Chapter 35: Q. 3 (page 1016)

A 2.0-cm-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 $15 c m$ 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

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

b. The position of the image is $20 cm$ 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 :

$S_{1} = 20 c m f_{1} = 10 c m$

Part (b) step 1: Given information

We need to calculate the image position and height.

Part (b) step 2: Simplify

Finding $S_{1}^{'}$ for the first lens:

$\begin{array}{r} \frac{1}{S_{1}} + \frac{1}{S_{1}^{'}} & = \frac{1}{f_{1}} \\ S_{1}^{'} & = \frac{f_{1} S_{1}}{S_{1} - f_{1}} \\ S_{1}^{'} & = \frac{(10 cm) (20 cm)}{20 cm - 10 cm} \\ S_{1}^{'} & = \frac{200 cm}{10 cm} \\ S_{1}^{'} & = 20 cm \end{array}$

Magnification for first lens is :

$m_{1} = - \frac{S_{1}^{'}}{S_{1}} = - \frac{20 cm}{20 cm} = - 1$

For the second lens $S_{2}^{'}$ :

$\begin{array}{r} \frac{1}{S_{2}} + \frac{1}{S_{2}^{'}} & = \frac{1}{f_{2}} \\ S_{2}^{'} & = \frac{f_{2} S_{2}}{S_{2} - f_{2}} \\ S_{2}^{'} & = \frac{(15 cm) (10 cm)}{10 cm - 15 cm} \\ S_{2}^{'} & = \frac{150 cm}{- 5 cm} \\ S_{2}^{'} & = - 30 cm \end{array}$

Magnification of the second lens is :

$m_{2} = - \frac{S'_{2}}{S_{2}} = - \frac{(- 30 c m)}{10 c m} = 3 c m$

Now, total magnification is:

$M = m_{1} \times m_{2} = (- 1) \times (3) = - 3$

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Short Answer

Step by step solution

Part (a) step 1: Given Information

Part (a) step 2: Simplify

Part (b) step 1: Given information

Part (b) step 2: Simplify

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