1. Chapter 14 Refraction of Light Hr Physics

2. Sorry this is late.

5. Snell's Law n i sinΘ i = n r sinΘ r index of refraction of first medium x sine of the angle of incidence = index of refraction of second medium x sine of the angle of refraction The index of refraction will vary for different wavelengths (Table 1 gives values for wavelength 589 nm). This is how a prism works.

6. Snell's Law Pracice Sunlight passes into a raindrop at an angle of 22.5˚ from the normal at one point on the droplet. What is the angle of refraction? (Hint: see Table 1 p.490)

7. Snell's Law Practice 16.7˚

8. Snell's Law Practice For an incoming ray of light in a vacuum wavelength 589 nm, fill in the unknown values: from(medium)to(medium)Θ i Θ r a. flint glasscrown glass25.0˚ ? b. air?14.5˚ 9.80˚ c. airdiamond31.6˚ ?

9. Snell's Law Practice a. 27.5˚ b. glycerine c. 12.5˚

10. Snell's Law Practice A ray of transparent light of vacuum wavelength 550 nm traveling in air enters a slab of transparent material. The incoming ray makes an angle of 40.0˚ with the normal, and the refracted ray makes an angle of 26.0˚ with the normal. Find the index of refraction of the transparent material. (Assume the index of refraction of air for light of wavelength 550 nm is 1.00).

14. Total Internal Reflection Total internal reflection occurs for all angles greater than the critical angle. We can find the critical angle by sin Θ c = n r n i n i

15. Critical Angle Practice Problems Glycerine is used to make soap and other personal care products. Find the critical angle for light traveling from glycerine (n = 1.473) into air.

16. Critical Angle Practice Problems 42.8˚

22. Solving Lens Problems As with mirrors M = h' = -d i h d r M is positive – image is upright and virtual M is negative – image is real and inverted

25. Sign Conventions forLenses +- preal objectreal object in front ofin back of the lensthe lens qreal imagevirtual image in back of in front of the lensthe lens fconverging diverging

26. Lenses Practice Problems An object is placed 20.0 cm in front of a converging lens of focal length 10.0 cm. Find the image distance and the magnification. Describe the image.

27. Lenses Practice Problems 20.0 cm, M = -1.00; real, inverted image

28. Lenses Practice Problems Sherlock Holmes examines a clue by holding his magnifying glass at arm's length and 10.0 cm away from an object. The magnifying glass has a focal length of 15.0 cm. Find the image distance and the magnification. Describe the image that he observes.

29. Lenses Practice Problems -30.0 cm, M = 3.00; virtual, upright image

30. Lenses Practice Problems An object is placed 20.0 cm in front of a diverging lens of focal length 10.0 cm. Find the image distance and the magnification. Describe the image.

31. Lenses Practice Problems -6.67 cm, M = 0.333; virtual, upright image