Finding focal point of concave lens using concave and convex

[TheCapacitor]

Homework Statement

[/B]
I was doing http://stao.ca/VLresources/sci-tie-data/lessons/1400_1499/DivergingLensExperimentDeta.pdf experiment. Let's look at this image:

Suppose we get the imaginary object at dv by convergence lens. And this object is like a real object for the concave lens. Then how do we get a real object on the screen? It is known that only virtual objects can be made on the screen by convex lenses, and the image should be before the real object and after the lens.

Homework Equations

[/B]
According to TLE (Thin lens equation) we can find the focal point, however I'm confused about what is real and imaginary here.

The Attempt at a Solution

I had tried hard to draw ray diagram for this one. No success in forming an imaginary image in dv or real image at dr.

 

[BvU]

Homework Statement

[/B]
Suppose we get the imaginary object at dv by convergence lens. And this object is like a real object for the concave lens. Then how do we get a real object on the screen? It is known that only virtual objects can be made on the screen by convex lenses, and the image should be before the real object and after the lens.

Homework Equations

[/B]
According to TLE (Thin lens equation) we can find the focal point, however I'm confused about what is real and imaginary here.

The Attempt at a Solution

I had tried hard to draw ray diagram for this one. No success in forming an imaginary image in dv or real image at dr.

It's called a virtual image, so I'd also call it a virtual object . But for the concave lens it indeed can be treated as a real object, but with a negative object distance in the formula. And that way you can get a positive image distance, so a real image ! Not from a concave lens on its own, but from a combination of convex and concave. You could casually say that the converging rays from the convex lens are diverged a little by the concave lens, thereby 'pushing' the image further away from the convex lens.

And now for the drawing: for the convex lens you already know:

• in parallel to the axis $\Rightarrow$ out through focal point
• in through focal point $\Rightarrow$ out parallel to the axis
• in through the center $\Rightarrow$ straight on

And sure enough it's the same for the convex lens, only the focal point is at the other side of the lens. See your figure 2 (which only shows one of the three) or 3 (which shows 2 of them -- can you draw the third one ? )

Your first step is to draw the image from the concave lens - as if the convex lens wasn't there. d_i in figure 4 gives you d_v in figure 5. As it says in Step3...

 

[TheCapacitor]

It's called a virtual image, so I'd also call it a virtual object . But for the concave lens it indeed can be treated as a real object, but with a negative object distance in the formula. And that way you can get a positive image distance, so a real image ! Not from a concave lens on its own, but from a combination of convex and concave. You could casually say that the converging rays from the convex lens are diverged a little by the concave lens, thereby 'pushing' the image further away from the convex lens.

And now for the drawing: for the convex lens you already know:

• in parallel to the axis $\Rightarrow$ out through focal point
• in through focal point $\Rightarrow$ out parallel to the axis
• in through the center $\Rightarrow$ straight on

And sure enough it's the same for the convex lens, only the focal point is at the other side of the lens. See your figure 2 (which only shows one of the three) or 3 (which shows 2 of them -- can you draw the third one ? )

Your first step is to draw the image from the concave lens - as if the convex lens wasn't there. d_i in figure 4 gives you d_v in figure 5. As it says in Step3...

Is this correct?
The object on the screen is real object for the concave lense and the object before the screen is virtual object for the concave lens.

 

[TheCapacitor]

It's called a virtual image, so I'd also call it a virtual object . But for the concave lens it indeed can be treated as a real object, but with a negative object distance in the formula. And that way you can get a positive image distance, so a real image ! Not from a concave lens on its own, but from a combination of convex and concave. You could casually say that the converging rays from the convex lens are diverged a little by the concave lens, thereby 'pushing' the image further away from the convex lens.

And now for the drawing: for the convex lens you already know:

• in parallel to the axis $\Rightarrow$ out through focal point
• in through focal point $\Rightarrow$ out parallel to the axis
• in through the center $\Rightarrow$ straight on

And sure enough it's the same for the convex lens, only the focal point is at the other side of the lens. See your figure 2 (which only shows one of the three) or 3 (which shows 2 of them -- can you draw the third one ? )

Your first step is to draw the image from the concave lens - as if the convex lens wasn't there. d_i in figure 4 gives you d_v in figure 5. As it says in Step3...

Is this correct?
The object on the screen is real object for the concave lense and the object before the screen is virtual object for the concave lens.

View attachment 105023

 

[BvU]

The focal point of the convex lens is not affected by the concave lens; you make it look different with the dashed lines.

Furthermore, you don't really construct the 'real' image from the convex lens (you only take the parallel rays; I can't see the size relation between object and image this way).

Back to the drawing board, and then we'll look at what the concave (the word converge lens must be a mistake ?) lens makes of this. And it doesn't make a real objecet but a real image on the screen. And not with a point size but with a specific nonzero size.

 

[TheCapacitor]

The focal point of the convex lens is not affected by the concave lens; you make it look different with the dashed lines.

Furthermore, you don't really construct the 'real' image from the convex lens (you only take the parallel rays; I can't see the size relation between object and image this way).

Back to the drawing board, and then we'll look at what the concave (the word converge lens must be a mistake ?) lens makes of this. And it doesn't make a real objecet but a real image on the screen. And not with a point size but with a specific nonzero size.

Yes, but the image created is created at the screen, where the rays meet, not where the dashed lines meet. The dashed line meeting is where image would have been if no concave lens.

 

[BvU]

I was doing http://stao.ca/VLresources/sci-tie-data/lessons/1400_1499/DivergingLensExperimentDeta.pdf experiment

Did you really observe an upright image ?
Your drawing gives the impression of just a bunch of random lines. Nothing matches the rules in post #2, the red lines are not in line with the black rays from the left. And I see nothing back from what you measured in Step 1.

Oops ?

 

[TheCapacitor]

Did you really observe an upright image ?
Your drawing gives the impression of just a bunch of random lines. Nothing matches the rules in post #2, the red lines are not in line with the black rays from the left. And I see nothing back from what you measured in Step 1.

Oops ?

I'm sorry, I tried drawing it multiple times, with no success. Acctually I'm sure the 2 top lines are not according your convention, but they are good. Third one is a mistake.

I found one diagram according to your conventions, but it looks really not accurate.
https://jeremynarag.wordpress.com/2016/02/26/determining-focal-lengths-of-concave-and-convex-lenses/
The rays that go from the concave lens are unchanged there :(

 

[BvU]

Fig. 3 there is kind of aa spoiler: they are doing your work! Convex lens imaging looks OK, for the concave lens I have to look again.
All their data is in the accompanying table. What are your results from preceding step ? And the settings for step 2 ?

 

[TheCapacitor]

Fig. 3 there is kind of aa spoiler: they are doing your work! Convex lens imaging looks OK, for the concave lens I have to look again.
All their data is in the accompanying table. What are your results from preceding step ? And the settings for step 2 ?

Let's stay outside the experiment and just try to get the right ray diagram. I think they have a mistake in the concave lens. If you can get a correct ray diagram or correct theirs, and explain me why they think the object on the screen is real and the one that we don't see is the virtual one, I would appriaciate it a lot.

 

[BvU]

Staying outside experiments is scientifically a very bad idea -- there's a lot of fact-free posts on PF to corroborate that. But never mind.

If you write 'I think they have a mistake in the concave lens' I can't help you further without more specifics from you.

[edit] ah, you mean in the spoiling link ? Do we really have to fix that too ?

real object for the concave lense is on the screen

Yes, the 'previously real' image from the concave lens solo on the previous (Step 1, d_i) location of the screen. It serves as an object for the concave lens. I suppose they call it virtual object because it's on the 'wrong' side of the concave lens.

In your experiment, if you have d_i and d, you have d_v.

Now, when I (finally) read the instructions, the aim of the experiment is to determine f_d, so it's a bit hard to make a correct ray diagram before working out that focal distance.