Question

A 2.0-cm-tall object is $20cm$to the left of a lens with a focal length of$10cm$A second lens with a focal length of $15cm$is$30cm$to the right of the first lens.

a. Use ray tracing to find the position and height of the image. Do this accurately using a ruler or paper with a grid, then make measurements on your diagram.

b. Calculate the image position and height. Compare with your ray-tracing answers in part a.

Short answer

a. The position and height of the image is given below in step.

b. The position of the image is $20\mathrm{cm}$away to the first lens.

Step-by-step solution

Part (a) step 1: Given Information 

We need to find the position and height of the image use ray tracing.

Part (a) step 2: Simplify

Consider :

$$\begin{gathered} S_1=20cm \\ {f}_1=10cm \end{gathered}$$

Part (b) step 1: Given information

We need to calculate the image position and height.

Part (b) step 2: Simplify

Finding$S_1^{\text{'}}$for the first lens:

$\begin{array}{rr}\frac{1}{S_1}+\frac{1}{S_1^{\text{'}}}& =\frac{1}{f_1}\\ S_1^{\text{'}}& =\frac{f_1{S}_1}{S_1-f_1}\\ S_1^{\text{'}}& =\frac{\left(10\text{cm}\right)\left(20\text{cm}\right)}{20\text{cm}-10\text{cm}}\\ S_1^{\text{'}}& =\frac{200\text{cm}}{10\text{cm}}\\ S_1^{\text{'}}& =20\text{cm}\end{array}$

Magnification for first lens is :

$m_1=-\frac{S_1^{\text{'}}}{S_1}=-\frac{20\text{cm}}{20\text{cm}}=-1$

For the second lens $S_2^{\text{'}}$:

$\begin{array}{rr}\frac{1}{S_2}+\frac{1}{S_2^{\text{'}}}& =\frac{1}{f_2}\\ S_2^{\text{'}}& =\frac{f_2{S}_2}{S_2-f_2}\\ S_2^{\text{'}}& =\frac{\left(15\text{cm}\right)\left(10\text{cm}\right)}{10\text{cm}-15\text{cm}}\\ S_2^{\text{'}}& =\frac{150\text{cm}}{-5\text{cm}}\\ S_2^{\text{'}}& =-30\text{cm}\end{array}$

Magnification of the second lens is :

$m_2=-\frac{S{\text{'}}_2}{S_2}=-\frac{\left(-30cm\right)}{10cm}=3cm$

Now, total magnification is:

$M=m_1\times m_2=\left(-1\right)\times \left(3\right)=-3$