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danago
Nov10-06, 10:27 PM
"A concave mirror of focal length 1m is used to magnify an object by a factor of 4. Where is the object in relation to the pole of the mirror?"

Heres what i did:

\begin{array}{c}
M = \frac{v}{u} \\
\therefore 4 = \frac{v}{u} \\
\therefore v = 4u \\
\end{array}



\begin{array}{c}
\frac{1}{f} = \frac{1}{u} + \frac{1}{v} \\
\therefore 1 = \frac{1}{u} + \frac{1}{{4u}} = \frac{5}{{4u}} \\
\therefore u = 1.25m \\
\end{array}


Where am i going wrong? According to my textbook, the answer is 0.75m.

Thanks in advance,
Dan.

Alt+F4
Nov10-06, 10:43 PM
You are forgeting that M = -di/do

Formula = (1/do + 1/di) = (1/f)

So 4do = - di

1/do - 1/4do = 1

1/4do - 4/4do = 1

3/4do = 1

do = .75

TheloniousMONK
Nov11-06, 02:27 AM
3/4do = 1

do = .75

Right.

3/(4do) = 1 so do = .75