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Rf 1) 2) 3) p.a. Which ray is NOT correct? R f 1) 3) p.a. Ray through center should reflect back on self. Which ray is NOT correct?

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Presentation on theme: "Rf 1) 2) 3) p.a. Which ray is NOT correct? R f 1) 3) p.a. Ray through center should reflect back on self. Which ray is NOT correct?"— Presentation transcript:

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2 Rf 1) 2) 3) p.a. Which ray is NOT correct?

3 R f 1) 3) p.a. Ray through center should reflect back on self. Which ray is NOT correct?

4 Cases for Concave Mirrors CONCAVE MIRROR: OBJECT BEYOND C Image type: REAL VIRTUAL NO IMAGE Image location: BETWEEN F & V BETWEEN F & C AT C BEYOND C BEHIND MIRROR Image size: SAME ENLARGED REDUCED Image orientation:ERECT INVERTED

5 Cases for Concave Mirrors CONCAVE MIRROR: OBJECT BEYOND C Image type: REAL VIRTUAL NO IMAGE Image location: BETWEEN F & V BETWEEN F & C AT C BEYOND C BEHIND MIRROR Image size: SAME ENLARGED REDUCED Image orientation:ERECT INVERTED

6 CONCAVE MIRROR: OBJECT AT C Image type: REAL VIRTUAL NO IMAGE Image location: BETWEEN F & V BETWEEN F & C AT C BEYOND C BEHIND MIRROR Image size: SAME ENLARGED REDUCED Image orientation:ERECT INVERTED

7 CONCAVE MIRROR: OBJECT AT C Image type: REAL VIRTUAL NO IMAGE Image location: BETWEEN F & V BETWEEN F & C AT C BEYOND C BEHIND MIRROR Image size: SAME ENLARGED REDUCED Image orientation:ERECT INVERTED

8 CONCAVE MIRROR: OBJECT BETWEEN C & F Image type: REAL VIRTUAL NO IMAGE Image location: BETWEEN F & V BETWEEN F & C AT C BEYOND C BEHIND MIRROR Image size: SAME ENLARGED REDUCED Image orientation:ERECT INVERTED

9 CONCAVE MIRROR: OBJECT BETWEEN C & F Image type: REAL VIRTUAL NO IMAGE Image location: BETWEEN F & V BETWEEN F & C AT C BEYOND C BEHIND MIRROR Image size: SAME ENLARGED REDUCED Image orientation:ERECT INVERTED

10 CONCAVE MIRROR: OBJECT AT F Image type: REAL VIRTUAL NO IMAGE Image location: BETWEEN F & V BETWEEN F & C AT C BEYOND C BEHIND MIRROR Image size: SAME ENLARGED REDUCED Image orientation:ERECT INVERTED

11 CONCAVE MIRROR: OBJECT AT F Image type: REAL VIRTUAL NO IMAGE Image location: BETWEEN F & V BETWEEN F & C AT C BEYOND C BEHIND MIRROR Image size: SAME ENLARGED REDUCED Image orientation:ERECT INVERTED

12 CONCAVE MIRROR: OBJECT BETWEEN V & F Image type: REAL VIRTUAL NO IMAGE Image location: BETWEEN F & V BETWEEN F & C AT C BEYOND C BEHIND MIRROR Image size: SAME ENLARGED REDUCED Image orientation:ERECT INVERTED

13 CONCAVE MIRROR: OBJECT BETWEEN V & F Image type: REAL VIRTUAL NO IMAGE Image location: BETWEEN F & V BETWEEN F & C AT C BEYOND C BEHIND MIRROR Image size: SAME ENLARGED REDUCED Image orientation:ERECT INVERTED

14 O I Mirror Equation c dodo didi d o = distance object is from mirror: Positive: object in front of mirror Negative: object behind mirror d i = distance image is from mirror: Positive: inverted image (in front of mirror) Negative: upright image (behind mirror) f = focal length mirror: Positive: concave mirror Negative: convex mirror (coming soon) f

15 The image produced by a concave mirror of a real object is: Mirror Equation: 1)Always Real 2)Always Virtual 3)Sometimes Real, Sometimes Virtual Concave mirror: f > 0 Real Object means in front of mirror: d o > 0 d i can be negative or positive!

16 Concave Mirror Where in front of a concave mirror should you place an object so that the image is virtual? Mirror Equation: 1)Close to mirror 2)Far from mirror 3)Either close or far 4)Not Possible When d o < f then d i <0 : virtual image. Virtual image means behind mirror: d i < 0 Concave mirror: f > 0 Object in front of mirror: d o > 0

17 O I Magnification Equation dodo dodo hoho Angle of incidence didi hihi Angle of reflection didi h o = height of object: Positive: always h i = height of image: Positive: image is upright Negative: image is inverted m = magnification: Positive / Negative: same as for h i < 1: image is reduced > 1: image is enlarged    

18 Solving Equations A candle is placed 6 cm in front of a concave mirror with focal length f=2 cm. Determine the image location. Rf p.a. Compared to the candle, the image will be: Larger Smaller Same Size

19 Solving Equations A candle is placed 6 cm in front of a concave mirror with focal length f=2 cm. Determine the image location. d i = + 3 cm (in front of mirror) Real Image! Compared to the candle, the image will be: Larger Smaller Same Size Rf p.a.

20 Magnification A 4 inch arrow pointing down is placed in front of a mirror that creates an image with a magnification of –2. What is the size of the image? 1)2 inches 2)4 inches 3)8 inches What direction will the image arrow point? 1)Up2) Down 4 inches

21 Magnification A 4 inch arrow pointing down is placed in front of a mirror that creates an image with a magnification of –2. What is the size of the image? 1)2 inches 2)4 inches 3)8 inches What direction will the image arrow point? 1)Up2) Down (-) sign tells us it’s inverted from object Magnitude gives us size. 4 inches

22 O Convex Mirror Rays c 1) Parallel to principal axis reflects ______________. 2) Through f, reflects ______________________. #2 3) Through center. #3 Image is: Virtual or Real Uprightor Inverted Reduced orEnlarged (always true for convex mirrors!) f #1 Complete the rays!

23 O Convex Mirror Rays c 1) Parallel to principal axis reflects through f. 2) Through f, reflects parallel to principal axis. #2 I 3) Through center. #3 Image is: Virtual (light rays don’t really cross) Upright (same direction as object) Reduced (smaller than object) (always true for convex mirrors!): f #1

24 CONVEX MIRROR Image type: REAL VIRTUAL NO IMAGE Image location: BETWEEN F & V BETWEEN F & C AT C BEYOND C BEHIND MIRROR Image size: SAME ENLARGED REDUCED Image orientation:ERECT INVERTED

25 CONVEX MIRROR Image type: REAL VIRTUAL NO IMAGE Image location: BETWEEN F & V BETWEEN F & C AT C BEYOND C BEHIND MIRROR Image size: SAME ENLARGED REDUCED Image orientation:ERECT INVERTED

26 Solving Equations A candle is placed 6 cm in front of a convex mirror with focal length f=-3 cm. Determine the image location. Determine the magnification of the candle. If the candle is 9 cm tall, how tall does the image candle appear to be?

27 Solving Equations A candle is placed 6 cm in front of a convex mirror with focal length f=-3 cm. Determine the image location. Determine the magnification of the candle. If the candle is 9 cm tall, how tall does the image candle appear to be? m = + 1/3 h i = + 3 cm Image is Upright! d i = - 2 cm (behind mirror) Virtual Image!

28 Where should you place an object in front of a convex mirror to produce a real image? 1)Object close to mirror 2)Object far from mirror 3)Either close or far 4)You can’t

29 Where should you place an object in front of a convex mirror to produce a real image? Mirror Equation: 1)Object close to mirror 2)Object far from mirror 3)Either close or far 4)You can’t d i is negative! f is negative d o is positive Real image means d i > 0 Convex mirror: f < 0 Object in front of mirror: d o > 0

30 Mirror Summary Angle of incidence = Angle of Reflection Principal Rays –Parallel to P.A.: Reflects through focus –Through focus: Reflects parallel to P.A. –Through center: Reflects back on self |f| = R/2

31 Light Doesn’t Just Bounce It Also Refracts! Reflected: Bounces (Mirrors!) Refracted: Bends (Lenses!) ii rr 11 22 n2n2 n1n1  i =  r n 1 sin(  1 )= n 2 sin(  2 )

32 Speed of light in medium Index of refraction Speed of light in vacuum so Index of Refraction 186,000 miles/second: it’s not just a good idea, it’s the law! always!

33 Snell’s Law n1n1 n2n2 When light travels from one medium to another the speed changes v=c/n and the light bends n 1 sin(  1 )= n 2 sin(  2 ) 11 22 1) n 1 > n 2 2) n 1 = n 2 3) n 1 < n 2 Compare n 1 to n 2.

34 n1n1 n2n2 Snell’s Law Practice normal A ray of light traveling through the air (n=1) is incident on water (n=1.33). Part of the beam is reflected at an angle  r = 60. The other part of the beam is refracted. What is  2 ? 11 rr Usually, there is both reflection and refraction!

35 n1n1 n2n2 Snell’s Law Practice normal A ray of light traveling through the air (n=1) is incident on water (n=1.33). Part of the beam is reflected at an angle  r = 60. The other part of the beam is refracted. What is  2 ? sin(60) = 1.33 sin(  2 )  2 = 40.6 degrees  1 =  r =  11 rr Usually, there is both reflection and refraction!


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