1. 2/5/16Oregon State University PH 212, Class #151 Snell’s Law This change in speed when light enters a new medium means that its wavefronts will bend, as follows: n 1 sin  1 = n 2 sin  2  1 = angle of incidence;  2 = angle of refraction If n 2 > n 1, light bends toward the normal. If n 2 < n 1, 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

2. 2/5/16Oregon State University PH 212, Class #152 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).

3. 2/5/163Oregon State University PH 212, Class #15

4. Figure 23.29A 2/5/164Oregon State University PH 212, Class #15

5. Figure 23.29B 2/5/165Oregon State University PH 212, Class #15

6. 2/5/16Oregon State University PH 212, Class #156 Harnessing Refraction: Lenses A simple slab of glass or other refractive material will transmit a ray of light (air-to-glass-to-air) with just a slight displacement—but with no “bending” of the ray from start to finish. That is, the first angle of incidence is equal to the final angle of refraction. With any shape other than a slab (i.e. other than two parallel surfaces for the ray to enter and exit), the above is not true. We use this fact and build various combinations of non-parallel surfaces— lenses—in order to bend light in useful ways:

7. Converging vs. Diverging Lenses Converging lens: Incident parallel rays converge at a focal point on the refractive (right) side. If designed to be used in air (or any medium whose n value is less than that of the lens material), a converging lens is thicker in the center than at the edges. Diverging lens: Incident parallel rays diverge, so they appear to originate at a focal point on the incident (left) side. If designed to be used in air (or any medium whose n value is less than that of the lens material), a diverging lens is thinner in the center than at the edges. The focal length is the distance from the lens to the focal point, as measured along the central axis. 2/5/167Oregon State University PH 212, Class #15

8. Figure 23.34 2/5/168Oregon State University PH 212, Class #15

9. 2/5/16Oregon State University PH 212, Class #159 Ray Tracing with Thin Lenses To see how a lens determines an image location, size and orientation (compared to the object), one easy way is to trace a few rays to see how their lines converge at the image location. Three such rays are especially useful: (1)From the object to the lens along a path parallel to the lens’s principal axis, then refracting to align with the focal point, F *. (2) From the object toward the focal point, F *, then refracting out parallel to the principal axis. (3) From the object, straight through the center of the thin lens. * For a converging lens, use the focal point on the left (incident) side; for a diverging lens, use the focal point on the right side.

10. Figure 23.41 2/5/1610Oregon State University PH 212, Class #15

11. 2/5/16Oregon State University PH 212, Class #1511 The Thin-Lens Equation The thin lens approximation holds for a lens whose thick- ness is small compared to its focal length. With that assumption, plus some geometry and algebra, we can use this calculation for lenses: 1/d 0 + 1/d i = 1/f Definitions and sign conventions: Object distance (d 0 ): + when object is left of the lens (real); Image distance (d i ): + if the image is right of the lens (real); – if the image is left of the lens (virtual). Focal length (f): + for a converging lens. – for a diverging lens.

12. 2/5/16Oregon State University PH 212, Class #1512 The Magnification Equation for Lenses The magnification equation for lenses is: m = h i /h o = –d i /d o where h o is the object height and h i is the image height (negative if the image is inverted with respect to the positive object height). The meaning of the sign of m: Assuming the object is upright: m is + if the image is upright. m is – if the image is inverted.

13. 2/5/16Oregon State University PH 212, Class #1513 1/d o + 1/d i = 1/f m = h i /h o = –d i /d o Practice: A lens has a focal length of –32 cm. An object is placed 19 cm in front of it (to its left). a.Calculate the image distance and magnification. b.Describe the image. (Real/virtual? Upright/inverted? Enlarged/reduced?)

14. 2/5/16Oregon State University PH 212, Class #1514 1/d o + 1/d i = 1/f m = h i /h o = –d i /d o Practice: A lens has a focal length of –32 cm. An object is placed 19 cm in front of it (to its left). a.Calculate the image distance and magnification. b.Describe the image. (Real/virtual? Upright/inverted? Enlarged/reduced?) d o = 19, and f = –32, so:1/d i = 1/f – 1/d o Thus: d i = (d o )(f)/(d o – f) = (19)(–32)/(19 - –32) = –11.9 cm The image distance is negative, so the image is virtual, upright and reduced—indeed, the only sort of image that a diverging lens (i.e. a lens with a negative f) can produce. Verify this with the magnification eqn.: m = –d i /d o = –(–11.9/19) = (+)0.627