Biomedical Engineering Reference
In-Depth Information
Sample Problem
Now that we've learned how to determine the equivalent power of a thick lens and
the location of its principal planes, let's try a problem. We'll work on the thick lens
problem that we solved previously in Chapter 5.
An object is located 25.00 cm
in front of a biconvex crown glass lens that has an front radius of curvature of
10.00 cm and a back radius of curvature of
5.00 cm (Fig. 5-6). The lens has a
thickness of 20.00 mm. Using an equivalent lens approach, locate the image with
respect to the posterior surface and calculate the magnification produced by the
lens. Is the image real or virtual? Is the image erect or inverted?
−
In Chapter 5, we solved this problem by taking a surface-by-surface approach.
Now, let's see how we can solve it using an equivalent lens strategy.
We previously determined the front (
F
1
) and back surface (
F
2
) powers to be
+
5.20 D
and
+
10.40 D, respectively. First, we determine the equivalent power as follows:
t
⎛
⎞
⎠
F
e
=
F
1
+
F
2
−
F
1
F
2
⎝
0.20 m
⎛
⎝
⎞
⎠
F
e
=
5.20 D
+
10.40 D
−
(5.20 D)(10.40 D)
1.52
F
e
= +
14.89 D
Next, we locate the primary principal plane as follows:
t
⎞
⎠
⎛
F
2
⎝
A
1
H
=
F
e
0.02
⎛
⎝
⎞
⎠
(10.40 D)
1.52
A
1
H
=
14.89 D
A
1
H
= +
0.0092 m or
+
0.92 cm
Similarly, we can locate the secondary principal plane
t
⎛
⎞
⎠
−
F
1
⎝
′
A
2
H
=
F
e
⎛
⎝
0.02
⎞
⎠
−
(5.20 D)
1.52
′
A
2
H
=
14.89 D
′
A
2
H
= −
0.0046 m or
−
0.46 cm
