1. Flat Mirrors Consider an object placed in front of a flat mirror
Object O is placed a distance p in front of the mirror (p = object distance)
Light reflecting from mirror appears to originate at point I
An image of object O is formed at a distance q behind the mirror (q = image distance)
It is a virtual image (light does not really converge there)
When light actually converges at an image point, it is a real image (can be projected on a screen)
We can use (at least 2) rays to determine the image orientation and position
Notice p = q for a flat mirror
Image is upright and virtual

2. Flat Mirrors The height h of the object equals the image height h’
Lateral magnification M:
General definition for any type of mirror
For flat mirrors, M = 1
“Magnification” can mean enlargement or reduction in size in optics
A flat mirror also produces an apparent left – right reversal
A waving left hand appears to be a waving right hand in the mirror

3. Example Problem #23.1
Does your bathroom mirror show you older or younger than your actual age? Compute an order-of-magnitude estimate for the age difference, based on data that you specify.
Solution:
Class interactive – Solution will be determined in class

4. Concave Spherical Mirrors
Consider light from an object O striking a spherical concave mirror
If rays diverge at small angles, they all reflect through same image point
Large diverging angles mean rays intersect principle axis at different points, resulting in a blurred image
Geometry of concave mirrors
(from yellow and blue similar triangles)

5. Concave Spherical Mirrors
Additional triangle geometry yields the mirror equation:
If an object is very far from the mirror, then 1 / p ≈ 0 and q ≈ R / 2
In this case, the image point is called the focal point F and the image distance q is called the focal length f = R / 2
Note that focal point  focus point
The mirror equation can be written in terms of focal length:

6. Convex Spherical Mirrors
Geometry of a convex mirror
We can use the same mirror equation as for concave mirrors, but we just need to use a particular sign convention for each quantity in the equation

7. Sign Conventions for Mirrors
Also, R > 0 (R < 0) for concave (convex) mirrors

8. Ray Diagrams for Concave Mirrors

9. Ray Diagrams for Convex Mirrors
Note that the sideview mirror on the passenger side of a car is a convex mirror (hence the warning given on the mirror)

10. Example Problem #23.16
A convex spherical mirror with a radius of curvature of 10.0 cm produces a virtual image one-third the size of the real object. Where is the object?
Solution (details given in class):
10.0 cm in front of the mirror

11. Example Problem #23.19
A spherical mirror is to be used to form an image, five times as tall as an object, on a screen positioned 5.0 m from the mirror.
Describe the type of mirror required.
Where should the mirror be positioned relative to the object?
Solution (details given in class):
Concave
1.0 m in front of the mirror

12. Images Formed by Refraction
Consider light from object O refracting at a spherical surface between transparent media
From Snell’s law and geometry:
Note that real images are formed on the side opposite that of the incident light

13. Images Formed by Refraction
If the refracting surface is flat, R   and we get:
Now the (virtual) image is on the same side of the surface as the object

14. Interactive Example Problem: Lens Design 101
Animation and solution details given in class.
(PHYSLET Physics Exploration 34.3, copyright Pearson Prentice Hall, 2004)

15. Example Problem #23.25
A transparent sphere of unknown composition is observed to form an image of the Sun on its surface opposite the Sun. What is the refractive index of the sphere material?
Solution (details given in class):
2.00

16. Thin Lenses A thin lens forms an image by refraction of light
Has 2 refracting surfaces
Examples of thin lenses shown at right
Lenses in group (a) converge parallel rays to a focal point on the opposite side of the lens
Lenses in group (b) diverge parallel rays so they appear to originate from a focal point on the same side of the lens
(converging lens)
(diverging lens)

17. Geometry of Thin Lenses
Magnification by a lens
Thin–lens equation
Lens maker’s equation
(same as for a mirror)
(same as for a mirror)
( f = focal length in air)
( n = index of refraction of the lens material)
(R1 (R2) = radius of curvature of front (back) surface)

18. Sign Conventions for Thin Lenses

19. Ray Diagrams for Thin Lenses
Converging lenses

20. Ray Diagrams for Thin Lenses
Diverging lenses

21. Example Problem #23.35
A certain LCD projector contains a single thin lens. An object 24.0 mm high is to be projected so that its image fills a screen 1.80 m high. The object-to-screen distance is 3.00 m. (a) Determine the focal length of the projection lens. (b) How far from the object should the lens of the projector be placed in order to form the image on the screen?
Solution (details given in class):
(b) 39.5 mm
(a) 39.0 mm

22. Interactive Example Problem: Building a Converging Lens
Animation and solution details given in class.
(PHYSLET Physics Problem 35.11, copyright Pearson Prentice Hall, 2004)

23. Combinations of Thin Lenses
If 2 thin lenses are used to form an image, use the following procedure:
The image produced by the first lens is calculated as though the second lens were not present
The image formed by the first lens is treated as the object for the second lens
The image formed by the second lens is the final image of the system
The overall magnification is the product of the magnifications of the separate lenses
This procedure can be extended to 3 or more lenses

24. Example Problem #23.40
An object is placed 20.0 cm to the left of a converging lens of focal length 25.0 cm. A diverging lens of focal length 10.0 cm is 25.0 cm to the right of the converging lens. Find the position and magnification of the final image.
Solution (details given in class):
Position = 9.26 cm in front of the 2nd lens
Magnification =

25. Example Problem #23.50
Front of mirror
Back of mirror
Back of lens
Front of lens
The object shown above is midway between the lens and the mirror. The mirror’s radius of curvature is 20.0 cm, and the lens has a focal length of –16.7 cm. Considering only the light that leaves the object and travels first towards the mirror, locate the final image formed by this system. Is the image real or virtual? Is it upright or inverted? What is the overall magnification of the image?
Solution (details given in class):
The image is virtual, upright, located 25.3 cm behind the mirror, with an overall magnification of

26. HW Problems #23.51, 23.59
#23.51:
#23.59:

27. Aberrations Spherical aberration Chromatic aberration
Light passing through lens at different distances from the principal axis is focused at different points
Apertures are used to help narrow the incoming beam of light
Chromatic aberration
Different wavelengths of light refracted by a lens focus at different points
Can be reduced by using combination of converging and diverging lenses