1. Chapter 33 Continued Properties of Light Law of Reflection
Law of Refraction or Snell’s Law
Chromatic Dispersion
Brewsters Angle

3. Dispersion: Different wavelengths have different velocities and therefore different indices of refraction. This leads to different refractive angles for different wavelengths. Thus the light is dispersed. The frequency does not change when n changes.

4. Snells Law
Red

5. Snells Law
blue

6. Fiber Cable
Total Internal Reflection Simulator Halliday, Resnick, Walker: Fundamentals of Physics, 7th Edition - Student Companion Site

7. Why is light totally reflected inside a fiber optics cable
Why is light totally reflected inside a fiber optics cable? Internal reflection

8. Total Internal Reflection

9. What causes a Mirage eye sky 1.09 1.09 1.08 1.08 1.07 1.07 1.06
Index of refraction
1.06
Hot road causes gradient in the index of refraction that increases
as you increase the distance from the road

10. Inverse Mirage Bend

11. Chromatic Dispersion

12. How does a Rainbow get made?
Primary rainbow
Secondary rainbow
Supernumeraries

13. Polarization by Reflection: Brewsters Law

14. Snells Law Example
47. In the figure, a 2.00-m-long vertical pole extends from the bottom of a swimming pool to a point 50.0 cm above the water. What is the length of the shadow of the pole on the level bottom of the pool?
Consider a ray that grazes the top of the pole, as shown in the diagram below. Here 1 = 35o, l1 = 0.50 m, and l2 = 1.50 m.
The length of the shadow is x + L.
x is given by
x = l1tan1 = (0.50m)tan35o = 0.35 m. 2 1 l2 l1 L x
air
water
shadow
L is given by
L=l2tan 
Use Snells Law to find 

15. Calculation of L
According to the law of refraction, n2sin2 = n1sin1. We take n1 = 1 and n2 = 1.33
L is given by 2 1 l2 l1 L x
air
water
shadow
The length of the shadow is L+x.
L+x = 0.35m m = 1.07 m.

16. Lecture 14 Images Chapter 34
Geometrical Optics
Fermats Principle
Law of reflection
Law of Refraction
Plane Mirrors and Spherical Mirrors
Spherical refracting Surfaces
Thin Lenses
Optical Instruments
Magnifying Glass, Microscope, Refracting telescope
Polling Questions

17. Geometrical Optics:Study of reflection and refraction of light from surfaces
The ray approximation states that light travels in straight lines
until it is reflected or refracted and then travels in straight lines again.
The wavelength of light must be small compared to the size of
the objects or else diffractive effects occur.

18. Law of Reflection
Mirror B A 1

19. Fermat’s Principle JAVA APPLET
Using Fermat’s Principle you can prove the
Reflection law. It states that the path taken
by light when traveling from one point to
another is the path that takes the shortest
time compared to nearby paths.
JAVA APPLET
Show Fermat’s principle simulator

20. Two light rays 1 and 2 taking different paths between points A and B and reflecting off a vertical mirror 1 2 A B
Plane Mirror
Use calculus - method
of minimization

21. Write down time as a function of y
and set the derivative to 0.

23. Plane Mirrors Where is the image formed
Mirrors and Lenses
Plane Mirrors Where is the image formed

24. Plane mirrors Angle of incidence Virtual side Real side Normal
Virtual image
Angle of reflection
i = - p
eye
Object distance = - image distance
Image size = Object size

25. Problem: Two plane mirrors make an angle of 90o
Problem: Two plane mirrors make an angle of 90o. How many images are there for an object placed between them?
mirror
eye
object 2 3 1
mirror

26. Using the Law of Reflection to make a bank shot
Assuming no spin
Assuming an elastic collision
No cushion deformation d d
pocket

27. i = - p magnification = 1
What happens if we bend the mirror?
Concave mirror.
Image gets magnified.
Field of view is diminished
Convex mirror.
Image is reduced.
Field of view increased.

28. Rules for drawing images for mirrors
Initial parallel ray reflects through focal point.
Ray that passes in initially through focal point reflects parallel
from mirror
Ray reflects from C the radius of curvature of mirror reflects along
itself.
Ray that reflects from mirror at little point c is reflected
symmetrically

30. Spherical refracting surfaces
Using Snell’s Law and assuming small
Angles between the rays with the central
axis, we get the following formula:

31. Thin Lenses

32. Apply this equation to Thin Lenses where the thickness is small compared to object distance, image distance, and radius of curvature. Neglect thickness.
Converging lens
Diverging lens

33. Thin Lens Equation Lensmaker Equation Lateral Magnification for a Lens
What is the sign convention?

34. Sign Convention Light r1 r2 i Real side - R Virtual side - Vp i
Real object - distance p is pos on V side (Incident rays are diverging)
Radius of curvature is pos on R side.
Real image - distance is pos on R side.
Virtual object - distance is neg on R side. Incident rays are converging)
Radius of curvature is neg on the V side.
Virtual image- distance is neg on the V side.

35. Rules for drawing rays to locate images from a lens
1) A ray initially parallel to the central axis will pass through the focal point.
2) A ray that initially passes through the focal point will emerge from the lens
parallel to the central axis.
3) A ray that is directed towards the center of the lens will go straight through the lens undeflected.

36. Real image ray diagram for a converging lens

37. Object Distance=20 cm and lens focal length is +10 cm
Find:
Image distance
Magnification
Height of image
Is image erect or inverted
Is it real or virtual
Draw the 3 rays

38. Object Distance =5 cm and focal length is + 10 cm
Find:
Image distance
Magnification
Height of image
Is image erect or inverted
Is it real or virtual
Draw the 3 rays

39. p=Object Distance =10 cm

40. Virtual image ray diagram for converging lens

41. Virtual image ray diagram for a diverging lens

42. . Example Virtual side Real side
24(b). Given a lens with a focal length f = 5 cm and object distance p = +10 cm, find the following: i and m. Is the image real or virtual? Upright or inverted? Draw the 3 rays. . F1 F2 p
Virtual side
Real side
Image is real,
inverted.

43. 24(e). Given a lens with the properties (lengths in cm) r1 = +30, r2 = -30, p = +10, and n = 1.5, find the following: f, i and m. Is the image real or virtual? Upright or inverted? Draw 3 rays.
Real side
Virtual side r1 . F1 F2 p r2
Image is virtual, upright.

44. 27. A converging lens with a focal length of +20 cm is located 10 cm to the left of a diverging lens having a focal length of -15 cm. If an object is located 40 cm to the left of the converging lens, locate and describe completely the final image formed by the diverging lens. Treat each lens
Separately. f1
Lens 1
Lens 2 f2 10 40
+20
-15

45. Since m = -i1/p1= - 40/40= - 1 , the image is invertedf1
Lens 1
Lens 2 f2 10 40
+20
-15 40 30
Ignoring the diverging lens (lens 2), the image formed by the
converging lens (lens 1) is located at a distance
Since m = -i1/p1= - 40/40= - 1 , the image is inverted
This image now serves as a virtual object for lens 2, with p2 = - (40 cm - 10 cm) = - 30 cm.

46. f1
Lens 1
Lens 2 f2 10 40
+20
-15 30
Thus, the image formed by lens 2 is located 30 cm to the left of lens 2. It is virtual (since i2 < 0).
The magnification is m = (-i1/p1) x (-i2/p2) = (-40/40)x(30/-30) =+1, so the image
has the same size orientation as the object.

47. Optical Instruments Magnifying lens Compound microscope
Refracting telescope

48. Chapter 34 Problem 32
In Figure 34-35, A beam of parallel light rays from a laser is incident on a solid transparent sphere of index of refraction n.
Fig
(a) If a point image is produced at the back of the sphere, what is the index of refraction of the sphere?
(b) What index of refraction, if any, will produce a point image at the center of the sphere? Enter 'none' if necessary.

49. Chapter 34 Problem 91
Figure 34-43a shows the basic structure of a human eye. Light refracts into the eye through the cornea and is then further redirected by a lens whose shape (and thus ability to focus the light) is controlled by muscles. We can treat the cornea and eye lens as a single effective thin lens (Figure 34-43b). A "normal" eye can focus parallel light rays from a distant object O to a point on the retina at the back of the eye, where processing of the visual information begins. As an object is brought close to the eye, however, the muscles must change the shape of the lens so that rays form an inverted real image on the retina (Figure 34-43c).

50. (a) Suppose that for the parallel rays of Figure 34-43a and Figure 34-43b, the focal length f of the effective thin lens of the eye is 2.52 cm. For an object at distance p = 48.0 cm, what focal length f' of the effective lens is required for the object to be seen clearly?
(b) Must the eye muscles increase or decrease the radii of curvature of the eye lens to give focal length f'?