The rays leaving the two-component optical system of FIGUREP$35.27$produce two distinct images of the$1.0cm$-tall object. what are the position (relative to the lens), orientation, and height of each image?

We have given that:

focal length of mirror$f_1=10cm$,

the distance from the mirror$S_1=5cm$.

We need to find a position(relative to the lens), orientation, and height of each image.

Firstly Let us find ${S_1}^{\text{'}}$:

$\frac{f_1{S}_1}{S_1-f_1}$ (Substitute values in equation.)

localid="1648896821479" $S{\text{'}}_1=\frac{\left(10cm\right)\left(5cm\right)}{5cm-10cm}$

localid="1648896899300" $S_1=\frac{50cm}{-5cm}$

localid="1648896910594" $S{\text{'}}_1=10cm$.

Here, ${S^{\text{'}}}_1$is the height of the first image, $f_1$is the focal length of the mirror and$S_1$is the distance from the mirror.

now let us findlocalid="1648896779745" $S{\text{'}}_2$:

localid="1648896921675" $S{\text{'}}_2=\frac{f_2{S}_2}{S_2-f_2}$ (substitute values in the equation.)

localid="1648896933269" $S{\text{'}}_2=\frac{\left(10cm\right)\left(20cm\right)}{20cm-10cm}$

localid="1648896942880" $S{\text{'}}_2=\frac{200cm}{10cm}$

localid="1648896955537" $S{\text{'}}_2=20cm$.

Here ${S^{\text{'}}}_2$is the height of the second image, $f_2$is the focal length of lens and $S_2$is the distance from the lens.