Thin lens and images

For a thin convex lens and paraxial rays the thin-lens equation 1/u + 1/v = 1/f fixes the image. The object distance u is always positive, the focal length f of a convex lens is positive, and the image distance v is positive for a real image on the far side and negative for a virtual image on the same side as the object. The lateral magnification is m = −v/u, so m < 0 means an inverted real image and m > 0 an upright virtual image. As the object approaches the lens the image moves, changes size and can switch between real and virtual; at u = f the outgoing rays are parallel and no finite image forms.

Questions

As an object moves closer to a convex lens, how does the position of its image change?

Why is the real image larger than the object when the object distance lies between f and 2f?

At u = f, why can no sharp image be formed on a screen placed anywhere?

Predict

Before moving anything, predict whether the image distance v grows or shrinks when u falls from 2f to 1.5f, and whether the image grows or shrinks. Write your prediction down and compare it with the experiment.

Try

Keep f = 0.10 m and h = 0.03 m and change only the object distance u. Try u = 2f, then u = 1.5f, then u = f, then u = 0.5f, recording v, m and whether the image is real or virtual each time. Watch the dashed lines: a real image forms where the actual rays cross, a virtual image where their backward extensions cross.

Observe

For u > 2f the image is inverted, real and diminished; at u = 2f it is inverted, real and the same size; for f < u < 2f it is inverted, real and enlarged. At u = f the outgoing rays are parallel and there is no finite image. For u < f the image is upright, virtual and enlarged, on the same side as the object. The closer u gets to f, the farther and larger the real image becomes.

Explain

How the equation fixes the image position

Rewrite 1/u + 1/v = 1/f as 1/v = 1/f − 1/u. As the object distance u decreases, 1/u increases, so 1/v decreases and the image distance v increases. That is why an object closer to the lens (but still beyond f) casts its real image farther away.

Why the magnification changes

The magnification is m = −v/u. Because u decreases while v increases, the ratio v/u grows, so |m| grows. The sign of m follows the sign of v: v > 0 gives m < 0, an inverted real image; v < 0 gives m > 0, an upright virtual image.

The special case u = f

When u = f, 1/f − 1/u = 0, so 1/v = 0 and v tends to infinity. The rays leaving the object emerge parallel to each other: they never cross on the far side, and their backward extensions never cross either, so no finite image is formed. Do not substitute a very large finite v here.

Check

With f = 0.10 m, confirm that u = 0.20 m gives v = 0.20 m and m = −1 (real, inverted, same size); u = 0.15 m gives v = 0.30 m and m = −2 (real, inverted, enlarged); and u = 0.05 m gives v = −0.10 m and m = 2 (virtual, upright, enlarged, on the object side). Check that 1/u + 1/v equals 1/f each time.

Explore further

A diverging (concave) lens has f < 0. The same equation 1/u + 1/v = 1/f still holds, but v is always negative: such a lens can only form a diminished, upright, virtual image. Derive this yourself. You can also compare with Pendulum and time to see how different experiments share the idea of describing change with a formula.