Chapter 35: Q. 2 (page 1016)

A converging lens with a focal length of $40 c m$ and a diverging lens with a focal length of $- 40 c m$ are $160 c m$ apart. A $2 c m$ tall object is $60 c m$ 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

Expert verified

a. The ray tracing to find the position and height of the image is given below.

b. The position of the image is $120 c m$ 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:

$S_{1} = 60 c m f_{8} = 40 c m$

Part (b) step 1: Given Information

We need to calculate the image position and height.

Part (b) step 2: Simplify

First, finding to $S'_{1}$ :

$\frac{1}{S'_{1}} = \frac{1}{f_{8}} - \frac{1}{S_{1}} \frac{1}{S'_{1}} = \frac{1}{40 c m} - \frac{1}{60 c m} \frac{1}{S'_{1}} = \frac{1}{120 c m} S'_{1} = 120 c m$

Now, finding the magnification of the first lens:

$m_{1} = \frac{S'_{1}}{S_{1}} m_{1} = - \frac{120 c m}{60 c m} = - 2$

For the second lens $S'_{2}$ :

$\frac{1}{S'_{2}} = \frac{1}{S_{2}} + \frac{1}{f_{2}} \frac{1}{S'_{2}} = \frac{1}{+ 40 c m} + \frac{1}{- 40 c m} S'_{2} = - 20 c m$

Next, magnification of the second lens :

$m_{2} = - \frac{S'_{2}}{S_{2}} m_{2} = - \frac{- 20 c m}{40 c m} = + 0.5$

At last, total magnification of lens :

$M = m_{1} \times m_{2} = (- 2) \times (0.5) = - 1$

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Short Answer

Step by step solution

Part (a) 1:Given Information

Part (a) step 2:Simplify

Part (b) step 1: Given Information

Part (b) step 2: Simplify

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