Question

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?

Short answer

The position (relative to the lens), orientation, and height of each image is$S\text{'}=10cmandS{\text{'}}_2=20cm$.

Step-by-step solution

Given information

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.

Simplify

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.

Graphical representation