Question

A $2.0cm$-tall object is $20cm$to the left of a lens with a focal length of$10cm$. A second lens with a focal length of$5cm$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 using ray tracing is given below in step.

b. The position of the image is $20cm$away from the first lens.

Step-by-step solution

Part (a) step 1: Given Information

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

Part (a) step 2: Simplify

Consider:

$$\begin{gathered} f_1=10cm \\ {f}_2=5cm \end{gathered}$$

Part (b) step 1: Given Information

We need to calculate the image position and height.

Part (b) step 2:  Simplify

For first lens:

$$\begin{gathered} \frac{1}{S_1}+\frac{1}{S_1^{\text{'}}}=\frac{1}{f_1} \\ \frac{1}{S_1^{\text{'}}}=\frac{1}{f_1}-\frac{1}{S_1} \\ \frac{1}{S_1^{\text{'}}}=\frac{1}{10}-\frac{1}{20} \\ \frac{1}{S_1^{\text{'}}}=\frac{1}{20} \\ {S}_1^{\text{'}}=20\text{cm}\to realimage \end{gathered}$$

Next, finding value for magnification $M_1$:

$M_1=-\frac{S_1^{\text{'}}}{S_1}--\frac{20}{20}=-1$

For second lens:

$$\begin{gathered} \frac{1}{S_2}+\frac{1}{S_2^{\text{'}}}=\frac{1}{f_2} \\ \frac{1}{S_2^{\text{'}}}=\frac{1}{f_2}-\frac{1}{S_2} \\ \frac{1}{S_2^{\text{'}}}=\frac{1}{5}-\frac{1}{10} \\ {S}_2^{\text{'}}=10\text{cm}\to realimage. \end{gathered}$$

Next, we have to find the value for magnification $M_2$:

$M_2=-\frac{S_2^{\text{'}}}{S_2}--\frac{10}{10}=-1$

Finally, finding the total magnification $M$:

$M=M_1{M}_2=\left(-1\right)\left(-1\right)=1$