Question

A $15cm$-focal-length converging lens is $20cm$to the right of a $7.0cm$-focal-length converging lens. A $1.0cm$-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

a. Value of L is $L=14cm$.

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

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:

$$\begin{gathered} f_1=7cm \\ {f}_2=15cm \end{gathered}$$

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" $$\begin{gathered} \frac{1}{S_1}+\frac{1}{S_2^{\text{'}}}=\frac{1}{f_2} \\ \frac{1}{S_2^1}=\frac{1}{f_2}-\frac{1}{S_2} \\ \frac{1}{S_2^{\text{'}}}=\frac{1}{15}-\frac{1}{-10} \\ \frac{1}{S_2^{\text{'}}}=\frac{1}{6} \\ {S}_2^{\text{'}}=6cm\to realimage \end{gathered}$$

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:

$$\begin{gathered} M=M_1{M}_2 \\ M=-\frac{S_1^{\text{'}}}{S_1}-\frac{S_1^{\text{'}}}{S_2} \\ M=\frac{S_1^{\text{'}}{S}_2^{\text{'}}}{S_1{S}_2} \\ M=\frac{\left(14\right)(-10)}{\left(14\right)\left(6\right)} \\ M=1.7cm \end{gathered}$$