# Thread: image of the moon

1. ## image of the moon

You wish to form an image of the moon, diameter 3500 km, a distance of 380 000 km away, with a reflecting telescope. If the image needs to be 4.5 cm in diameter, find the required focal length of the mirror in the telescope.
Formulas are
$\frac{1}{f}=\frac{1}{di}+\frac{1}{do}$
$\frac{hi}{ho}=\frac{-di}{do}$
F stands for focal point, di is distance to the image, do is distance to the object, hi is height of the image and ho is height of the object.

$\frac{4.5*10^-5}{3500}=\frac{-di}{380000}$
$f=-4.88*10^-3$
$\frac{1}{f}=\frac{1}{380000}+\frac{1}{-4.88*10^-3}$
$f= -4.88*10^-3 km$
is this correct???

2. Originally Posted by iamanoobatmath
Formulas are
$\frac{1}{f}=\frac{1}{di}+\frac{1}{do}$
$\frac{hi}{ho}=\frac{-di}{do}$
F stands for focal point, di is distance to the image, do is distance to the object, hi is height of the image and ho is height of the object.

$\frac{4.5*10^-5}{3500}=\frac{-di}{380000}$
$f=-4.88*10^-3$
$\frac{1}{f}=\frac{1}{380000}+\frac{1}{-4.88*10^-3}$
$f= -4.88*10^-3 km$
is this correct??? Yes
Since $d_O$ is very large the image must be placed very near the focus. Therefore $d_i = f$.