1. 4.Example: Multiple mirrors and virtual images: how far away are the virtual images? 1m3m VII. Mirrors

2. 4.Example: Multiple mirrors and virtual images: how far away are the virtual images? 2m3m

3. 4.Example: Multiple mirrors and virtual images: how far away are the virtual images? 6m2m

4. 4.Example: Multiple mirrors and virtual images: how far away are the virtual images? 6m2m8m

5. 4.Example: Multiple mirrors and virtual images: how far away are the virtual images? 2m8m10m16m 18m

6. d) Example: A telescope is designed to take images of distant galaxies. If the diameter of the telescope’s spherical mirror is 40 cm, where should a detector be placed? *Step 1: Stars ~ infinitely far away *Step 2: Use mirror equation: 1/s + 1/s’ = 2/R => 0 + 1/f = 2/R, so f = R/2 = 10 cm. VII. Mirrors

7. A.Refraction at a spherical surface 1.Images a)Object distance, Image distance, and Radius of curvature s n 1 < n 2 s’ 11 22 R    r  Exterior angle of a triangle:  1 =   =    VIII. Lenses

8. A.Refraction at a spherical surface 1.Images a)Object distance, Image distance, and Radius of curvature s n 1 < n 2 s’ 11 22 R    r  Snell’s Law: n 1 sin  1 =n 2 sin    VIII. Lenses

9. A.Refraction at a spherical surface 1.Images a)Object distance, Image distance, and Radius of curvature s n 1 < n 2 s’ 11 22 R    r  Tangents: tan  = r/(s +  ). tan  = r/(s’ -  ). tan  = r/(R -  ). VIII. Lenses

10. A.Refraction at a spherical surface 1.Images a)Object distance, Image distance, and Radius of curvature s n 1 < n 2 s’ 11 22 R    r  Small angle approximation: tan  ~  (  << 1). sin  ~  VIII. Lenses

11. A.Refraction at a spherical surface 1.Images a)Object distance, Image distance, and Radius of curvature s n 1 < n 2 s’ 11 22 R    r Rewrite Snell’s Law: n 1  1 ~ n 2  2, so  2 = (n 1 /n 2 )(  ). VIII. Lenses

12. A.Refraction at a spherical surface 1.Images a)Object distance, Image distance, and Radius of curvature s n 1 < n 2 s’ 11 22 R    r * Use exterior law: (n 1  n 2  ) = (n 2 - n 1 )  VIII. Lenses

13. A.Refraction at a spherical surface 1.Images a)Object distance, Image distance, and Radius of curvature s n 1 < n 2 s’ 11 22 R    r * Small angles: tan  ~  ~ r/s (  << s). tan  ~  ~ r/s’ (  <<s’). tan  ~  ~ r/R (  << R). VIII. Lenses

14. *Put it all together! n 1 /s + n 2 /s’ = (n 2 - n 1 )/R.(VIII.A.2) A.Refraction at a spherical surface 1.Images a)Object distance, Image distance, and Radius of curvature s n 1 < n 2 s’ 11 22 R    r VIII. Lenses

15. A.Refraction at a spherical surface 1.Images b)Magnification s n 1 < n 2 s’ 11 22 h’ tan  1 = h/s. tan  2 = -h’/s’. M = h’/h = -s’tan  2 /(stan  1 ). h VIII. Lenses

16. * Use Snell’s Law & small angle M = h’/h ~ -s’  2 /(s  1 ) = -n 1 s’/(n 2 s). (VIII.A.3) A.Refraction at a spherical surface 1.Images b)Magnification s n 1 < n 2 s’ 11 22 h’ h VIII. Lenses

17. 3.Example: An insect is centrally embedded in a spherical globule of amber. The index of refraction for the amber is 2 and the diameter of the globule is 4 cm. How far from the surface does the insect appear? n 1 /s + n 2 /s’ = (n 2 -n 1 )/R; 2/(2 cm) + (1/s’) = (-1)1/(2 cm); 1/s’ = 1/(4 cm) - 1/(1 cm) = -3/4 cm -1. s’ = -4/3 = -1.333 cm: virtual image VIII. Lenses Magnification: M = -n 1 s/n 2 s’ = -2*(-4/3)/1*2 = 4/3

18. B.Refraction at a flat surface 1. Let R => infinity. Sphere => plane n 1 /s = -n 2 /s’, or s’ = -(n 2 /n 1 )s.(VIII.B.1) n 1 > n 2. VIII. Lenses

19. B.Refraction at a flat surface 2.Example: A frog sits at the bottom of a reflecting pool. The pool is.5 m deep with n = 4/3. Where does the frog appear? s’ = -(n 2 /n 1 )s = -(1/[4/3])(0.5 m) = 3/8 m. n 1 > n 2. VIII. Lenses

20. C.Thin Lenses 2.Ray tracing: single lens VIII. Lenses

21. C.Thin Lenses 2.Ray tracing: double lens VIII. Lenses

22. C.Thin Lenses 6.Example: What is the image distance for a thin glass lens with a focal length of 4/3 cm and an object at 10 cm? 1/s’ =1/f - 1/s = 1/(4/3 cm) - 1/10cm =.65 cm -1. Thus, s’ = +1.5 cm. How is the image magnified? M = -s’/s = -1.5/10; M = -.15 VIII. Lenses

23. 7.Example: What is the image distance and magnification for the following pair of lenses: Lens 1: R 1 = infinity, R 2 = -10 cm, n = 1.5 Lens 2: R 1 = 10 cm, R 2 = -10 cm, n = 1.5 The lenses are separated by 50 cm, and the object is at a distance of 45 cm from the first lens. Step 1: Find the focal lengths 1/f1 = (1/2)(0 - 1/(-10cm) => f1 = 20 cm. 1/f2 = (1/2)(1/10cm - 1/(-10cm)) => f2 = 10 cm. VIII. Lenses

24. 7.Example:What is the image distance and magnification for the following pair of lenses: Lens 1: R 1 = infinity, R 2 = -10 cm, n = 1.5 Lens 2: R 1 = 10 cm, R 2 = -10 cm, n = 1.5 The lenses are separated by 50 cm, and the object is at a distance of 45 cm from the first lens. Step 2: Find image distance for lens 1: 1/s’ 1 = 1/f1 - 1/s 1 = 1/20cm - 1/45cm => s’ 1 = 36 cm. VIII. Lenses

25. 7.Example:What is the image distance and magnification for the following pair of lenses: Lens 1: R 1 = infinity, R 2 = -10 cm, n = 1.5 Lens 2: R 1 = 10 cm, R 2 = -10 cm, n = 1.5 The lenses are separated by 50 cm, and the object is at a distance of 45 cm from the first lens. Step 3: Find image magnification for lens 1: M1 = -s’ 1 = -36 cm/45 cm = -0.8 VIII. Lenses

26. 7.Example: What is the image distance and magnification for the following pair of lenses: Lens 1: R 1 = infinity, R 2 = -10 cm, n = 1.5 Lens 2: R 1 = 10 cm, R 2 = -10 cm, n = 1.5 Step 4: Find image distance for lens 2 using image 1 as object: 1/s’ 2 = 1/10 cm - 1/14 cm => s’ 2 = 35 cm. The image is to the right of the second lens. VIII. Lenses

27. 7.Example:What is the image distance and magnification for the following pair of lenses: Lens 1: R 1 = infinity, R 2 = -10 cm, n = 1.5 Lens 2: R 1 = 10 cm, R 2 = -10 cm, n = 1.5 Step 5: Find total magnification M2 = -s’ 2 /s 2 = -35 cm/14 cm = -2.5. Total magnification: M = M1* M2 = (-0.8)*(-2.5) = 2 VIII. Lenses