Divergent thin lens producing real images

Question

The question is as in the title, can a divergent lens produce a real image, when backed by a convergent lens?

Which are the conditions to be respected?

This was a homework and I have received the solution but I think is wrong.

Basing this on research already done online, the only possibility looks like that a divergent lens should have as object a virtual image. Is it true?

EDIT: In the case of a convergent lens to the left, and a divergent lens to the right, with the light coming from the left, how would be?

Answer 1

I'm not certain what "backed by a convergent lens" means in this context.

A divergent lens by itself cannot form a real image, since a divergent lens has a negative focal distance. Use the thin lens equation:

$$\frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{f}$$

Since $f < 0$ and $d_o > 0$ by convention, $(\text{positive}) + \frac{1}{d_i} = (\text{negative})$ implies that $d_i < 0$, making the image a virtual one.

By combining a divergent lens with a convergent one, however, a real image can be formed in a variety of cases. For example, consider a setup in which an object is positioned to the left of a divergent lens of focal length $f_d$, followed to the right by a convergent lens of focal length $f_c < f_d$. This system would converge parallel rays of light into forming a real image.

Another example consists of a converging lens to the left of a diverging one. If the focus of the converging lens is located to the right of the diverging lens but to the left of the diverging lens' focus, a real image will form. Draw out a diagram for this setup, and see if you can determine why the image will be real.

Answer 2

Any discussion of the type of image a lens can form (real or virtual) must include information about the type of object that is being used.

A divergent lens, by itself, can form only a virtual image of a real object.

if we pre-condition the light from the object by passing it through a converging lens, then the resulting intermediate image can be a real or virtual object for the diverging lens.

Suppose that we have an object and a converging lens, such that there would be a real image located 20 cm from the converging lens. If we put another, diverging lens in the middle of this gap, 10 cm from the image and the converging lens, the real image becomes a virtual object for the second lens, which could result in a real image from the diverging lens...

Answer 3

You have two good fundamental answers, but a slightly different take on your answer is that yes, this is done all the time in optical systems to compensate for various distortions, aberrations and errors. A good example is an achromatic doublet, where a convex (converging) lens of one material is put in direct contact with a concave (diverging) lens along a surface of common curvature. Of course, the overall effect is converging. The refractive index as a function of wavelength is such that both materials is different, but they are chosen so that both materials the same index at some wavelength in the middle of the "achromatic band". Therefore, at that wavelength, the contacting surface has no optical effect at all. However, at wavelengths higher than this middle wavelength the refractive index, the converging lens has a higher index than the diverging lens, so that the surface between them has a slight nett converging effect. At wavelength shorter than the middle, the diverging lens has the higher index, and so the surface is nett diverging. This engineered variation in optical power is used to offset the overall variation of the system's optical power with wavelength, i.e. it is a mechanism to correct for chromatic aberration.