When Light Encounters Matter

1. When Light Encounters Matter Electromagnetic waves are disturbances that transmit energy, just like other waves. In EM waves, the disturbance is to space itself, and the oscillators that do the original “disturbing” (and the oscillators at the other end that are ultimately “disturbed”) are charged particles—matter. And that matter can respond in several ways when receiving the light: It can reflect it (absorb, then re-emit it) back from the incident surface. This can be diffuse (scattered) from a rough surface or specular (mirror-like) from a smooth surface. It can refract it (absorb, then re-emit repeatedly) through the material. This can be diffuse (scattered) transmission through a translucent material or specular transmission through a transparent material. It can absorb the light altogether—as thermal energy. This is an opaque material. OSU PH 212, Before Class #14

2. Figure 23.9 OSU PH 212, Before Class #14

3. OSU PH 212, Before Class #14

4. “Slowed” (Delayed) Light: The Index of Refraction The speed of light, c (≈ 3.00 x 108 m/s), is its speed of direct travel in a vacuum. When light refracts through matter, it’s essentially absorbed by each charged particle it encounters, then re-emitted. Each such absorption/re-emission takes some time—the particle has to “get up to speed”—vibrate with the same frequency as the light. Thus, any refractive (transparent) material—that’s any material whose molecules will re-transmit rather than absorb the disturbance—has a characteristic slower speed at which it will do this re-transmitting: v = c/n,where n is the index of refraction of that material. So n = 1 for a vacuum; all other n values are greater than 1. (The text lists some common materials and their refractive indexes.) OSU PH 212, Before Class #14

5. Snell’s Law This change in speed when light enters a new medium means that its wavefronts will bend, as follows: n1sin1 = n2sin2 1 = angle of incidence; 2 = angle of refraction If n2 > n1, light bends toward the normal. If n2 < n1, light bends away from the normal. Implications: Apparent depths/distances Displacements by a transparent slab Total internal reflection (gems, fiber optics) Dispersion (prisms and rainbows) Lenses OSU PH 212, Before Class #14

6. A light ray goes from air to water. The angle of incidence is 45o. What is the angle of refraction? • A light ray goes from water to air. The angle of incidence is 45o. What is the angle of refraction? • Again, to generalize (fill in the blanks here): • - If the index increases (across the boundary) • than the angle gets __________. • - If the index decreases (across the boundary) • than the angle gets __________. OSU PH 212, Before Class #14

7. A light ray goes from air to water. The angle of incidence is 45o. What is the angle of refraction? • A light ray goes from water to air. The angle of incidence is 45o. What is the angle of refraction? • Again, to generalize (fill in the blanks here): • - If the index increases (across the boundary) • than the angle gets smaller. • - If the index decreases (across the boundary) • than the angle gets larger. OSU PH 212, Before Class #14

8. Practice: A ray of light traveling through air (n ≈ 1.00) strikes an aquarium (nglass = 1.500; nwater = 1.333) at an angle of incidence of 60°. At what angle would the ray enter the water after passing through the glass? At what angle would the ray re-emerge into the air on the other side? OSU PH 212, Before Class #14

9. Practice: A ray of light traveling through air (n ≈ 1.00) strikes an aquarium (nglass = 1.500; nwater = 1.333) at an angle of incidence of 60°. At what angle would the ray enter the water after passing through the glass? Air to glass: nairsin(60°) = nglasssin(glass) Glass to water: nglasssin(glass) = nwatersin(water) So: nairsin(60°) = nwatersin(water) or: water = sin-1[(nair/nwater)sin(60°)] = 40.5° At what angle would the ray re-emerge into the air on the other side? Water to glass: nwatersin(40.5°) = nglasssin(glass) Glass to air: nglasssin(glass) = nairsin(air) So: nwatersin(40.5°) = nairsin(air) or: air = sin-1[(nwater/nair)sin(40.5°)] = 60.0° OSU PH 212, Before Class #14

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11. OSU PH 212, Before Class #14

12. Total Internal Reflection Suppose a ray of light is traveling through water and strikes the surface (ready to emerge into air) at an angle of 50°. At what angle does it emerge? Use Snell’s Law—if you can.… When light is going from a more refractive to a less refractive medium (so n1 > n2), any angle of incidence greater than a certain critical angle results in total internal reflection: sinc = n2/n1 In other words: The critical angle is the incidence angle that produces a refraction angle of 90°. Among other useful results, total internal reflection makes possible the sparkle of gem-stones and the transmission of light via fiber optic cables. Read pages 969-970 and study the figures. OSU PH 212, Before Class #14

13. Dispersion: Refraction Depends on Wavelength As it turns out, the speed of transmission of light waves through a transparent substance depends on the wave-length being transmitted. The shorter wavelengths (higher frequencies and energies) are slowed a little more than the longer wavelengths (lower frequencies and energies). It simply takes a little more time to get those re-transmitting molecules vibrating faster (i.e. higher frequency). Thus, with visible light, for example, n is higher for blue light than red light. So blue light is bent slightly more when traveling between substances with differing refractive indexes. This produces the effects we know for prisms (and rainbows use both dispersion and internal reflection). OSU PH 212, Before Class #14

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