What is the difference between a real image and a virtual image?

A real image is formed where light rays actually converge — you can project it onto a screen. A virtual image is formed where rays appear to diverge from — you cannot project it, but you can see it by looking through the lens (like in a magnifying glass or your eye looking at a reflected image in a mirror).

Why does a convex lens sometimes produce a virtual image?

When the object is placed inside the focal length (u < f), the refracted rays diverge and never actually cross — they appear to come from a point behind the lens. This is exactly how a magnifying glass works: the object is close to the lens, producing an enlarged, virtual, upright image.

What is the power of a lens and how is it related to focal length?

The power of a lens is P = 1/f, measured in dioptres (D) when f is in metres. A lens with f = 25 cm = 0.25 m has power P = +4 D. Eye prescriptions are written in dioptres — a −2 D prescription means you need a concave lens of focal length −0.5 m to correct myopia (short-sightedness).

How is the lens formula related to mirrors?

The mirror formula is identical in form: 1/v + 1/u = 1/f, using the same sign convention. Concave mirrors converge light like convex lenses; convex mirrors diverge it like concave lenses. The physics is different (reflection vs refraction) but the geometry produces the same formula.

Lens Formula Calculator — Interactive Ray Diagram and Image Formation

Q: What is the power of a lens and how is it related to focal length?

The power of a lens is P = 1/f, measured in dioptres (D) when f is in metres. A lens with f = 25 cm = 0.25 m has power P = +4 D. Eye prescriptions are written in dioptres — a −2 D prescription means you need a concave lens of focal length −0.5 m to correct myopia (short-sightedness).

Q: How is the lens formula related to mirrors?

The mirror formula is identical in form: 1/v + 1/u = 1/f, using the same sign convention. Concave mirrors converge light like convex lenses; convex mirrors diverge it like concave lenses. The physics is different (reflection vs refraction) but the geometry produces the same formula.

The lens formula relates the focal length of a lens to the object distance and image distance. It underpins all of optics — from your eye and spectacle lenses to microscopes, telescopes, cameras, and projectors. Understanding how lenses form images, and whether those images are real or virtual, upright or inverted, is central to optics and vision science.

Use the interactive ray diagram below to see image formation change as you move the object closer or further from the lens.

Image distance v

40.0 cm (real)

Magnification m

-1.00 (inverted)

Image height

3.0 cm

Focal length |f|20 cm

Object distance u40 cm

The Lens Formula

Using the real-is-positive sign convention:

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

where $u$ is always negative (object is to the left of the lens).

Equivalently written as:

$\frac{1}{f} = \frac{1}{v} + \frac{1}{u'} \quad \text{where } u' = |u|$

Symbol	Quantity	Sign convention
f	Focal length	+ = convex (converging), − = concave (diverging)
u	Object distance	always negative (object on left)
v	Image distance	+ = real image (right), − = virtual image (left)

Magnification

$m = -\frac{v}{u}$

$m > 0$ : upright image
$m < 0$ : inverted image
$|m| > 1$ : image is larger than object (magnified)
$|m| < 1$ : image is smaller than object (diminished)

Worked Examples

Worked Example

Example 1 — Object beyond 2f (convex lens)

A convex lens has focal length f = 15 cm. An object is placed 45 cm from the lens. Find the image distance and magnification.

Using $u = -45$ cm, $f = +15$ cm:

$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{15} + \frac{1}{-45} = \frac{3}{45} - \frac{1}{45} = \frac{2}{45}$

$v = \frac{45}{2} = 22.5 \text{ cm}$

Since $v > 0$ , the image is real and on the opposite side of the lens.

Using the Cartesian magnification formula $m = v/u$ :

$m = \frac{v}{u} = \frac{+22.5}{-45} = -0.5$

Since $m < 0$ , the image is inverted. Since $|m| = 0.5 < 1$ , the image is diminished.

The image is real, inverted, and half the size of the object — exactly what you expect when the object is beyond $2f$ of a converging lens.

Worked Example

Example 2 — Object inside focal length (convex lens)

The same lens (f = 15 cm) with object placed 10 cm away (inside the focal length).

Using $u = -10$ cm, $f = +15$ cm:

$\frac{1}{v} = \frac{1}{15} + \frac{1}{-10} = \frac{2}{30} - \frac{3}{30} = -\frac{1}{30}$

$v = -30 \text{ cm}$

Since $v < 0$ , the image is virtual — it appears on the same side as the object (this is how a magnifying glass works).

$m = \frac{v}{u} = \frac{-30}{-10} = +3$

Since $m > 0$ , the image is upright. Since $|m| = 3 > 1$ , it is magnified — this is exactly how a magnifying glass works.

Frequently Asked Questions

Related Concepts

How Do 3D Glasses Work? →

Polarisation and stereoscopic vision — the optics of 3D cinema.

What is a Mirage? →

Atmospheric refraction bending light — the same optics as a lens, applied to air.

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