Question

In FIGURE P$35.30$, what are the position, height, and orientation of the final image? Give the position as a distance to the right or left of the lens.

Short answer

The position, height, and orientation of the final image is given below.

Step-by-step solution

Given information.

We need to find out the position, height, orient-tion, and a distance from right or left lens.

Simplify

As we can get the values from the diagram:

$$\begin{gathered} f_1=10cm \\ {f}_2=-30cm \\ h=1cm \\ {S}_1=5cm \end{gathered}$$

Here, $S_1$is the distance from lens, $f_1$is the focal length of lens, $h$is the height of the object,$f_2$is the focal length of the mirror and ${S^{\text{'}}}_1$ is the distance of the image from the lens.

First, Let us find the$S_1^{\text{'}}$

$$\begin{gathered} s_1^{\text{'}}=\frac{f_1{S}_1}{S_1-f_1} \\ {S}_1^{\text{'}}=\frac{\left(10\right)\left(5\right)}{5-10} \\ {S}_1^{\text{'}}=\frac{50}{-5} \\ {S}_1^{\text{'}}=-10cm \end{gathered}$$

Calculation.

Now, Let us find out $S_2^{\text{'}}$

$$\begin{gathered} S_2^{\text{'}}=\frac{f_2{S}_2}{S_2-f_2} \\ {S}_2^{\text{'}}=\frac{(-30)\left(15\right)}{15-(-30)} \\ {S}_2^{\text{'}}=\frac{-450}{45} \\ {S}_2^{\text{'}}=-10cm \end{gathered}$$

Finally, Let us find $S_3^{\text{'}}$, Since$f_1=f_3=10cm$

$$\begin{gathered} S_3^{\text{'}}=\frac{f_3{S}_3}{S_3-f_3} \\ {S}_3^{\text{'}}=\frac{\left(10\right)\left(15\right)}{15-10} \\ {S}_3^{\text{'}}=\frac{150}{5} \\ {S}_3^{\text{'}}=30cm \end{gathered}$$

Finally, we get$h^{\text{'}}$from next relation:

$$\begin{gathered} h^{\text{'}}=h\frac{S_1^{\text{'}}}{S_1}\frac{S_2^{\text{'}}}{S_2}\frac{S_3^{\text{'}}}{S_3} \\ {h}^{\text{'}}=(1.0cm)\frac{-10cm}{5.0cm}\frac{-10cm}{15cm}\frac{30cm}{15cm} \\ {h}^{\text{'}}=2.7cm \end{gathered}$$.