Question

An object 1.25 cm high is held 2.9 cm from a person’s cornea, and its reflected image is measured to be 0.171cm high.

Part (a) What is the magnification?

Part (b) What is the image distance in cm?

Part (c) Find the radius of curvature in cm of the convex mirror formed by the cornea. $sig.gif?tid=2M68-BC-2E-46-8A73-19928$

Answer 1

Part (a) What is the magnification?

Magnification, $M=\frac{h_{i}}{h_{o}}=-\frac{i}{o}$

$M=\frac{h_{i}}{h_{o}}$

$M=\frac{0.171cm}{1.25cm}$

ANSWER: ${\color{Red} M=0.1368}$

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Part (b) What is the image distance in cm?

Magnification, $M=\frac{h_{i}}{h_{o}}=-\frac{i}{o}$

$M=-\frac{i}{o}$

$0.1368=-\frac{i}{2.9cm}$

$-0.1368*2.9cm=i$

ANSWER: ${\color{Red} i=-0.39672cm}$

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Part (c) Find the radius of curvature in cm of the convex mirror formed by the cornea. $sig.gif?tid=2M68-BC-2E-46-8A73-19928$

First, find the focal length

Use formula $\frac{1}{f}=\frac{1}{o}+\frac{1}{i}$

$\frac{1}{f}=\frac{1}{2.9cm}-\frac{1}{0.39672cm}$

$\frac{1}{f}=\frac{1}{2.9}-\frac{1}{0.39672}=\frac{0.39672-2.9}{2.9*0.39672}=\frac{-2.50328}{2.9*0.39672}$

$f=-\frac{2.9*0.39672}{2.50328}$

${\color{Red} f=-0.46cm}$

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Radius of curvature, $C=2f=-2*0.46cm$

ANSWER: ${\color{Red} C=-0.92cm}$

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