Chapter 35: Q. 5 (page 1017)

A $2.0$ -tall object is $20 c m$ to the left of a lens with a focal length of $10 c m$ . A second lens with a focal length of $- 5 c m$ is $30 c m$ 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

Expert verified

a. The position and height of the image using ray tracing are given below in step.

b. The position of the image is $20 c m$ 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:

$f_{1} = 10 c m f_{2} = - 5 c m$

Part (b) step 1: Given Information

We need to calculate the image position and height.

Part (b) step 2: Simplify

For the first lens:

$\frac{1}{S_{1}} + \frac{1}{S'_{1}} = \frac{1}{f_{1}} \frac{1}{S'_{1}} = \frac{1}{f_{1}} - \frac{1}{S_{1}} \frac{1}{S'_{1}} = \frac{1}{10} - \frac{1}{20} S'_{1} = 20 c m \to r e a l i m a g e$

Next, finding the value for magnification $M_{1}$ :

$M_{1} = - \frac{S'_{1}}{S_{1}} = - \frac{20}{20} = - 1$

For the second lens:

$\frac{1}{S_{2}} + \frac{1}{S'_{2}} = \frac{1}{f_{2}} \frac{1}{S'_{2}} = \frac{1}{f_{2}} - \frac{1}{S_{2}} \frac{1}{S'_{2}} = \frac{1}{- 5} - \frac{1}{10} S'_{2} = - 3.3 c m \to r e a l i m a g e$