Geometrical optics. Photometry. Quantum nature of light > Optical systems > Problem 13.3.7

Statement

$13.3.7.$ The image of an object on the frosted glass of the camera when photographing from a distance of 15 m reaches a height of 30 mm, and from a distance of 9 m - a height of 51 mm. Find the focal length of the lens.

Solution

Let's define some magnitudes:
$s_1$ = 15 m, first distance between object and camera
$y_1'$ = 30 mm, size of image on camera at first take
$s_2$ = 9 m, second distance between object and camera
$y_2'$ = 51 mm, size of image on camera at second take

It's known that
$\frac{1}{s_1}+\frac{1}{s_1'}=\frac{1}{f}$ (1)
and the magnification is
$\frac{y_1'}{y}=\frac{s_1'}{s_1}$ (2)
and for second take, magnification is
$\frac{y_2'}{y}=\frac{s_2'}{s_2}$ (3)
and relation between distances also is given by:
$\frac{1}{s_2}+\frac{1}{s_2'}=\frac{1}{f}$ (4)
where $y$ is real size of the object and $s_i'$ ($i$ = 1,2) are the image-camera distance for each take.
Equaling (2) and (3):
$\frac{y_1's_1}{s_1'}=\frac{y_2's_2}{s_2'}$ (5)
Equaling (1) and (4):
$\frac{1}{s_1}+\frac{1}{s_1'}=\frac{1}{s_2}+\frac{1}{s_2'}$ (6)
Separating $s_2'$ from (5):
$s_2' = \frac{y_2's_2s_1'}{y_1's_1}$ (7)
Putting (7) into (6):
$\frac{1}{s_1'} = \frac{y_2'(s_1-s_2)}{s_1(y_2's_2-y_1's_1)}$ (8)
Substituting (8) into (1), and developing
$f = \frac{y_2's_2-y_1's_1}{y_2'-y_1'} = 3/7\; \rm{m}$

Answer

$f$ = 3/7 m