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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?. Case 5: p < f. 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. q < 0. Real or Virtual
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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? Case 5: p < f 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 q < 0 Real or Virtual Magnified or Reduced Up-right or Upside-down M = -q/p = 3
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!!!
Refraction and Lenses Optical illusion:the pencil is not bent at the air-water boundary. caused by non-trivial passage of light rays.
Refraction Details, 1 • Light may refract into a material where its speed is lower • The angle of refraction is less than the angle of incidence • The ray bends toward the normal
Refraction Details, 2 • Light may refract into a material where its speed is higher • The angle of refraction is greater than the angle of incidence • The ray bends away from the normal
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 • The index of refraction, n, of a medium can be defined
Index of Refraction, cont • For a vacuum, n = 1 • For other media, n > 1 • n is a unitless ratio
Frequency Between Media • As light travels from one medium to another, its frequency does not change • Both 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 same
Index of Refraction Extended • The frequency stays the same as the wave travels from one medium to the other • v = ƒ λ • The ratio of the indices of refraction of the two media can be expressed as various ratios
q1 q1 n = c/v :index of refraction v: speed of light in a medium q2 v1 = c/n1, v2 = c/n2 n1sinq1 = n2sinq2 Snell’s Law All three beams (incident, reflected, and refracted) are in one plane. n > 1
Index of Refraction n depends on l. Dispersion
q1 q1 q1> q2 q2 water
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 • Ray 5 shows internal reflection
Critical Angle • A particular angle of incidence will result in an angle of refraction of 90° • This angle of incidence is called the critical angle
Critical Angle, cont • 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 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 q2 n2 n1 (> n2) q1 n1sin(q1) = n2sin(q2) Total internal reflection when q2 = 90 sin(qc) = n2/n1 < 1 qc: critical angle
How could fish survive from spear fishing? Fish vision qf = 2qc qc = sin-1(1/1.33) = 49
Q. What is the critical angle for a glass to air surface if the Index of refraction for glass is 1.5. sinqc = na/ng = 1.0/1.5 = 0.667 qc = 42
q1 q1 q2 v1 = c/n1, v2 = c/n2 n1sinq1 = n2sinq2 air water
0 A fish swims below the surface of the water. Suppose an observer is looking at the fish straight above the fish. The observer sees • the fish at a greater depth than it really is. • the fish at the same depth. • the fish at a smaller depth than it really is. • no fish due to total internal reflection.
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? nIsinqI = nIIsinqII qII > qI nI > nII I nIIsinqII = nIIIsinqIII II qIII > qII nII > nIII III nI > nII > nIII
In glass n (red) ≈ 1.51 n (purple) ≈ 1.53