Inconsistent Expiremental Data of Two-Lens System

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Homework Help Overview

The discussion revolves around calculating the focal length of a diverging lens in a two-lens system, which includes a converging lens of known focal length. Participants are analyzing experimental data related to the image distances and object distances in the setup.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the method of calculating the final image in a multi-lens system and question the validity of their results based on the expected behavior of diverging lenses. There is confusion regarding the sign conventions for focal lengths and image distances, particularly in relation to virtual images.

Discussion Status

Some participants have offered insights into the nature of the lens setup, suggesting it resembles a Galilean telescope, which can produce virtual images. Others are exploring the calculations and the implications of treating the image from the first lens as an object for the second lens, while questioning the assumptions made about the lens types and their configurations.

Contextual Notes

There are indications that the lenses may not be in contact, which could affect the calculations. Additionally, participants are grappling with the implications of their experimental data, which appears to contradict established expectations about the behavior of diverging lenses.

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Homework Statement



There is a problem in which we are supposed to calculate the focal length of a diverging lens by putting it in series with a converging lens of known focal length and measuring the resulting image.

Distances of objects all from a single common reference point:

Object: 100cm
Convex Lens: 55.9cm
Image from Convex Lens (this was found by removing diverging lens): 18.85
Diverging Lens: 31.5cm
Final Image: 6.5cm

Previously calculated focal length of converging lens has yielded answers between 19.7 and 20.4, so I will use 20.


Homework Equations



1/f = 1/o + 1/i

My understanding is that f is negative if it's a diverging lens or mirror and the image distance is negative if it's a virtual image (meaning it's right side up).


The Attempt at a Solution



This problem drives me nuts because everything I've heard in class and found on the internet says that you can calculate the final image in a multi-lens system by finding the image of the first lens and then using that as the object for the second lens, and also that the image of a diverging lens ALWAYS right side up and smaller. However this experimental data would indicate that the diverging lens magnified the image of the converging lens, which is supposed to be impossible.

Math: When you put the first image distance and object distance in terms of the converging lens and use the lens equation you get that the focal length of the lens is 20.13, which is consistent with my expectation. However when I do this with the diverging lens I get that the focal length is 25.6. I thought the focal length was supposed to be negative for diverging lenses? When I try to draw the ray diagram I have to do it like a converging lens to make it work. Should I conclude that we actually used 2 converging lenses rather than a diverging lens? That would be a very obnoxious result but I guess the guy who set it up could have switched the lenses. Please try this yourself to see if I'm doing it right.

Data in terms of converging lens:
Object: 44.1cm
Image 1: 37.05cm

Data in terms of converging lens:
Image 1: 12.65cm
Image 2: -25cm
 
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Your lens setup is known as the Galilean telescope. It does magnify objects, and it does produce virtual images.
 
So is the second lens divergent or convergent? If it's divergent, what did I do wrong?
 
It is not clear how you compute the focal length. The equation you mentioned is valid only for lenses that are in contact with each other, but your data seem to indicate they are at a distance.

I think you should show your calculation entirely to avoid any confusion.
 
The way I tried to do it was to calculate the image for only the first lens and then to use that as the object for the second lens. So I used the equation for just one lens twice. I talked to my TA and he says that the distance for image 1 from the diverging lens is actually negative because it is like a "virtual object" because the first lens made it, which explains the odd ray diagram technique that was necessary to produce my image.

The TA said however that you should trace both lenses together. One line can come from the ray that comes out parallel from the converging lens, which travels away from the focus of the diverging lens when it passes through. Then you can use the ray that goes through the focus at the backside of the converging lens by drawing a parallel ray that goes from the diverging lens' focus, which means it will come out out of the diverging lens parallel. Apparently parallel rays cross each other in the plane of the focus, so you can use the new ray you created out of the incoming focus of the diverging lens to find where the original ray that came out of the converging lens crosses it when it passes through the diverging lens. That gives you the path of a second line, which will give you your image. I know this is hard to see just by type. Sorry :(.
 

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