Find Focal Length of Diverging Lens | Physics Homework Help

[kent davidge]

Homework Statement

When an object is placed at the proper distance to the left of a converging lens, the image is focused on a screen 30.0 cm to the right of the lens. A diverging lens is now placed 15.0 cm to the right of the converging lens, and it is found that the screen must be moved 19.2 cm farther to the right to obtain a sharp image. What is the focal length of the diverging lens?

Homework Equations

The Attempt at a Solution

(sorry my poor english) I don't know how to solve this problem using calculus. Then I've used the trick shown below and I've found approximately the correct answer (the error was caused by an incorrect alignment in the sketch). My doubt is if it's correct to use this method for finding the solution. Since I got the right answer, I think it is valid. But also, I was unable to draw the image in the second situation, it would be at the location of the blue dot... but the rays does not intercept each other.

 

[haruspex]

It's just that you've made the second lens too powerful.
First, redraw it so that the second lens is vertical and centred on the same horizontal axis. Make it taller if necessary.
Take the lower ray as it was for the single lens, and just deflect it very slightly lower as it passes through the second lens. You should now find it intercepts the upper ray a little further to the right.

 

[kent davidge]

oh ok
how can I find the focal length without drawing that diagram?

 

[haruspex]

oh ok
how can I find the focal length without drawing that diagram?

See the equations at https://en.m.wikipedia.org/wiki/Lens_(optics)#Compound_lenses

 

[kent davidge]

I'll check these equations.
See my new sketch. Is it now correct?

 

[haruspex]

I'll check these equations.
See my new sketch. Is it now correct?

No, that's worse!
Go back to your previous diagram. Straighten up the second lens. Remove the dashed line and the line below it. Continue the lower ray from the object straight through the second lens as a dashed line. This should make it the same as the lower ray in the single lens drawing, meeting the upper ray at the same point as before. This shows where the image used to be.
Where the dashed line starts in the second lens, draw a solid line at a slightly steeper angle. This should also meet the upper ray, but further away. This shows the new image.

 

[kent davidge]

ohh it works
but the focal point should lie in the same axis as the lens, it's not so?

 

[haruspex]

ohh it works
but the focal point should lie in the same axis as the lens, it's not so?

Yes, that's the diagram, but what you have labelled as the focal point isn't. That point is the apex of the image from the single lens. The focal point is where rays parallel to the axis converge. Consequently, yes, it will always lie on the axis.
You happen to have placed the second lens almost at the focal point of the first. That might be misleading in trying to figure out the combined focal point. But I don't think you need to find that. Use the equation at the link I posted to relate the focal length of the second lens to where the new image is.

 

[UchihaClan13]

See the equations at https://en.m.wikipedia.org/wiki/Lens_(optics)#Compound_lenses

You can't use the thin lens separated by a distance formula
It's only applicable if the rays are parallel and come from infinity
Not for other cases

 

[UchihaClan13]

the image need not be point sized right??
And it needn't necessarily form on the focus?
Then just use the virtual object consideration

 

[UchihaClan13]

Did you get the final answer as -26.71 cm??
@kent davidge
If yes then i'll tell you my method

 

[UchihaClan13]

Okay so consider the image formed by the converging lens to be a virtual object for the diverging lens
Then just find out u,v w.r.t the diverging lens and plug them in the lens formula
Which will give you your required focal length?
UchihaClan13

 

[haruspex]

You can't use the thin lens separated by a distance formula
It's only applicable if the rays are parallel and come from infinity
Not for other cases

The formula I linked to only claims to give the new focal length, so in that sense the formula is applicable. But you are right that it does not provide a formula for the new image location. Your method (image of image) is definitely the way to go.

 

[kent davidge]

haruspex I understand it now! Thank you for your kind attention.
UchihaClan13 oh yes. Before I see your post I've worked on this problem several times and I got -26.71 cm by using the method you've mentioned.

haruspex and UchihaClan13, here is my final result:

and using the diagram, I found that 1 / f₂ = (1 / s'₂) + (1 / i₂). It was very gratifying...

 

[UchihaClan13]

The formula I linked to only claims to give the new focal length, so in that sense the formula is applicable. But you are right that it does not provide a formula for the new image location. Your method (image of image) is definitely the way to go

i completely agree with you on that
But what's the use of finding the focal length of a combination of lenses if you're being asked for say,something else?(If it's for only the focal length,then yes use it by all means just like the thin lenses in contact method)

Anyways it's correct
And @kent davidge
I am glad you used the correct method after your conceptually incorrect tries/methods
:)
UchihaClan13