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Ex. 26.2 A concave mirror has a 30 cm radius of curvature. If an object is placed 10 cm from the mirror, where will the image be found? f = R/2 = 15 cm,

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Presentation on theme: "Ex. 26.2 A concave mirror has a 30 cm radius of curvature. If an object is placed 10 cm from the mirror, where will the image be found? f = R/2 = 15 cm,"— Presentation transcript:

1 Ex. 26.2 A concave mirror has a 30 cm radius of curvature. If an object is placed 10 cm from the mirror, where will the image be found? f = R/2 = 15 cm, p = 10 cm 1/p + 1/q = 1/f  1/10 + 1/q = 1/15 3/30 + 1/q = 2/30 1/q = -1/30 q = -30 cm Real or Virtual Magnified or Reduced Up-right or Upside-down q < 0 M = -q/p = 3 Case 5: p < f

2 Q. An upright image that is one-half as large as an object is needed to be formed on a screen in a laboratory experiment using only a concave mirror with 1 m radius of curvature. If you can make this image, I will give you $10. If you can’t you should pay me $10. Deal or no deal? Why? 1/p + 1/q = 1/f = 2/R > 0 M = -q/p = ½ > 0 should be a real image: q > 0 M = -q/p cannot be positive, if q > 0. No deal!!!

3 Refraction and Lenses Optical illusion: the pencil is not bent at the air-water boundary. caused by non-trivial passage of light rays.

4 Refraction Details, 1 Light may refract into a material where its speed is lower Light may refract into a material where its speed is lower The angle of refraction is less than the angle of incidence The angle of refraction is less than the angle of incidence The ray bends toward the normalThe ray bends toward the normal

5 Refraction Details, 2 Light may refract into a material where its speed is higher Light may refract into a material where its speed is higher The angle of refraction is greater than the angle of incidence The angle of refraction is greater than the angle of incidence The ray bends away from the normalThe ray bends away from the normal

6 The Index of Refraction When light passes from one medium to another, it is refracted because the speed of light is different in the two media When light passes from one medium to another, it is refracted because the speed of light is different in the two media The index of refraction, n, of a medium can be defined The index of refraction, n, of a medium can be defined

7 Index of Refraction, cont For a vacuum, n = 1 For a vacuum, n = 1 For other media, n > 1 For other media, n > 1 n is a unitless ratio n is a unitless ratio

8 Frequency Between Media As light travels from one medium to another, its frequency does not change As light travels from one medium to another, its frequency does not change Both the wave speed and the wavelength do changeBoth the wave speed and the wavelength do change The wavefronts do not pile up, nor are created or destroyed at the boundary, so ƒ must stay the sameThe wavefronts do not pile up, nor are created or destroyed at the boundary, so ƒ must stay the same

9 Index of Refraction Extended The frequency stays the same as the wave travels from one medium to the other The frequency stays the same as the wave travels from one medium to the other v = ƒ λ v = ƒ λ The ratio of the indices of refraction of the two media can be expressed as various ratios The ratio of the indices of refraction of the two media can be expressed as various ratios

10 Snell’s Law 11 11 22 v: speed of light in a medium n = c/v : index of refraction v 1 = c/n 1, v 2 = c/n 2 n 1 sin  1 = n 2 sin  2 All three beams (incident, reflected, and refracted) are in one plane. n > 1

11 material n = c/v for = 589 nm vacuum1.00 air1.00029 water1.33 ice1.31 typical glass 1.52 polycarbonate1.59 diamond2.42 Index of Refraction n depends on. Dispersion

12 water 11 11 22  1 >  2

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14 Total Internal Reflection Total internal reflection can occur when light attempts to move from a medium with a high index of refraction to one with a lower index of refraction Total internal reflection can occur when light attempts to move from a medium with a high index of refraction to one with a lower index of refraction Ray 5 shows internal reflectionRay 5 shows internal reflection

15 Critical Angle A particular angle of incidence will result in an angle of refraction of 90° A particular angle of incidence will result in an angle of refraction of 90° This angle of incidence is called the critical angleThis angle of incidence is called the critical angle

16 Critical Angle, cont For angles of incidence greater than the critical angle, the beam is entirely reflected at the boundary For angles of incidence greater than the critical angle, the beam is entirely reflected at the boundary This ray obeys the Law of Reflection at the boundaryThis ray obeys the Law of Reflection at the boundary Total internal reflection occurs only when light attempts to move from a medium of higher index of refraction to a medium of lower index of refraction Total internal reflection occurs only when light attempts to move from a medium of higher index of refraction to a medium of lower index of refraction

17 Total internal reflection n 1 (> n 2 ) n2n2 11 22 n 1 sin( 1 ) = n 2 sin( 2 ) Total internal reflection when  2 = 90 sin( c ) = n 2 /n 1  c : critical angle < 1

18 How could fish survive from spear fishing? Fish vision  f = 2 c  c = sin -1 (1/1.33) = 49

19 n core n clad >

20 Q. What is the critical angle for a glass to air surface if the Index of refraction for glass is 1.5. sin  c = n a /n g = 1.0/1.5 = 0.667  c = 42

21 11 11 22 water air 1()1()1()1() 2()2()2()2() 107.5 2014.9 3022.0 4028.8 5035.0 6040.5 7044.8 8047.6 sin 1 /sin 2 1.33 1.33 1.33 1.33 1.33 1.33 1.33 1.33 v 1 = c/n 1, v 2 = c/n 2 n 1 sin  1 = n 2 sin  2

22 A fish swims below the surface of the water. Suppose an observer is looking at the fish straight above the fish. The observer sees 1. the fish at a greater depth than it really is. 2. the fish at the same depth. 3. the fish at a smaller depth than it really is. 4. no fish due to total internal reflection.

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24 Q.There are three layers of different media as shown in the figure. A beam of light bends as shown in the figure when it passes through the media. What can we say about the materials? I II III n I sin  I = n II sin  II n II sin  II = n III sin  III  II >  I  n I > n II  III >  II  n II > n III n I > n II > n III

25 wavelength (nm) n 361 (near UV) 1.539 434 (dark blue) 1.528 486 (green) 1.523 589 (yellow) 1.517 656 (red) 1.514 768 (dark red) 1.511 1200 (IR) 1.505 2000 (far IR) 1.497 Dispersion of Index of Refraction for Glass

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28 In glass n (red) ≈ 1.51 n (purple) ≈ 1.53

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30 http://science.howstuffworks.com/


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