Lectures_Lect27

# Lectures_Lect27 - Lecture 27 Mirrors Physics 212 Physics...

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Unformatted text preview: Lecture 27: Mirrors Physics 212 Physics 212 Lecture 27, Slide 1 Reflection Angle of incidence = Angle of reflection i = r i r Physics 212 Lecture 27, Slide 2 Flat Mirror All you see is what reaches your eyes You think object's location is where rays appear to come from. Flat Mirror r i Object All rays originating from peak will appear to come from same point behind mirror! Image 12 Physics 212 Lecture 27, Slide 3 Flat Mirror (1) Draw first ray perpendicular to mirror 0 = i = r (3) Extend the reflected rays behind the mirror (4) Lines appear to intersect a distance d behind mirror. This is the image location. Virtual Image: No light actually gets here r i (2) Draw second ray at angle. i = r d d Physics 212 Lecture 27, Slide 4 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 Physics 212 Lecture 27, Slide 5 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 6 You will also get images from curved mirrors: Physics 212 Lecture 27, Slide 7 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: Note: analogous to "converging lens" Real object can produce real image 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 8 Aside: For a spherical mirror, R = 2f R center of sphere sometimes labeled "C" 2f f Physics 212 Lecture 27, Slide 9 Recipe for finding image: 1) Draw ray parallel to axis 2) Draw ray through focus reflection goes through focus reflection is parallel to axis al norm object 2f image f 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 i = r and will intersect the two rays shown at the image point. Physics 212 Lecture 27, Slide 10 S > 2f image is: real inverted smaller f>0 s>0 s' > 0 1 1 1 + = S S f S M =- S object 2f image f S S' f Physics 212 Lecture 27, Slide 11 Checkpoint 1a The diagram above shows three light rays reflected off a concave mirror. Which ray is NOT correct? A B C C is not correct as it does not go through the focal point. Physics 212 Lecture 27, Slide 12 S = 2f image is: real inverted same size 1 1 1 + = S S f S M =- S object image f 2f S' S f Physics 212 Lecture 27, Slide 13 2f > S > f image is: real inverted bigger 1 1 1 + = S S f S M =- S 2f image object f f S' S Physics 212 Lecture 27, Slide 14 f > S > 0 rays no longer meet in front of the mirror but they do meet behind the mirror f object image (virtual) f S Physics 212 Lecture 27, Slide 15 f > S > 0 image is: virtual upright bigger f>0 s>0 s' < 0 1 1 1 + = S S f S M =- S f object image (virtual) f S S'<0 Physics 212 Lecture 27, Slide 16 Checkpoint 1b The diagram above shows three light rays reflected off a concave mirror. The image is A. Upright and reduced B. Upright and enlarged C. Inverted and reduced D. Inverted and enlarged Physics 212 Lecture 27, Slide 17 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: Note: analogous to "diverging lens" Real object will produce virtual image f Physics 212 Lecture 27, Slide 18 S > 0 image is: virtual upright smaller f<0 s>0 s' < 0 1 1 1 + = S S f S M =- S object image (virtual) f<0 S>0 S'<0 Physics 212 Lecture 27, Slide 19 Executive Summary Mirrors & Lenses: S > 2f 2f > S > f f > S > 0 real inverted smaller real inverted bigger virtual upright bigger f > 0 c o nc ave ( c o nve r g ing ) f c o nve r g ing f 1 1 1 + = S S f f < 0 S M =- S S > 0 virtual upright smaller ( d ive r g ing ) c o nve x f d ive r g ing f Physics 212 Lecture 27, Slide 20 It's always the same: 1 1 1 + = 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: positive if object is "upstream" of lens S' : positive if image is "downstream" of lens f: positive if converging lens Mirrors sign conventions S: positive if object is "upstream" of mirror S' : positive if image is "upstream" of mirror f: positive if converging mirror (concave) Physics 212 Lecture 27, Slide 21 Checkpoint 2a The image produced by a concave mirror of a real object is A. Always upright B. Always inverted C. Sometimes upright and sometimes inverted 2 f > S > f im ag e is : r e al inve r t e d b ig g e r 1 1 1 + = S S f S M =- S f > S > 0 r ay s no lo ng e r m e e t in f r o nt o f t h e m ir r o r b ut t h e y d o m e e t b e h ind t h e m ir r o r 2f im ag e ob j e c t f f ob j e c t im ag e (vir t ual) f S' S f S If the object is behind the focal length it will reflect an inverted image. If the object is in front of the focal length it will produce a virtual upright image. Physics 212 Lecture 27, Slide 22 Checkpoint 2b The image produced by a convex mirror of a real object is A. Always upright B. Always inverted C. Sometimes upright and sometimes inverted S > 0 im ag e is : vir t ual upr ig h t s m alle r ob j e c t im ag e ( vir t ual) f <0 S >0 S '<0 It's like the back of a spoon, or one of those mirrors in the corner of a convenience store. Physics 212 Lecture 27, Slide 23 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 24 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? y R= 5 0 x (-20,-15) A) f =50cm B) f = 25cm C) f = 50cm D) f = 25cm For a spherical mirror | f | = R/2 = 25cm. 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 25 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 x coordinate of the image? Calculation y R= 5 0 f = -25 cm x (-20,-15) A) 11.1 cm B) 22.5 cm C) 11.1 cm D) 22.5cm Mirror equation 1 1 1 = - 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) Physics 212 Lecture 27, Slide 26 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 x = 11.1 cm f = -25 cm x (-20,-15) What is the y coordinate of the tip of the image? A) 11.1 cm B) 10.7 cm C) 9.1 cm D) 8.3cm Magnification equation M =- S S s = 20 cm s' = 11.1 cm M = 0.556 yimage = 0.556 yobject = 0.556*(15 cm) = 8.34 cm Physics 212 Lecture 27, Slide 27 ...
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