Inconsistent Expiremental Data of Two-Lens System

  • Thread starter Thread starter ShamelessGit
  • Start date Start date
  • Tags Tags
    Data System
Click For Summary
The discussion revolves around calculating the focal length of a diverging lens in a two-lens system involving a converging lens. Experimental data suggests that the diverging lens magnifies the image from the converging lens, contradicting established principles that state diverging lenses produce smaller, right-side-up images. The calculations yield a focal length of 25.6 for the diverging lens, raising questions about its classification as a diverging lens. A teaching assistant advises that the image distance for the diverging lens should be treated as negative, indicating a virtual object, and suggests tracing both lenses together for accurate results. The complexity of the ray diagrams and calculations highlights potential confusion in multi-lens systems.
ShamelessGit
Messages
38
Reaction score
0

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
 
Physics news on Phys.org
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 :(.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
996
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 11 ·
Replies
11
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K