Lect27 - Lecture 27: Mirrors Physics 212 35 30 25 20 15 10...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Lecture 27: Mirrors Physics 212 35 30 25 20 15 10 5 Confused Confident Avg = 3.3 Physics 212 Lecture 27, Slide 1 Physics 0 Your comments • I think "Man in the Mirror" by Michael Jackson should be played while giving this lecture! What do you say? • I didn't do this. I was too focused on the test that we had tonight. Which wasn't too bad by the way. • I am off to take a fizax exam! • i just took an exam leave me in peace brother. • I conquered that exam!!! • Please have mercy on those of us who are majoring in other subjects aside from electrical engineering. • Einstein, Newton and Pascal decide to play hide and seek. Einstein is it, closes his eyes, counts to 10 then opens them. Pascal is no where to be seen. Newton is sitting right in front of Einstein, with a piece of chalk in his hand. He's sitting in a box drawn on the ground, a meter to a side. Einstein says "Newton, you're terrible, I've found you!" Newton says "No no, Einy. You've found one Newton per square meter. You've found Pascal!" Physics 212 Lecture 27, Slide 2 Physics Reflection Angle of incidence = Angle of reflection θi = θ r θi θr That’s all of the physics – everything else is just geometry! Physics 212 Lecture 27, Slide 3 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 4 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: No light actually gets here θr θi d d Physics 212 Lecture 27, Slide 5 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 BB Physics 212 Lecture 27, Slide 6 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 eye then you can’t see it Physics 212 Lecture 27, Slide 7 Physics You will also get images from curved mirrors: Physics 212 Lecture 27, Slide 8 Physics Concave: Consider the case where the shape of the mirror is such 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: These mirrors are often sections of 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… f Physics 212 Lecture 27, Slide 9 Physics Aside: For a spherical mirror, R = 2f 2f R 2f center of sphere sometimes labeled “C” f Physics 212 Lecture 27, Slide 10 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 object 2f image f You now know the position of the same point on the image Physics 212 Lecture 27, Slide 11 Physics S > 2f 2f image is: real inverted smaller 111 += S S′ f S′ M =− S object 2f image f S S’ f Physics 212 Lecture 27, Slide 12 Physics S = 2f image is: real inverted same size 111 += S S′ f S′ M =− S object f image 2f S’ S f Physics 212 Lecture 27, Slide 13 Physics 2f > S > f image is: image real inverted bigger 111 += S S′ f S′ M =− S 2f image object f f S’ S Physics 212 Lecture 27, Slide 14 Physics f >S>0 rays no longer meet rays in front of the mirror but they do meet behind the mirror f object image (virtual) f S Physics 212 Lecture 27, Slide 15 Physics f >S>0 image is: virtual upright bigger 111 += S S′ f S′ M =− S image (virtual) f object f S S’<0 Physics 212 Lecture 27, Slide 16 Physics Convex: Consider the case where the shape of the mirror is such that light rays parallel to the axis of the mirror are all “focused” to a common spot a distance f behind the mirror: f Physics 212 Lecture 27, Slide 17 Physics S >0 image is: virtual upright smaller 111 += S S′ f object image (virtual) f<0 S>0 S’<0 Physics 212 Lecture 27, Slide 18 Physics Executive Summary – Mirrors & Lenses: S > 2f 2f 2f > S > f f >S>0 real inverted smaller real inverted bigger virtual upright bigger (converging) concave f converging f 111 += S S′ f S′ M =− S S >0 virtual upright smaller (diverging) convex diverging f f Physics 212 Lecture 27, Slide 19 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: S: S’ : f: positive if object is “upstream” of lens positive if image is “downstream” of lens positive if converging lens Mirrors sign conventions positive if object is “upstream” of mirror positive if image is “upstream” of mirror positive if converging mirror (concave) Physics 212 Lecture 27, Slide 20 Physics Summary Table Convex Converging Lens f + Real or Virtual Concave Converging Mirror + Real or Virtual Concave Diverging Lens – Always Virtual Convex Diverging Mirror – Always Virtual Image Sign of S’ Magnifica tion Sign of M + or – Larger or Smaller + or – Upright or Inverted + or – Larger or Smaller + or – Upright or Inverted Always – Always Smaller Always + Upright Always – Always Smaller Always + Upright Physics 212 Lecture 27, Slide 21 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 22 Physics Preflight 4, 5 80 60 40 20 0 80 60 40 20 0 Physics 212 Lecture 27, Slide 23 Physics Preflight 7 BB 2f > S > f image is: image real inverted bigger 111 += S S′ f S′ M =− S f >S>0 rays no longer meet rays in front of the mirror but they do meet behind the mirror 2f image object f f object image (virtual) f S’ S 50 f S If the object is behind the focal length it will reflect an inverted image. 40 30 20 10 0 If the object is in front of the focal length it will produce a virtual upright image. Physics 212 Lecture 27, Slide 24 Physics Preflight 9 BB S >0 image is: virtual upright smaller object object image (virtual) 70 60 f<0 S>0 S’<0 50 40 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 25 Physics Calculation An arrow is located in front of a convex spherical mirror of radius R = 50cm. The tip of the arrow is located at (-20cm,-15cm). (-20,-15) y R= 5 0 x 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 26 Physics Calculation An arrow is located in front of a convex spherical mirror of radius R = 50cm. The tip of the arrow is located at (-20cm,-15cm). What is the focal length of the mirror? A) f =50cm B) f = 25cm C) f = -50cm D) f = -25cm y R=5 0 BB x (-20,-15) For a spherical mirror | f | = R/2 = 25cm. R/2 Rule for sign: Positive on side of mirror where light goes after hitting mirror y R f <0 f = - 25 cm Physics 212 Lecture 27, Slide 27 Physics An arrow is located in front of a convex spherical mirror of radius R = 50cm. The tip of the arrow is located at (-20cm,-15cm). Calculation y R=5 0 f = -25 cm (-20,-15) x What is the x coordinate of the image? A) 11.1 cm Mirror equation B) 22.5 cm 111 =− s′ f s C) -11.1 cm D) -22.5cm BB 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) Physics 212 Lecture 27, Slide 28 Physics Calculation An arrow is located in front of a convex spherical mirror of radius R = 50cm. The tip of the arrow is located at (-20cm,-15cm). y R=5 0 x = 11.1 cm 11.1 f = -25 cm (-20,-15) x What is the y coordinate of the tip of the image? A) -11.1 cm Magnification equation B) -10.7 cm M =− S′ S C) -9.1 cm D) -8.3cm BB s = 20 cm s’ = -11.1 cm M = 0.556 yimage = 0.55 yobject = 0.556*(-15 cm) = -8.34 cm Physics 212 Lecture 27, Slide 29 Physics ...
View Full Document

Ask a homework question - tutors are online