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A concave mirror of focal length $8cm$ forms an inverted image of an object placed at a certain distance. If the image is twice as large as the object, what is the distance of the object and the image from the pole of the mirror?

I started off with the relation: $\frac{h_o}{h_i} = \frac{d_o}{d_i}$ since the image is inverted, $h_i = -2h_o$, and $\frac{h_o}{h_i} = \frac{1}{-2}$.

$\implies \frac{d_o}{d_i} = \frac{1}{-2} \implies -2d_o = d_i$.

Using the mirror formula $\frac1f = \frac{1}{d_i} + \frac{1}{d_o}$ and substituting $d_i$ for $-2d_o$, $\frac1f = \frac{1}{-2d_o} + \frac{1}{d_o}$

$\implies \frac1f = \frac{1 + -2}{-2d_o} = \frac{1}{2d_o} \implies d_o = \frac{f}{2}$

Since $f = -8cm$ (using Cartesian sign convention),

$d_o = -4cm$, but how can a real image be formed when the object is placed at less than the focal length?

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Your first equation:

$$ \frac{h_o}{h_i} = \frac{d_o}{d_i} $$

is wrong. It should be:

$$ \frac{h_o}{h_i} = -\frac{d_o}{d_i} $$

To justify this, consider that with the object and image outside the focal point $d_o/d_i$ will be positive, but because the image is inverted $h_o/h_i$ is negative. You need the negative sign in the equation to get equality.

With the negative sign you'll get $d_i = 2d_o$. Substitiute this in to the mirror equation and you'll get the correct answer for $d_o$, which is is obviously outside the focal length.

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  • $\begingroup$ But, I considered that, because I said that $\frac{h_o}{h_i}$ is negative and that $-2h_o = h_i$. $\endgroup$
    – Gerard
    Jul 21, 2013 at 5:19
  • $\begingroup$ @user1305192: yes, but you then concluded that $-2d_o = d_i$ and this is wrong because you missed out the minus sign. You should have concluded that $2d_o = d_i$. $\endgroup$
    – John Rennie
    Jul 21, 2013 at 6:07
  • $\begingroup$ I understand. But what's the reason behind this? We can't just arbitrarily change signs to satisfy the equation, there has to be a logical reason. $\endgroup$
    – Gerard
    Jul 21, 2013 at 16:21
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    $\begingroup$ @user1305192: Sign conventions in the lens/mirror equation are a bit of a mess. For example when analysing lenses it's common to take both $d_o$ and $d_i$ as positive even when they're on opposite sides of the lens, and if you do this you require the negative sign in the magnification equation. Really it's up to the student to use the correct magnification equation for the sign convention they choose. $\endgroup$
    – John Rennie
    Jul 22, 2013 at 5:59

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