Chapter 35: Q. 32 (page 1018)

A $15 c m$ -focal-length converging lens is $20 c m$ to the right of a $7.0 c m$ -focal-length converging lens. A $1.0 c m$ -tall object is distance L to the left of the first lens.
a. For what value of L is the final image of this two-lens system halfway between the two lenses?
b. What are the height and orientation of the final image?

Short Answer

Expert verified

a. Value of L is $L = 14 c m$ .

b. The height and orientation of the final image is $M = - 1.7 c m$ .

Step by step solution

Part (a) Step 1: Given Information.

We need to find out the object is distance L of the first lens.

Part (a) Step 2: Calculation.

Since:

$f_{1} = 7 c m f_{2} = 15 c m$

Here, $f_{1}$ and $f_{2}$ is the focal length of first and second lens respectively.

Firstly, Let us find the initial position of the object for the second lens:

role="math" localid="1650112585802" $\frac{1}{S_{1}} + \frac{1}{S_{2}^{'}} = \frac{1}{f_{2}} \frac{1}{S_{2}^{1}} = \frac{1}{f_{2}} - \frac{1}{S_{2}} \frac{1}{S_{2}^{'}} = \frac{1}{15} - \frac{1}{- 10} \frac{1}{S_{2}^{'}} = \frac{1}{6} S_{2}^{'} = 6 c m \to r e a l i m a g e$

Here, $S_{2}$ and $f_{2}$ is distance of the object from the lens and the focal length of the second lens respectively.

Part (b) Step 1: Given Information.

We need to find the height and orientation of the final image.

Part (b) Step 2: Calculation.

Let us find Magnification:

$M = M_{1} M_{2} M = - \frac{S_{1}^{'}}{S_{1}} - \frac{S_{1}^{'}}{S_{2}} M = \frac{S_{1}^{'} S_{2}^{'}}{S_{1} S_{2}} M = \frac{(14) (- 10)}{(14) (6)} M = 1.7 c m$

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Short Answer

Step by step solution

Part (a) Step 1: Given Information.

Part (a) Step 2: Calculation.

Part (b) Step 1: Given Information.

Part (b) Step 2: Calculation.

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