 “We need to take a vote on who will win the We '05 NCAA title game rematch tonight? 20 A) Illinois B) UNC” A) 10 Let’s Do It ! 05 40 BB 0 Confused Confident Physics 212 Lecture 27, Slide 3 Physics Some Exam Stuff • Exam Tomorrow Night at 7:00 – – – – Covers material in Lectures 19 - 26 Bring your ID: Rooms determined by discussion section (see link) Conflict exam at 5:15 – sign up in your gradebook if you need to If you have conflicts with both of these, you should have heard from Prof Beck about scheduling https://my.physics.illinois.edu/undergrad/examprep/set_session.asp?crsnbr=212&examnum=3 Physics 212 Lecture 27, Slide 4 Physics Another Comment “Now it all started four days ago when my friends and I went to visit my friends grandmother for Thanksgiving dinner. It was a long drive to this grandmothers house, at least three interstate changes, then this song came on the radio. Now I was busy driving and keeping the public at large safe from 2000 lbs of minivan I was driving but I noticed that this same song seemed to have played on the radio the nearly the entire duration of the trip but that's not what I came to write about. I came to write about Thanksgiving dinner. At said grandmother's house, we had a Thanksgiving dinner that couldn't be beat, didn't get up until 7 o'clock. On the drive home, a song came on the radio. This song had a familiar ring to it and after the first ten minutes or so I realized it was indeed the same song that was on before and I caught on that the guy in the song was talking about littering on Thanksgiving day. That sounded awfully familiar, but that's not what I am talking about. I am talking about my birthday which was the next day. I got a 3 dollar toaster for being so lucky to have my birthday on black friday. I had some time to think about that song. Now there was only one of two things I could have done and the first one was I could have just forgotten the song and gone on with my life which wasn't very likely and wasn't what I expected or I could have just looked up the song and got it out of my system which is what I expected but when I looked online for the song there was a third possibility which I hadn't counted upon and I recognized the album cover from physics lecture. It was Arlo Guthrie's Alice's Restaurant Massacre.” Physics 212 Lecture 27, Slide 5 Physics Reflection Angle of incidence = Angle of reflection θi = θr θi θr That’s all of the physics – everything all everything else is just geometry! else Physics 212 Lecture 27, Slide 6 Physics Flat Mirror • All you see is what reaches your eyes – You think object’s location is where rays appear to come from. Flat Mirror θr θi Object All rays originating from peak will appear to come from same point behind mirror! Image 12 Physics 212 Lecture 27, Slide 7 Physics Flat Mirror (1) Draw first ray perpendicular to mirror 0 = θi = θr (2) Draw second ray at angle. θi = θr (3) Lines appear to intersect a distance d behind mirror. This is the image location. Virtual Image: Virtual No light actually gets here No θr θi d d Physics 212 Lecture 27, Slide 8 Physics ACT A woman is looking at her reflection in a flat vertical mirror. The lowest part of her body she can see is her knee. BB If she stands closer to the mirror, what will be the lowest part of her reflection she can see in the mirror. A. Above her knee B. Her knee C. Below her knee Physics 212 Lecture 27, Slide 9 Physics ACT A woman is looking at her reflection in a flat vertical mirror. The lowest part of her body she can see is her knee. If she stands closer to the mirror, what will be the lowest part of her reflection she can see in the mirror. A. Above her knee B. Her knee C. Below her knee If the light doesn’t get to your get eye then you can’t see it eye Physics 212 Lecture 27, Slide 10 Physics You will also get images from curved mirrors: Physics 212 Lecture 27, Slide 11 Physics Concave: Consider the case where the shape of the mirror is such Consider that light rays parallel to the axis of the mirror are all “focused” to a common spot a distance f in front of the mirror: to Note: analogous to “converging lens” Real object can produce real image These mirrors are often sections of spheres sections (assumed in this class). (assumed For such “spherical” For mirrors, we assume all mirrors, angles are small even though we draw them big to make it easy to see… to f Physics 212 Lecture 27, Slide 12 Physics Aside: For a spherical mirror, R = 2f 2f R 2f 2f f center of sphere sometimes labeled “C” sometimes Physics 212 Lecture 27, Slide 13 Physics Recipe for finding image: 1) Draw ray parallel to axis reflection goes through focus 2) Draw ray through focus reflection is parallel to axis norm object 2f 2f al f image normal You now know the position of the same point on the image Note: any other ray from tip of arrow will be reflected according to θi = θr and will intersect the two rays shown at the image point. Physics 212 Lecture 27, Slide 14 Physics image is: real inverted smaller S > 2f 2f object f>0 s>0 s’ > 0 111 += S S′ f S′ M =− S f 2f 2f image S S’ f Physics 212 Lecture 27, Slide 15 Physics S = 2f object 111 += S S′ f S′ M =− S image is: real inverted same size f image 2f 2f f S’ S Physics 212 Lecture 27, Slide 16 Physics 2f > S > f image is: image real inverted bigger object 2f 2f 111 += S S′ f S′ M =− S f image f S’ S Physics 212 Lecture 27, Slide 17 Physics f >S>0 rays no longer meet rays in front of the mirror object f but they do meet but do meet behind the mirror behind image (virtual) f S Physics 212 Lecture 27, Slide 18 Physics image is: virtual upright bigger f >S>0 f>0 s>0 s’ < 0 object f 111 += S S′ f S′ M =− S image (virtual) f S S’<0 Physics 212 Lecture 27, Slide 19 Physics Convex: Consider the case where the shape of the mirror is such Consider that light rays parallel to the axis of the mirror are all “focused” to a common spot a distance f behind the mirror: to Note: analogous to “diverging lens” Real object will produce virtual image f Physics 212 Lecture 27, Slide 20 Physics S >0 image is: virtual upright smaller f<0 s>0 s’ < 0 111 += S S′ f S′ M =− S image (virtual) object f<0 S>0 S’<0 Physics 212 Lecture 27, Slide 21 Physics Executive Summary – Mirrors & Lenses: Executive S > 2f 2f real inverted smaller 2f > S > f real inverted bigger f >S>0 f>0 concave virtual upright bigger (converging) converging f f S′ M =− S 111 += S S′ f f<0 S >0 virtual upright smaller convex (diverging) diverging f f Physics 212 Lecture 27, Slide 22 Physics It’s always the same: 111 += S S′ f S′ M =− S You just have to keep the signs straight: s’ is positive for a real image f is positive when it can produce a real image Lens sign conventions S: S’ : f: positive if object is “upstream” of lens positive of positive if image is “downstream” of lens positive positive if converging lens Mirrors sign conventions S: S’ : f: positive if object is “upstream” of mirror positive of positive if image is “upstream” of mirror positive positive if converging mirror (concave) Physics 212 Lecture 27, Slide 23 Physics Preflight 2 BB 60 C is not correct as it does not go through the focal point. 50 40 30 20 10 0 Physics 212 Lecture 27, Slide 24 Physics Preflight 4, 5 80 80 60 60 40 40 20 20 0 0 Physics 212 Lecture 27, Slide 25 Physics Preflight 7 BB 2f > S > f object 2f 2f 111 += S S′ f S′ M =− S image is: image real inverted bigger f >S>0 f rays no longer meet rays in front of the mirror object f but they do meet but do meet behind the mirror behind image (virtual) image f S’ f S S 50 If the object is behind the focal length it will reflect an inverted image. 40 30 20 If the object is in front of the focal length it will produce a virtual upright image. 10 0 Physics 212 Lecture 27, Slide 26 Physics Preflight 9 BB S >0 image is: virtual upright smaller image (virtual) object object 70 60 f<0 50 40 S>0 S’<0 30 20 10 It's like the back of a spoon, or one of those mirrors in the corner of a convenience store. 0 Physics 212 Lecture 27, Slide 27 Physics Calculation An arrow is located in front of a convex An spherical mirror of radius R = 50cm. 50cm. The tip of the arrow is located The at (-20cm,-15cm). y R=5 0 x (-20,-15) Where is the tip of the arrow’s image? • Conceptual Analysis • Mirror Equation: 1/s + 1/s’ = 1/f • Magnification: M = -s’/s • Strategic Analysis • Use mirror equation to figure out the x coordinate of the image • Use the magnification equation to figure out the y coordinate of the tip of the image Physics 212 Lecture 27, Slide 28 Physics Calculation An arrow is located in front of a convex An spherical mirror of radius R = 50cm. 50cm. The tip of the arrow is located The at (-20cm,-15cm). y R=5 0 BB x (-20,-15) What is the focal length of the mirror? A) f =50cm B) f = 25cm C) f = -50cm D) f = -25cm For a spherical mirror | f | = R/2 = 25cm. For R/2 Rule for sign: Positive on side of mirror where light goes after hitting mirror y R f = - 25 cm f <0 Physics 212 Lecture 27, Slide 29 Physics Calculation y R=5 0 An arrow is located in front of a convex An spherical mirror of radius R = 50cm. 50cm. The tip of the arrow is located The at (-20cm,-15cm). x f = -25 cm (-20,-15) What is the x coordinate of the image? A) 11.1 cm Mirror Mirror equation equation B) 22.5 cm C) -11.1 cm D) -22.5cm BB 111 =− s′ f s fs s = 20 cm s− f f = -25 cm (−25)(20) ′= s = -11.1 cm 20 + 25 s′ = Since s’ < 0 the image is virtual (on the “other” side of the mirror) the Physics 212 Lecture 27, Slide 30 Physics Calculation An arrow is located in front of a convex An spherical mirror of radius R = 50cm. 50cm. The tip of the arrow is located The at (-20cm,-15cm). y R=5 0 x = 11.1 cm 11.1 x f = -25 cm (-20,-15) What is the y coordinate of the tip of the image? A) -11.1 cm B) -10.7 cm Magnification Magnification equation equation C) -9.1 cm S′ S s = 20 cm s’ = -11.1 cm D) -8.3cm BB M =− M = 0.556 yimage = 0.55 yobject = 0.556*(-15 cm) = -8.34 cm object 0.556*( 15 Physics 212 Lecture 27, Slide 31 Physics ...
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