P 2 a q 1 20 cm 30 cm 10 cm where distances are now

p
2
= a - q
1
= (20 cm) - (30 cm) = - 10 cm,
where distances are now measured from the
second lens, whose focal length is f
2
= 25 cm :
1
p
2
+
1
q
2
=
1
f
2
.
This yields
q
2
=
p
2
f
2
017 (part 2 of 3) 10 points
What is the magnification of the second lens?
Explanation:
The magnification of the first lens is given
by
M
1
= -
q
1
p
1
= -
30 cm
15 cm
= - 2 .
The total magnification M of the two lenses
is the product of M
1
and M
2
:
M = M
1
M
2
= ( - 2) (0 . 714286)
= - 1 . 42857 .
Hence, the final image is real, inverted, and
enlarged.
keywords:
019 (part 1 of 1) 10 points
An object 18 cm high is placed 10 cm in
front of a convex mirror with a focal length of
- 5 . 8 cm.
What is the image height?

Drews, Scott – Homework 12 – Due: Nov 26 2006, midnight – Inst: Ditmire 9
The magnification is given by M = -
i
o
.
Hence, the image height is given by
H = h M = (18 cm) -
- 3 . 67089 cm
10 cm
¶
= 6 . 6076 cm .
keywords:
020 (part 1 of 3) 10 points
A concave mirror has a focal length f =
21 cm.
Determine the conditions on the object po-
sition p for the image to be erect at q .
M to be positive is that
p < f .
021 (part 2 of 3) 10 points
Determine the object position p for the image
to be erect and 4 times the size of the object.
9. f < p
In order for the image to be erect, the mag-
nification M = -
q
p
must be positive. ( p =
object distance, q = image distance). Us-
ing the mirror equation, we can solve for q in
terms of p and f .
1
p
+
1
q
=
1
f
, so
1
q
=
1
f
-
1
p
⇒
1
q
=
p - f
f p
.
Now, plugging into the equation for M we
obtain
M = -
q
p
=
f
f - p
.
So the condition for M to be positive is that
p < f .
021 (part 2 of 3) 10 points
Determine the object position p for the image
to be erect and 4 times the size of the object.
What is the radius of curvature of the mirror?
Assume the mirror is spherical.
keywords:
023 (part 1 of 3) 10 points
Hint: Construct a ray diagram.

Drews, Scott – Homework 12 – Due: Nov 26 2006, midnight – Inst: Ditmire 10
Given: A real object is located at “ p
1
=
11
6
f ” to the left of a convergent lens with a
focal length f as shown in the figure below.
0
11
6
f f
( × f )
The image is
1. real, erect, and smaller.
2. virtual, inverted, and smaller.
3. virtual, erect, and larger.
4. virtual, erect, and smaller.
5. virtual, inverted, and larger.
6. real, inverted, and larger. correct
7. real, inverted, and smaller.
8. real, erect, and larger.
Explanation:
Basic Concepts:
1
p
+
1
q
=
1
f
m =
h
0
h
= -
q
p
Converging Lens f > 0
∞ > p > f f < q < ∞ 0 > m > -∞
f > p > 0 -∞ < q < 0 ∞ > m > 1
Solution: