Question

A concave mirror (f₁ = 13.0 cm) and a convex mirror (f₂ = −5.80 cm) are facing each other and are separated by a distance of 35.0 cm. An object is placed between the mirrors and is 17.5 cm from each mirror. Consider the light from the object that reflects first from the concave mirror and then from the convex mirror. What is the distance of the image (d_i2) produced by the convex mirror?

Answer 1

The object for the convez mirror is the image reflected by the concave mirror. The image of the concave mirror is given by

$\frac{1}{s_1}+\frac{1}{s_1'}=\frac{1}{f_1}$

$\frac{1}{s_1'}=\frac{1}{f_1}-\frac{1}{s_1}$

$\frac{1}{s_1'}=\frac{1}{13.0cm}-\frac{1}{17.5cm}$

$s_1'=50.55cm$

That is the image for the convex mirror, but the distanc from the S'1 to the convex mirror is

$s_2=35cm-s'_1$

$s_2=35cm-50.55cm$

$s_2=-15.55cm$..........is negative because is at the back of the convex mirror.

Then we aplly the same mirror equation

$\frac{1}{s_2'}=\frac{1}{f_2}-\frac{1}{s_2}$

$\frac{1}{s_2'}=\frac{1}{-5.8cm}-\frac{1}{-15.55cm}$

$s_2'=-9.25cm$.........The image is at the back of the mirror

The distance from the mirror is $|s_2'|=9.25cm$