Question

s O h C -h' F I s' (a) Similar triangles, so h'/h = -s'/s Similar triangles, so -h'/h = (s' - f)/f O h C s' - f F h -h' I f s' (b)

          s
O
h
C
-h'
F
I
s'
(a)
Similar triangles, so h'/h = -s'/s
Similar triangles, so -h'/h = (s' - f)/f
O
h
C
s' - f
F
h
-h'
I
f
s'
(b)

Added by Chris K.

Elementary and Intermediate Algebra

Marvin L. Bittinger, David J.… 4th Edition

Chapter 15

Instant Answer

Solved by Expert Mark Scythian

01/29/2025

Step 1

We have an object at point \( O \) with height \( h \), a mirror with center \( C \), and focal point \( F \). The object distance is \( s \), the image distance is \( s' \), and the image height is \( h' \). Show more…

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Thanks for your feedback!

I would like to prove that C = R = 2f using diagrams such as these, and by (i) adding a ray that passes through the center (and therefore reflects back on itself); and (ii) constructing similar triangles using the height of the object and the height of the image as the vertical sides. I can set up the similar triangles and create a variety of ratios; I can get as far as showing there are two congruent triangles whose bases span the distance from C to f. But I can't quite complete the proof.