A Two-Lens System

Part A = 31.8 cm

Part B = 0.236 cm

Part C = Upright

Part D = 19.42 cm

Part E = The image is a different size and oriented differently.

Solution Below:

_{1}= +10.0 cm, and the other at x = +20.0 cm with focal length f

_{2}= +8.0 cm . An object 1.00 centimeter tall is placed at x = -50.0 cm.

Part A

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

Start by finding V_{2}:

v_{2} = 1/(1/f_{2} – 1/u_{2})

v_{2} = 11.8 cm

Add v_{2} to the position of the second lens gives x = 31.8 cm

31.8 cm

Part B

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

Use the formula v_{1} * v_{2} /(u_{1} u_{2})

This gives a height of 0.235 cm

Note: MasteringPhysics wants an answer of 0.236 cm

0.236 cm

Part C

Upright

Part D

_{s}at x = 0. We want to choose this new lens so that it produces an image

at the same location as before.

What is the focal length of the new lens at the origin?

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

The object is 50 cm from the new lens and has an image distance of 31.76 cm. This gives an answer of 19.42 cm.

19.42 cm

Part E

_{s}the same size as the image formed by the compound lens system? Does it have the same orientation?

The object is 50 cm from the new lens and has an image distance of 31.76 cm. This gives an answer of 19.42 cm.

The image is a different size and oriented differently.

The answer to B is 4/17 cm = 0.235 cm and not 400/17 cm = 23.5 cm. However, the part about mastering physics wanting 0.236 cm is true, but the fraction seems to work just fine.

Yes indeed- thanks for the catch. If the fraction works, that’s great!

At part B the answer is 0.236 cm, not 23.6

Good catch, sorry for the confusion!