Geometrical optics. Photometry. Quantum nature of light > Rectilinear propagation and reflection of light > Problem 13.1.17

Statement

$13.1.17.$ The actual image of an object in a concave mirror is three times the size of the object. After the object was moved 80 cm away from the mirror, its image became half the size of the object. Find the focal length of the mirror.

Solution

Firstly, the magnification of the object is given by:
$k = \frac{s'}{s}$ (1)
where $s'$ is the image-mirror distance and $s$ is the object-mirror distance, with $k$ = 3.
And the relation for these distances is
$\frac{1}{s} +\frac{1}{s'} = \frac{1}{f}$ (2)
When object is moved a distance $d$ away from the mirror, the magnification changes
$m = \frac{s''}{s+d}$ (3)
with $m$ = 0.5.
And the relation for these distances is
$\frac{1}{s+d}+\frac{1}{s''}=\frac{1}{f}$ (4)
From (1) and (2), eliminating term $s'$,
$\frac{s}{f} = 1+\frac{1}{k}$ (5)
From (3) and (4), eliminating term $s''$,
$\frac{s}{f}=1+\frac{1}{m}-\frac{d}{f}$ (6)
Equaling (5) and (6), and separating $f$,
$f = \frac{mkd}{k-m} = 48\; \rm{cm}$

Answer

$f$ = 48 cm