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## Homework Statement

*An outdoor security camera has a simple lens with focal object length fo = 6.00 cm and perfectly focuses the image of a bird onto its detector located 9.00 cm away from the lens. The animal then flies away from the lens and the new focal distance is 8.00 cm. (a) How far did the bird move? (b) What is the new magnification?*

## Homework Equations

$\begin{array}{l}

\[\frac{1}{f} = \frac{1}{s} + \frac{1}{{s'}}\]

\end{array}$

## The Attempt at a Solution

**This is my solution:**$\begin{array}{l}

\frac{1}{{{f_0}}} = \frac{1}{s} + \frac{1}{{s'}}\\

\frac{1}{{{s_1}}} = \frac{1}{{{f}}} - \frac{1}{{s'}} = \frac{1}{6} - \frac{1}{9} = \frac{1}{{18}}\\

{s_1} = 18cm\\

\frac{1}{{{s_2}}} = \frac{1}{{{f}}} - \frac{1}{{s'}} = \frac{1}{6} - \frac{1}{8} = \frac{1}{{24}}\\

{s_2} = 24cm\\

\Delta {s_o} = {s_2} - {s_1} = 24cm - 18cm = 6cm

\end{array}$

As my textbook offers limited worked solutions for this topic I have looked around and it seems that the focal length does not change on a 'simple lens', even if it is bi-convex. It is s and s prime that changes. Or am I misunderstanding the initial question? The wording is somewhat vague.

**This is the solution I was presented with:**f1= 6.00 cm

s’1 = s’2= 9 cm ( f > s’ >2f ) + image is real, inverted, and reduced (for both)

f2 = 8.00 cm

s1 = ?

s2 = ?

Ds0= s2- s1

A camera should mean a bi-convex lens.

$\begin{array}{l}

\frac{1}{{{f_0}}} = \frac{1}{s} + \frac{1}{{s'}}\\

\frac{1}{{{s_1}}} = \frac{1}{{{f_1}}} - \frac{1}{{s'}} = \frac{1}{6} - \frac{1}{9} = \frac{1}{{18}}\\

{s_1} = 18cm\\

\frac{1}{{{s_2}}} = \frac{1}{{{f_2}}} - \frac{1}{{s'}} = \frac{1}{8} - \frac{1}{9} = \frac{1}{{72}}\\

{s_2} = 72cm\\

\Delta {s_o} = {s_2} - {s_1} = 72cm - 18cm = 54cm

\end{array}$

The bird flew 54 cm or just over half a metre from the camera.