2. Physics 1502: Lecture 29 Today’s Agenda Announcements: –Midterm 2: Monday Nov. 16 … –Homework 08: due Friday Optics –Index of Refraction

4. Prisms 11 22 Entering For air/glass interface, we use n(air)=1, n(glass)=n 33 44 Exiting

5. LIKE SO!In second rainbow pattern is reversed

8. Total Internal Reflection –Consider light moving from glass (n 1 =1.5) to air (n 2 =1.0) ie light is bent away from the normal. as  1 gets bigger,  2 gets bigger, but  2 can never get bigger than 90  !! In general, if sin  1  sin  C  (n 2 / n 1 ), we have NO refracted ray; we have TOTAL INTERNAL REFLECTION. For example, light in water which is incident on an air surface with angle  1 >  c = sin -1 (1.0/1.5) = 41.8  will be totally reflected. This property is the basis for the optical fibers used in communication. incident ray reflected ray refracted ray 22 11 rr GLASS AIR n2n2 n1n1 2

9. ACT 1: Critical Angle... air n =1.00 glass n =1.5 air n =1.00 cc water n =1.33 glass n =1.5 water n =1.33 cc Case I Case II An optical fiber is cladded by another dielectric. In case I this is water, with an index of refraction of 1.33, while in case II this is air with an index of refraction of 1.00. Compare the critical angles for total internal reflection in these two cases a)  cI >  cII b)  cI =  cII c)  cI <  cII

10. ACT 2: Fiber Optics air n =1.00 glass n =1.5 air n =1.00 cc water n =1.33 glass n =1.5 water n =1.33 cc Case I Case II The same two fibers are used to transmit light from a laser in one room to an experiment in another. Which makes a better fiber, the one in water ( I ) or the one in air ( II ) ? a)  Water b)  Air

11. Problem You have a prism that from the side forms a triangle of sides 2cm x 2cm x 2  2cm, and has an index of refraction of 1.5. It is arranged (in air) so that one 2cm side is parallel to the ground, and the other to the left. You direct a laser beam into the prism from the left. At the first interaction with the prism surface, all of the ray is transmitted into the prism. a)Draw a diagram indicating what happens to the ray at the second and third interaction with the prism surface. Include all reflected and transmitted rays. Indicate the relevant angles. b)Repeat the problem for a prism that is arranged identically but submerged in water.

12. Solution At the first interface  =0 o, no deflection of initial light direction. At 2nd interface  =45 o, from glas to air ? Critical angle: sin(  c )=1.0/1.5 =>  c = 41.8 o < 45 o Thus, at 2nd interface light undergoes total internal reflection At 3rd interface  =0 o, again no deflection of the light beam A) Prism in air B) Prism in water (n=1.33) At the first interface  =0 o, the same situation. At 2nd interface now the critical angle: sin(  c )=1.33/1.5 =>  c = 62 o > 45 o Now at 2nd interface some light is refracted out the prism n 1 sin(  1 ) = n 2 sin(  2 ) => at  2 = 52.9 o Some light is still reflected, as in A) ! At 3rd interface  =0 o, the same as A)

13. o i f h’ h R   h o-R R-i o i &

14. Nothing New! For the next few lectures we will be studying geometric optics. You should be comforted by the fact that you already know the underlying fundamentals of what is going on. –Namely, you know how light propagates in situations in which the length scales are much greater than the light’s wavelength. incident ray reflected ray 11 rr n1n1 refracted ray 22 n2n2 We will use these laws to understand the properties of mirrors (perfect reflection) and lenses (perfect refraction). We will also discover properties of combinations of lenses which will allow us to understand such applications as microscopes, telescopes and eyeglasses. –Reflection: –Refraction:

15. Plane Mirror Flat Mirror Object Virtual Image     i o REAL VIRTUAL o = -i

16. Flat Mirror Images of Extended Objects Plane Mirror Extended Object Virtual Image i o REAL VIRTUAL o = -i h h’ Magnification: M = h’/ h = 1

19. Multiple Reflection 180 o MIRROR 2 MIRROR 1 Object Image 6 6 Image 4 4 Image 2 2 Image 1 1 Image 3 3 Image 5 5

20. Lecture 29, ACT 1 Let’s now consider a curved mirror. We start with CONVEX mirror. –Where do the rays which are reflected from the convex mirror shown intersect? (a) to left of (b) to right of (c) they don’t intersect 7A

21. Lecture 29, ACT 2 What is the nature of the image of the arrow? (a) Inverted and in front of the mirror (b) Inverted and in back of the mirror (c) Upright and in back of the mirror 7B

22. Concave Spherical Mirrors We start by considering the reflections from a concave spherical mirror in the paraxial approximation (ie small angles of incidence close to a single axis): First draw a ray (light blue) from the tip of the arrow through the center of the sphere. This ray is reflected straight back since the angle of incidence = 0.   Note that the two rays intersect in a point, suggesting an inverted image. To check this, draw another ray (green) which comes in at some angle  that is just right for the reflected ray to be parallel to the optical axis. R   Now draw a ray (white) from the tip of the arrow parallel to the axis. This ray is reflected with angle  as shown. Note that this ray intersects the other two at the same point, as it must if an image of the arrow is to be formed there. Note also that the green ray intersects the white ray at another point along the axis. We will call this point the focal point ( ).

25. The Mirror Equation We will now transform the geometric drawings into algebraic equations: R     object  h image o i from triangles, eliminating , Now we employ the small angle approximations: Plugging these back into the above equation relating the angles, we get: Defining the focal length f = R/2, This eqn is known as the mirror eqn. Note that there is no mention of  in this equation. Therefore, this eqn works for all , ie we have an image!

26. Magnification We have derived the mirror eqn which determines the image distance in terms of the object distance and the focal length: What about the size of the image? How is h’ related to h?? From similar triangles: Now, we can introduce a sign convention. We can indicate that this image is inverted if we define its magnification M as the negative number given by:   R h o h’ i

27. More Sign Conventions Consider an object distance s which is less than the focal length:   h’ i Ray Trace: Ray through the center of the sphere (light blue) is reflected straight back. R h o f We call this a virtual image, meaning that no light from the object passes through the image point. Proof left to student: This situation is described by the same mirror equations as long as we take the convention that images behind the mirror have negative image distances s’. ie: In this case, i 0, indicating that the image is virtual (i 0). Ray parallel to axis (red) passes through focal point f. These rays diverge! ie these rays look they are coming from a point behind the mirror.

28. Concave-Planar-Convex What happens as we change the curvature of the mirror? –Plane mirror: »R =  IMAGE: virtual upright (non-inverted) h’  h  o i f IMAGE: virtual upright (non-inverted) –Convex mirror: »R < 0

29. Lecture 29, ACT 3 In order for a real object to create a real, inverted enlarged image, a) we must use a concave mirror. b) we must use a convex mirror. c) neither a concave nor a convex mirror can produce this image.

33. Mirror – Lens Definitions Some important terminology we introduced last class, –o = distance from object to mirror (or lens) – i = distance from mirror to image o positive, i positive if on same side of mirror as o. –R = radius of curvature of spherical mirror –f = focal length, = R/2 for spherical mirrors. –Concave, Convex, and Spherical mirrors. –M = magnification, (size of image) / (size of object) negative means inverted image R     object  h image o i