Question

A converging lens with a focal length of $40cm$and a diverging lens with a focal length of $-40cm$are $160cm$apart. A $2cm$tall object is $60cm$in front of the converging 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 ray tracing to find the position and height of the image is given below.

b. The position of the image is$120cm$away from the first lens.

Step-by-step solution

Part (a) 1:Given Information 

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

Part (a) step 2:Simplify

Consider:

$$\begin{gathered} S_1=60cm \\ {f}_8=40cm \end{gathered}$$

Part (b) step 1: Given Information

We need to calculate the image position and height.

Part (b) step 2: Simplify

First, finding to $S{\text{'}}_1$:

$$\begin{gathered} \frac{1}{S{\text{'}}_1}=\frac{1}{f_8}-\frac{1}{S_1} \\ \frac{1}{S{\text{'}}_1}=\frac{1}{40cm}-\frac{1}{60cm} \\ \frac{1}{S{\text{'}}_1}=\frac{1}{120cm} \\ S{\text{'}}_1=120cm \end{gathered}$$

Now, finding the magnification of the first lens:

$$\begin{gathered} m_1=\frac{S{\text{'}}_1}{S_1} \\ {m}_1=-\frac{120cm}{60cm}=-2 \end{gathered}$$

For the second lens $S{\text{'}}_2$:

$$\begin{gathered} \frac{1}{S{\text{'}}_2}=\frac{1}{S_2}+\frac{1}{f_2} \\ \frac{1}{S{\text{'}}_2}=\frac{1}{+40cm}+\frac{1}{-40cm} \\ S{\text{'}}_2=-20cm \end{gathered}$$

Next, magnification of the second lens :

$$\begin{gathered} m_2=-\frac{S{\text{'}}_2}{S_2} \\ {m}_2=-\frac{-20cm}{40cm}=+0.5 \end{gathered}$$

At last, total magnification of lens :

$M=m_1\times m_2=\left(-2\right)\times \left(0.5\right)=-1$