Can a real image be formed by a virtual image?

[Samuelriesterer]

That is, take the virtual image as the object for a second lens?

 

[M Quack]

Yes, that is quite common practice, e.g. in microscopes.

 

[Samuelriesterer]

I thought microscopes form the real image from the objective lens, then the viewing lens forms a virtual image from that real image. I was looking for the reverse.

 

[Drakkith]

I thought microscopes form the real image from the objective lens, then the viewing lens forms a virtual image from that real image. I was looking for the reverse.

Sure. You can place a diverging lens as the first lens in an optical system and then use a focusing lens to focus the light into a real image.

 

[sophiecentaur]

That is, take the virtual image as the object for a second lens?

Why not? It happens in your EYE, when you are wearing specs for short sight. or long sight. The description of Real of Virtual image is really only relevant when considering the overall behaviour of an optical system.
The problem here seems to be the old one of Classification getting in the way of Understanding. Who givesa about the name you can give an image within an optical system? When you are ready for it, study the systems for calculating the result of a complicated system of lenses (sign conventions vary from system to system`). Google "sign conventions" and see what you get.

 

[jtbell]

Example: Object at x = 0, lens #1 at x = 40, with focal length f₁ = 20. The image is at x = 80, right?

(Note "x" means position along the x-axis, i.e. optical axis, not object or image distance in the thin-lens equation.)

Now put lens #2 at x = 50, with focal length f₂ = 20. The image produced by lens #1 is the virtual object for lens #2, with object distance -30. Where is the image?

 

[Samuelriesterer]

Where is the image?

I find this topic entirely confusing so can you verify:

$s'_1 = \frac{f_1 s_1}{s_1 - f_1} = \frac{(20)(40)}{40-20} = 40$
$s_2 = (50-40) - s'_1 = 10 - 40 = -30$
$s'_2 = \frac{f_2 s_2}{s_2 - f_2} = \frac{(-20)(-30)}{-20+30} = -60$

 

[jtbell]

You need to distinguish carefully between the position x of an object, image, or lens, and the distance between a lens and an object (s) or an image (s'). A distance is the difference between two positions. It helps to draw a diagram.

$s'_1 = \frac{f_1 s_1}{s_1 - f_1} = \frac{(20)(40)}{40-20} = 40$

This is the distance of image #1 from lens #1. Lens #1 is located at x = 40. What is x for image #1?

$s_2 = (50-40) - s'_1 = 10 - 40 = -30$

This actually is the correct value for s₂, but I prefer to think of it as the difference between the position (x) of lens #2 and the position (x) of image #1, which is the object for lens #2.

$s'_2 = \frac{f_2 s_2}{s_2 - f_2} = \frac{(-20)(-30)}{-20+30} = -60$

I specified the focal length of lens #2 as 20, not -20.

 

[Samuelriesterer]

What is x for image #1?

The x position for image #1 is x = 80.

I specified the focal length of lens #2 as 20, not -20.

Ok:

$s'_2 = \frac{f_2 s_2}{s_2 - f_2} = \frac{(20)(-30)}{20+30} = 12$

So the final image distance is 12, and its x position is: x = 62?

 

[jtbell]

You got it!

So for lens #2, you have a virtual object, but a real image.

 

[Samuelriesterer]

Can you please clarify because it looks like the first image formed by the first lens is real because the object is outside the focal length.

 

[M Quack]

But the rays never get to the first image, because they are intercepted by the second lens first. So if you put a screen at the position of the first image (with the second lens in place) you will not see the image.

In any case, the distinction between real and virtual image is not very helpful.

 

[Samuelriesterer]

But the rays never get to the first image, because they are intercepted by the second lens first. So if you put a screen at the position of the first image (with the second lens in place) you will not see the image.

In any case, the distinction between real and virtual image is not very helpful.

OK I think I get it now. Yea it seems misleading that the virtual image is the object for the second lens because the rays never make it there. I guess I was picturing something different as if rays eminated from the virtual image, which of course is impossible. That is why I was struggling with the possibility of this scenerio.

 

[jtbell]

it looks like the first image formed by the first lens is real because the object is outside the focal length.

Yes, the image formed by the first lens is real. However, it acts as a virtual object for the second lens. "Real" and "virtual" are relative to each individual lens.

One way to keep things straight is to look at whether the light is converging or diverging as it enters or leaves the lens. If the light entering a lens is diverging, the object is real, as far as that lens is concerned. If the light entering is converging, the object is virtual. If the light leaving a lens is converging, the image is real, as far as that lens is concerned. If the light leaving is diverging, the image is virtual.

Notice what happens if we leave the original object and lens #1 where they are, in our example, but move lens #2 to x = 100, beyond the image formed by lens #1. Now that image becomes a real object for lens #2, because the light that was converging towards the first image has passed through it and is now diverging when it reaches lens #2.

 

[Samuelriesterer]

the image is real, as far as that lens is concerned

What does this mean because a converging lens can still form a virtual object if the object is within the focal length.