1. Phy2005
Applied Physics II
Spring 2017
Announcements:

2. Exam 2 NB: remember course grading scale!!
mean was 7.8/12 = 65% std. dev. 1.92
solutions were posted Thurs. morning
grades posted by tomorrow
please come to ofc. hours or make appt. if you are concerned about your performance
NB: remember course grading scale!!
S = 48 (T1+T2)/(2Tmax) + 32 (E/Emax)
+ 20 (HITT-20% low)/HITTmax

3. Hardest questions <50% correct

4. qi qr qi = qr Last time p is position of object q is position of image
Law of reflection
Last time
p > q
Spherical mirror
p is position of object
q is position of image
f is focal point
image is real if putting a
screen at the pt. would
form an actual image
otherwise, it’s virtual
image is either erect or
inverted (as in this fig.)
object
image
focus

5. Last time: Real vs. virtual image

6. For a small object, f = R/2 (spherical mirror)
Mirror Equation
1/p + 1/q = 1/f
For a small object, f = R/2 (spherical mirror)
1/p + 1/q = 2/R
Alert!!
Be careful with the sign!!
Negative means that it is behind the mirror!!
p can never be negative (why?)
negative q means the image is formed behind the mirror
VIRTUAL
How about f?

7. Focal point inside the mirror
For a concave mirror: f > 0
Convex mirror:
Focal point inside the mirror
f < 0
1/p + 1/q = 1/f < 0 : q should be negative.

8. All images formed by a convex mirror are VIRTUAL.
1/p + 1/q = 1/f < 0 : q should be negative.
All images formed by a convex mirror are VIRTUAL.
Magnification, M = -q/p
Negative M means that the image is upside-down.
For real images, q > 0 and M < 0 (upside-down).

9. Ex. 26.1 An object is placed at the center of curvature of a
mirror. Where is the image formed? Describe the image.
1/p + 1/q = 1/f f = R/2
Object is at the center: p = R
1/q = 1/f – 1/p
= 2/R – 1/R
= 1/R
q = R > 0 (Real Image)
M = -q/p = -R/R = -1
No magnification but upside-down

10. Upright or Upside-down
Ex A concave mirror has a 30 cm radius of curvature.
If an object is placed 10 cm from the mirror, where will the
image be found?
f = R/2 = 15 cm, p = 10 cm
1/p + 1/q = 1/f  1/10 + 1/q = 1/15
3/30 + 1/q = 2/30
1/q = -1/30
q = -30 cm
Case 5: p < f
Real or Virtual
Magnified or Reduced
Upright or Upside-down
q < 0
M = -q/p = 3

11. Q. An upright image that is one-half as large as an object is
needed to be formed on a screen in a laboratory experiment
using only a concave mirror with 1 m radius of curvature.
If you can make this image, I will give you $10. If you can’t
you should pay me $10. Deal or no deal? Why?
1/p + 1/q = 1/f = 2/R > 0
M = -q/p = ½ > 0
should be a real image: q > 0
M = -q/p cannot be positive, if q > 0.
No deal!!!

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13. Q1 A 2-cm high object is placed 120cm in front of a concave mirror whose radius of curvature is 60cm. Where is the image located, and what is its character?
20cm, upright
20 cm, inverted
40 cm, upright
40 cm, inverted
A concave mirror cannot form an image