What is the magnitude of the focal length

f

in the small angle approximation? Assume
the index of refraction of the lens is 2.0 and of
the medium is 1.0.
1.

f

=
1
4
a
2.

f

= 2
a
3.

f

=
3
4
a
4.

f

=
3
2
a
correct
5.

f

=
5
2
a
6.

f

=
a
7.

f

= 4
a
8.

f

=
1
3
a
9.

f

=
1
2
a
10.

f

= 3
a
Explanation:
R
1
is the
x
coordinate of the center of the
spherical surface
A
;
i.e.
,
R
1
=

R
A

=

3
a
.
Similarly,
R
2
=

a
.
The lens maker’s formula gives
1
f
= (
n

1)
parenleftbigg
1
R
1

1
R
2
parenrightbigg
= (2

1)
parenleftbigg
1

3
a

1

a
parenrightbigg
=
2
3
a
f
=
3
2
a .
037(part2of3)10.0points
The sign of the focal length is
1.
negative
2.
positive
correct
Explanation:
From the previous part, we see that
f >
0.
038(part3of3)10.0points
A
B
a
3
a
O
C
A
C
B
What will be the magnitude and the sign of
the focal length when the lens is flipped over?
1.
The magnitude decreases by more than a
factor 2, the sign is unchanged.
2.
The magnitude increases by less than a
factor 2, the sign is opposite.
3.
The magnitude increases by less than a
factor 2, the sign is unchanged.
mittal (im5936) – reflection refraction imaging HW – yeazell – (58010)
13
4.
The magnitude increases by more than a
factor 2, the sign is unchanged.
5.
The magnitude stays the same, the sign
is opposite.
6.
The magnitude decreases by less than a
factor 2, the sign is unchanged.
7.
The magnitude decreases by less than a
factor 2, the sign is opposite.
of the lens?
Assume the object is on the
lefthand side of the lens.
8.
The magnitude increases by more than a
10
.
3 cm
10 cm
What is the magnitude of the focal length
of the lens?
Assume the object is on the
lefthand side of the lens.
factor 2, the sign is opposite.
9.
The magnitude stays the same, the sign
is unchanged.
correct
10.
The magnitude decreases by more than a
factor 2, the sign is opposite.
Explanation:
When the lens is flipped over,
R
1
= +

R
B

, R
2
= +

R
A

1
f
= (
n

1)
parenleftbigg
1
R
1

1
R
2
parenrightbigg
= (2

1)
parenleftbigg
1
a

1
3
a
parenrightbigg
=
2
3
a
f
=
3
2
a .
Digression:
Notice that this is a negative sign
due to the interchange of

R
1

and

R
2

. This
compensates for the sign change in
R
1
and
R
2
, so
f
stays the same.
039(part1of3)10.0points
A lens has an index of refraction of 1
.
34.
The lens has a radius of curvature of 10
.
3 cm
on its lefthand side and a radius of curvature
of 10 cm on its righthand side.
Explanation:
Let :
n
= 1
.
34
,
R
1
= 10
.
3 cm
,
and
R
2
=

10 cm
.
From the perspective of the object, the ra
dius of curvature of a lens surface is
positive
if the center of curvature is
behind
a surface;
otherwise it is
negative
.
Applying the lens maker’s equation,
1
f
= (
n

1)
parenleftbigg
1
R
1

1
R
2
parenrightbigg
=
(
n

1) (
R
2

R
1
)
R
1
R
2
f
=
R
1
R
2
(
n

1) (
R
2

R
1
)
=
(10
.
3 cm)(

10 cm)
(1
.
34

1) [

10 cm

(10
.
3 cm)]
= 14
.
9232 cm
,
with magnitude
14
.
9232 cm
.
040(part2of3)10.0points
What is the sign of the focal length?
1.
negative
2.
Cannot be determined.
3.
positive
correct
Explanation:
mittal (im5936) – reflection refraction imaging HW – yeazell – (58010)
14
Since the lens is thicker on the axis than
near its outer edge, the focal length
f
is posi
tive and the lens is convergent.