Question

A compound lens system consists of two converging lenses, one at x=−20.0cm with focal length f1=+10.0cm, and the other at x=+20.0cm with focal length f2=+8.00cm. (Figure 1)An object 1.00 centimeter tall is placed at x=−50.0cm. What is the location of the final image produced by the compound lens system? Give the x coordinate of the image.

Express your answer in centimeters, to three significant figures or as a fraction.

Answer 1

The final image is formed at
`\(x = 31.8 \text{ cm}\)`

How to determine he location of the final image produced by the compound lens system?

First Lens:

Given:

`\(f_1\)` (focal length of the first lens) = 10 cm

`\(d_o\)` (object distance) = -20 - (-50) = 30 cm

Using the lens equation:

`\[ (1)/(d_i) + (1)/(d_o) = (1)/(f_1) \]`

Solving for
`\(d_i\)`:

`\[ (1)/(d_i) + (1)/(30) = (1)/(10) \]`

`\[ \Rightarrow d_i = 15 \text{ cm} \]`

The first lens forms the image at a distance of 15 cm to its right. Therefore, the position of the image formed by the first lens is
`\(x = -20 + 15 = -5 \text{ cm}\)`.

This image acts as the object for the second lens.

Second Lens:

Given:

`\(f_2\)` (focal length of the second lens) = 8 cm

`\(d_o\)` (object distance) = 20 - (-5) = 25 cm

Using the lens equation:

`\[ (1)/(d_i) + (1)/(d_o) = (1)/(f_2) \]`

Solving for
`\(d_i\)`:

`\[ (1)/(d_i) + (1)/(25) = (1)/(8) \]`

`\[ \Rightarrow d_i = 11.8 \text{ cm} \]`

The second lens forms the image at a distance of 11.8 cm to its right. Therefore, the position of the image formed by the second lens is
`\(x = 20 + 11.8 = 31.8 \text{ cm}\)`.

Hence, the final image is formed at
`\(x = 31.8 \text{ cm}\)`.

See image for missing part of the question.