Focal Length - please check my work and tell me what's wrong.

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Homework Help Overview

The discussion revolves around a two-lens system involving a diverging lens and a converging lens. The original poster presents a problem where a small object is positioned 25.0 cm from the diverging lens, and the converging lens is located 30.0 cm to the right, with the system forming a real inverted image 17.0 cm to the right of the converging lens. The focus is on determining the focal length of the diverging lens based on the given distances and lens types.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • The original poster attempts to apply the lens formula for both lenses but initially uses an incorrect focal length for the converging lens. Participants raise questions about the accuracy of the focal length used and the implications of negative distances in the context of focal lengths.

Discussion Status

Some participants have pointed out errors in the calculations and assumptions, particularly regarding the focal length of the converging lens. The discussion reflects an ongoing exploration of the problem, with participants questioning the validity of the results based on the definitions of focal lengths for diverging lenses.

Contextual Notes

There is a noted confusion regarding the focal lengths and the interpretation of negative distances, particularly in relation to the nature of diverging lenses. The problem context includes specific distances and lens types that are critical to the calculations being discussed.

vucollegeguy
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Homework Statement


A small object is 25.0 cm from a diverging lens as shown in the figure . A converging lens with a focal length of 11.4 cm is 30.0cm to the right of the diverging lens. The two-lens system forms a real inverted image 17.0 cm to the right of the converging lens.

The Attempt at a Solution


1/f = 1/o + 1/i at every step

conv lens

1/12 = 1/o + 1/17.0

o = 40.8

This means the second object, which is the first image, was 40.8 cm to the left of the conv lens.

This means the second object was 40.8 - 30 = 10.8 cm to the left of the div lens

So the image distance for the div lens is -10.8 and

1/f = 1/25.0 + 1/(-10.8)

1/f = 0.04 - 0.0926 = -0.05259

f = -19.0 cm is the focal length of the div lens
 
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Hello vucollegeguy,

vucollegeguy said:

Homework Statement


A small object is 25.0 cm from a diverging lens as shown in the figure . A converging lens with a focal length of 11.4 cm is 30.0cm to the right of the diverging lens. The two-lens system forms a real inverted image 17.0 cm to the right of the converging lens.

The Attempt at a Solution


1/f = 1/o + 1/i at every step

conv lens

1/12 = 1/o + 1/17.0

o = 40.8

Where do you get a length of 12 cm? :rolleyes: The problem statement says that the converging lens has a focal length of 11.4 cm. You might want to correct this before you move on. 'Turns out it makes a significant difference in the end.
 
I changed 12 to 11.4. My final answer came out to be -5.6 but distance can't be negative. Is my answer correct?
 
vucollegeguy said:
I changed 12 to 11.4. My final answer came out to be -5.6 but distance can't be negative. Is my answer correct?

Correct me if I'm wrong, but I thought you were solving for the focal length of the diverging lens. Focal lengths of diverging lenses are negative by definition.
 

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