θ =θ i r n sinθ = n sinθ

Transcription

1 θ i = θ r n = 1 sinθ1 n2 sin θ 2

2 Index of Refraction Speed of light, c, in vacuum is 3x10 8 m/s Speed of light, v, in different medium can be v < c. index of refraction, n = c/v. frequency, f, does not change in wave eqn. of v = f λ, wavelength, λ, depends on medium, λ = v/f = c/nf = λ 0 /n In some media, n, depends on f, this is called dispersion.

3 Waves, wave front, rays (Y&F section 33.2) Plane waves moving in +x direction r E r B ( x, y, z, t) = E0 yˆ cos( kx ωt) ( x, y, z, t) = B z ( kx ωt) 0 ˆ cos Wave front is a surface of constant phase E field Wave front (y-z plane surface of constant phase) y Ray propagation z x

4 Huygen s Principle (Y&F section 33.7) Huygen s principle; a wave front can be a source of secondary wavelets the spread out in all directions at the speed of propagation in the medium. The envelope of leading edges forms a wave front. This principle was stated by Huygen in 1678, it is derivable from Maxwell s eqn. It is a geometrical description of ray propagation. vt Plane wave example; Secondary wavelets create another wave front (plane)

5 Reflection from Huygen s Principle Consider wave fronts, separated by vt, the incident wave fronts in contact with the surface will create a wavelets according to Huygen s Principle and leds to another reflected wave front. Result is θ i = θ r reflected wave front Incident wave front vt vt vt ray diagram vt θ r θ i θ i θ r θ i = θ r Reflection Law

6 Snell s Law (law of refraction) Incident ray θ a normal Wave front Medium a, v a =c/n a Medium b, v b =c/n b θ b Refracted Wave front θ b θ a Refracted ray Angles are usually given relative to normal to plane n a sinθ a = n b sinθ b Snell s Law

7 Refraction from Huygen s Principle Now the speed changes, from medium a to medium b, so the Speed may change and the wavefront spacing differs. L sinθ a = v a t = ct/n a L sinθ b = v b t= ct/n b Medium a, v a =c/n a Medium b, v b =c/n b θ b L v b t v b t Wave front L v a t θ a va t Wave front n a sinθ a = n b sinθ b Snell s Law

8 Preflight 24 2) A ray of light passes from into water with an angle of incidence of 30 o. Which of the following quantities does not change as the light enters the water. Mark all correct answers. a) Wavelength b) frequency c) speed of propagation d) direction of propagation.

9 EXAMPLE, from into glass; θ a Suppose we have light in (n=1) incidence on glass (n=1.55) at an angle θ a =45 deg. What is the angle of the refracted light, θ b? θ b n sinθ = n (1)sin ( 45 ) = (1)sin sinθb = 1.55 θ = 27 b a glass (1.55)sinθ ( 45 ) sinθ = b b

10 EXAMPLE, from glass into ; Suppose we have light in glass (n=1.55) incident into at an angle θ a =30 deg. What is the angle of the refracted light, θ b? n glass sinθ = n (1.55)sin sinθ θ b b = = 51 a ( 30 ) = (1.55)sin sinθ (1)sinθ ( 45 ) b b = θ a θ b

11 Lecture 24, ACT 1 Which of the following ray diagrams could represent the passage of light from through glass and back to? (n =1 and n glass =1.5) (a) (b) (c) glass glass glass

12 Lecture 24, ACT 1 Which of the following ray diagrams could represent the passage of light from through glass and back to? (n =1 and n glass =1.5) (a) (b) (c) glass θ 1 θ 2 glass glass The behavior of these rays is determined from Snell s Law: n = 1 sinθ1 n2 sin Since n(glass) > n(), sinθ (glass) < sinθ (). Therefore, moving from to glass, ray will bend toward normal. this eliminates (a). Moving from glass to, ray will bend away from normal. this eliminates (c). As a matter of fact, the final angle in must be equal to the initial angle in!! θ 2

13 Suppose you are stranded on an island with no food. You see a fish in the water. Where should you aim your spear to hit the fish? ANSWER; do not aim directly at the apparent position of the fish. Aim at the inside of the fish.

14 Suppose in the previous question instead of a spear you had a high power laser to simultaneously kill and cook the fish (in the water). Where should you aim the laser?? ANSWER; aim directly at apparent fish position as the laser beam will refract to the correct fish position.

15 Total Internal Reflection Consider light moving from glass (n 1 =1.5) to (n 2 =1.0) n 1 incident ray θ 1 θ r reflected ray GLASS sinθ sinθ 2 1 > 1 = n n 2 1 θ 2 > θ 1 n 2 θ 2 refracted ray AIR I.e., light is bent away from the normal. as θ 1 gets bigger, θ 2 gets bigger, but θ 2 can never get bigger than 90!! 2 In general, if sin θ 1 > (n 2 / n 1 ), we have NO refracted ray; we have TOTAL INTERNAL REFLECTION. For example, light in water which is incident on an surface with angle θ 1 > θ c = sin -1 (1.0/1.33) = 48.8 will be totally reflected. This property is the basis for the optical fiber communication.

16 Preflight 24 4) The path of light is bent as it passes from medium 1 to medium 2. Compare the indexes of refraction in the two mediums. a) n 1 > n 2 b) n 1 = n 2 Snell s Law: n 1 sinθ 1 = n 2 sinθ 2 Here, θ 2 >θ 1 implies n 2 < n 1 c) n 1 < n 2 5) A light ray travels in a medium with n 1 and completely reflects from the surface of a medium with n 2. The critical angle depends on: a) n 1 only b) n 2 only c) n 1 and n 2 Critical angle occurs when θ 2 = 90 o Therefore, sinθ critical = n 2 /n 1

17 Total Internal Reflection Total internal reflection occurs when θ>θ c and provides 100% reflection. This has better efficiency than silvered mirror. Examples of devices using Critical Angle Prism Binoculars Fiber Optics Fiber optics is extremely important for high speed Internet and digital data transfer at long distances. Many companies (Lucent) have laid fiber over long Distances to provide internet service.

18 I2 (academic) network of fiber connections 10 gigabit/sec connections

19 Lecture 24, ACT 2: Critical Angle An optical fiber is surrounded by another dielectric. In case I this is water, with an index of refraction of 1.33, while in case II this is with an index of refraction of Compare the critical angles for total internal reflection in these two cases Case I Case II θ c θ c water n =1.33 glass n =1.5 water n =1.33 n =1.00 glass n =1.5 n =1.00 a) θ ci >θ cii b) θ ci =θ cii c) θ ci <θ cii

20 Lecture 24, ACT 2: Critical Angle An optical fiber is surrounded by another dielectric. In case I this is water, with an index of refraction of 1.33, while in case II this is with an index of refraction of Compare the critical angles for total internal reflection in these two cases a) θ ci >θ cii b) θ ci =θ cii c) θ ci <θ cii n 2 n 1 n 1 >n 2 Case I Case II θ c θ c water n =1.33 glass n =1.5 water n =1.33 n =1.00 glass n =1.5 n =1.00 Since n 1 >n 2 TIR will occur for θ > critical angle. Snell s law says sinθ c =n 2 /n 1. If n 2 =1.0, then θ c is as small as it can be. So θ ci >θ cii.

21 Dispersion: n = n(ω) The index of refraction depends on frequency, due to the presence of resonant transition lines. For example, ultraviolet absorption bands in glass cause a rising index of refraction in the visible, i.e., n(higher ω) > n(lower ω): n red = 1.52 n blue = 1.53 Index of refraction 1.54 white light frequency ultraviolet absorption bands prism Split into Colors

22 Rainbows

23 Rainbows