Question

An object which is 5.0cm high is placed 15.0cm left of a convex lens with a focal length of 10.cm. A second lens which is placed 48cm to the right of the first lens and has focal length of 10cm. Determine the location of the final image, the magnification of the final image, and the height of the final image.

Answer 1

$\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}\\s'=\frac{s_1f_1}{s_1-f_1}$

Put value

$s'=\frac{15\times10}{15-10}=30cm\\real$

Magnification

$m_1=\frac{s'_1}{s_1}=-\frac{30}{15}=-2$ inverted

$s_2=-48cm$

$s'_2=\frac{s_2f_2}{s_2-f_2}\\s'_2=\frac{-48\times10}{-48-10}=8.275cm$

I₂= real

$$m_2=-\frac{s'_2}{s_2}=\frac{-8.275}{-48}=0.1724cm$$

Total magnification

$$M=m_1m_2={-2}\times0.1724=-0.3448cm$$