1. Physics 7E
Prof. D. Casper

2. Admin Chapter 34 HW is due a week from today (Monday, 7 am) Reading
Due to my mistake, it has only been available since 7am today – sorry!
Reading
Wednesday: Chapter 35.1 – 35.3
Next Monday: Chapter 35.4 – 35.5
Week 9 10
Finals
Monday
Tuesday
Wednesday
Thursday
Friday
34.5 – 34. 8
35.1 – 35.3
Thanksgiving
35.4 – 35.5 Ch. 34 HW Due
Ch. 36 Lecture
Ch. 35 HW Due
Ch. 36 HW Due
Final (8am)

3. Concave Mirror: Ray Tracing

4. Convex Mirror: Ray Tracing
1 𝑠 + 1 𝑠 ′ = 1 𝑓 → 1 𝑠 ′ = 1 𝑓 − 1 𝑠
if 𝑓<0 and 𝑠>0, then 𝑠 ′ <0
Real object always forms virtual image in convex mirror

5. Concave Mirror: Real and Virtual Images
In these examples: 𝑅=20 cm, 𝑓=10 cm
𝑠=30 cm: 𝑠 ′ = 1 10 → 𝑠 ′ =15 cm 𝑚=− 𝑠 ′ 𝑠 =− 1 2 (real, inverted image)
𝑠=20 cm: 𝑠 ′ = 1 10 → 𝑠 ′ =20 cm 𝑚=− 𝑠 ′ 𝑠 =−1 (real, inverted image)
𝑠=10 cm: 𝑠 ′ = 1 10 → 𝑠 ′ →∞ (no image)
𝑠=5 cm: 𝑠 ′ = 1 10 → 𝑠 ′ =−10 cm 𝑚=− 𝑠 ′ 𝑠 =+2 (virtual, upright image)

6. A. real and larger than the object.
Q34.3
An object is placed 4.0 m away from a concave mirror of focal length +1.0 m. The image formed by the mirror is
A. real and larger than the object.
B. real and smaller than the object.
C. real and the same size as the object.
D. virtual and larger than the object.
E. virtual and smaller than the object.
Answer: B

7. A. real and larger than the object.
An object is placed 4.0 m away from a concave mirror of focal length +1.0 m. The image formed by the mirror is
A. real and larger than the object.
B. real and smaller than the object.
C. real and the same size as the object.
D. virtual and larger than the object.
E. virtual and smaller than the object.

8. A. real and larger than the object.
Q34.7
An object is placed 2.0 m away from a convex mirror of focal length –1.0 m. The image formed by the mirror is
A. real and larger than the object.
B. real and smaller than the object.
C. real and the same size as the object.
D. virtual and larger than the object.
E. virtual and smaller than the object.
Answer: E

9. A. real and larger than the object.
An object is placed 2.0 m away from a convex mirror of focal length –1.0 m. The image formed by the mirror is
A. real and larger than the object.
B. real and smaller than the object.
C. real and the same size as the object.
D. virtual and larger than the object.
E. virtual and smaller than the object.

10. Summary of Mirror Images
For a REAL object:
Plane Mirror
Image always virtual and upright (| 𝑠 ′ |=𝑠,𝑚=1)
Convex Mirror
Image always virtual and upright (| 𝑠 ′ |<𝑠, 0<𝑚<1)
Concave Mirror
If 𝑠<𝑓: Image virtual and upright (𝑚>1)
If 𝑠>𝑓: Image real and inverted (𝑚<0)
If 𝑠=𝑓, no image

11. Spherical Refracting Surface: Paraxial Approximation
𝑛 𝑎 𝑠 + 𝑛 𝑏 𝑠 ′ = 𝑛 𝑏 − 𝑛 𝑎 𝑅
𝑚= 𝑦 ′ 𝑦 =− 𝑛 𝑎 𝑠 ′ 𝑛 𝑏 𝑠
Center on side with outgoing light: 𝑅>0
Center not on side with outgoing light: 𝑅<0

12. Lens Properties Thicker in middle focal length 𝑓>0
Thinner in middle
focal length 𝑓<0

13. Thin Lens Equation 1 𝑠 + 1 𝑠 ′ = 1 𝑓
1 𝑠 + 1 𝑠 ′ = 1 𝑓
Sign conventions are all-important!
s > 0: Object on side with incoming light
s’ > 0: Image on side with outgoing light
f > 0: Converging lens
Lateral magnification:
𝑚= 𝑦 ′ 𝑦 =− 𝑠 ′ 𝑠

14. Ray-tracing: Converging Lens

15. Ray tracing: Diverging Lens

16. Multiple Lens Systems For multiple lenses and/or mirrors, you take
one element at a time.
The image of one element is the object of the next
In the diagram, we have two converging lenses with
𝑓 1 =8.0 cm and 𝑓 2 =6.0 cm a distance 36.0 cm apart
Find the image of O in Lens A (ignoring Lens B for now):
𝑠 1 ′ = 1 8 → 𝑠 1 ′ =24 cm
What is the object distance for Lens B?
𝑠 2 =36 cm−24 cm=12 cm
𝑠 2 ′ = 1 6 → 𝑠 2 ′ =12 cm
𝑚= 𝑚 1 × 𝑚 2 =(−2)×(−1)=+2

17. “Virtual” Objects
In the previous example, the image of Lens A was “in front” of Lens B (on the side the light entered Lens B) It is possible to have the opposite situation
Suppose both lenses have f = 10 cm, they are 10 cm apart, and the object is 20 cm from Lens A ( 𝑠 1 =20)
The image in Lens A will then be 20 cm to the right
of Lens A (and 10 cm to the right of B)
Since it is not on the side the light enters B, the object distance for Lens B is -10 cm
1 − 𝑠 2 ′ = 1 10 → 𝑠 2 ′ =+20 cm
Image of Object In Lens A
Image of Object In Lens B
Object
Virtual Object For Lens B
Lens A
Lens B

18. Lens-maker’s Equation
The focal length of a lens can be worked out by treating the image of its front face as the object for its back face See the book for a full derivation
The focal length depends on the
curvature of the two surfaces, and
the index for refraction of the material
For a thin lens, the result is:
1 𝑓 = 𝑛− 𝑅 1 − 1 𝑅 2