- #1
ewang
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- Homework Statement
- An ophthalmometer* is used by optometrists to measure the curvature of a patient’s cornea by shining an illuminated shape on the cornea and measuring the size of the reflected image relative to the object. For one patient, the object was 24 cm from their cornea and the image height was 0.0156 times the height of the object. What is the radius of this patient’s cornea?
- Relevant Equations
- f = R/2
1/f = 1/i + 1/o
m = -i/o
Since magnification is 0.0156, I have:
m = -i/o
0.0156 = -24cm/o
o = -1538.46 cm
1/f = 1/i + 1/o
1/f = 1/24 cm - 1/1538.36 cm
f = 24.38 cm
R = 48.76 cm
However, when I look up the average corneal radius, the google results show ~5 mm. Did I do something wrong?
m = -i/o
0.0156 = -24cm/o
o = -1538.46 cm
1/f = 1/i + 1/o
1/f = 1/24 cm - 1/1538.36 cm
f = 24.38 cm
R = 48.76 cm
However, when I look up the average corneal radius, the google results show ~5 mm. Did I do something wrong?