The lens formula 1/f = 1/u + 1/v is one of those equations that many students memorize without truly understanding where it comes from. But here's the thing: once you see the derivation using similar triangles and ray diagrams, the formula becomes obvious rather than mysterious. You'll stop second-guessing yourself about which distances go where, and lens problems will make intuitive sense.
What the Symbols Mean (and Why Sign Conventions Matter)
Before we derive anything, let's be crystal clear about notation:
- f = focal length of the lens
- u = object distance (from lens to object)
- v = image distance (from lens to image)
The sign convention we'll use is the "real is positive" convention for a convex lens: distances measured in the direction of actual light travel are positive. For our derivation, we'll work with a real object forming a real image, so all distances will be positive. This keeps the geometry straightforward and helps you understand the fundamental relationship before dealing with virtual images and negative values.
Setting Up the Ray Diagram
The lens formula derivation relies on a carefully constructed ray diagram for a convex (converging) lens. We need three key elements:
- An object of height
h₁placed at distanceufrom the lens - The principal axis running through the optical center of the lens
- Two specific rays that we can trace with certainty
The two rays we'll use are:
- Ray 1: A ray parallel to the principal axis that refracts through the far focal point F
- Ray 2: A ray passing through the optical center O that continues undeviated
Where these rays intersect, we get our image of height h₂ at distance v from the lens. This setup creates two pairs of similar triangles, which is the geometric key to our derivation.
Finding the First Pair of Similar Triangles
Look at the triangle formed by the object, the principal axis, and the ray passing through the optical center. This triangle has height h₁ (the object height) and base u (the object distance).
Now look at the corresponding triangle on the image side: it has height h₂ (the image height) and base v (the image distance).
These triangles are similar because they share the same angle at the optical center. For similar triangles, corresponding sides are proportional:
h₁/u = h₂/v
Rearranging this gives us our magnification relationship:
h₂/h₁ = v/u
This ratio v/u is the magnification m. Keep this relationship in mind—we'll need it shortly.
Finding the Second Pair of Similar Triangles
Now focus on the ray that enters parallel to the principal axis and refracts through the focal point F. Consider the small triangle at the focal point: it has height equal to the lens height at that point, and its base extends from the lens to the focal point (distance f).
But here's the clever bit: this small triangle is similar to the triangle formed by the entire image. Both triangles share the same angle at F, making them similar.
For a ray at height h above the axis at the lens position, the small triangle has height h and base f. The larger triangle has height h₂ and base (v - f).
Using similar triangles again:
h/f = h₂/(v - f)
We can also form another pair using the object side. The ray that starts from the top of the object and travels parallel to the axis has height h₁ at distance u. This creates similar triangles with:
h/f = h₁/(u - f)
Combining the Relationships to Derive the Lens Formula
From our two similar triangle equations, we can write:
h₂/(v - f) = h/f and h₁/(u - f) = h/f
This means: h₂/(v - f) = h₁/(u - f)
But we already know from our first pair of similar triangles that h₂/h₁ = v/u, so h₂ = h₁(v/u).
Substituting this into our equation:
h₁(v/u)/(v - f) = h₁/(u - f)
The h₁ cancels out:
(v/u)/(v - f) = 1/(u - f)
Cross-multiplying:
v(u - f) = u(v - f)
vu - vf = uv - uf
-vf = -uf
uf = vf
Wait, that doesn't look right! Let's reconsider. Actually, the correct approach is to recognize that:
v/(v - f) = u/(u - f) is incorrect. Let me use the proper similar triangles relationship.
From the focal point geometry: h₂/h₁ = (v - f)/f from one pair of triangles, and h₂/h₁ = f/(u - f) from considering the object side differently. Actually, the cleanest approach uses:
Triangle at focus: h₁/(u) = lens height at distance from focus/f, and noting that by similar triangles h₁/u = (v-f)/f × 1/something... Let me restart with the standard derivation:
From similar triangles through the optical center: h₂/h₁ = v/u
From similar triangles using the focal point: h₂/(v-f) = h₁/f for the triangle at F on the image side.
Therefore: h₂/h₁ = (v-f)/f = v/f - 1
But also h₂/h₁ = v/u, so:
v/u = v/f - 1
v/u = (v - f)/f
vf = u(v - f)
vf = uv - uf
uf = uv - vf
uf = v(u - f)
Dividing both sides by uvf:
u/(uv) = v/(uvf) × (u-f)/1...
Actually, let's divide by fvu:
uf/(fvu) = uv/(fvu) - uf/(fvu)
Let me use the standard approach cleanly: From vf = uv - uf
Divide everything by uvf:
vf/(uvf) = uv/(uvf) - uf/(uvf)
1/u = 1/f - 1/v
Rearranging:
1/f = 1/u + 1/v
And there it is—the lens formula!
Worked Example: Finding Image Position
Let's put the formula to work with a concrete example. A convex lens has a focal length of 15 cm. An object is placed 30 cm from the lens. Where will the image form?
Given:
- f = 15 cm
- u = 30 cm
- v = ?
Solution:
Using 1/f = 1/u + 1/v:
1/15 = 1/30 + 1/v
1/v = 1/15 - 1/30
Finding a common denominator:
1/v = 2/30 - 1/30 = 1/30
v = 30 cm
The image forms 30 cm on the other side of the lens. Because u = v in this case, the magnification is m = v/u = 1, meaning the image is the same size as the object. This makes sense: when an object is placed at 2f (twice the focal length), the image also appears at 2f on the opposite side.
Another example: If we place the object at 20 cm from the same lens (f = 15 cm):
1/15 = 1/20 + 1/v
1/v = 1/15 - 1/20 = 4/60 - 3/60 = 1/60
v = 60 cm
The image forms much farther away at 60 cm, with magnification m = 60/20 = 3 (three times larger). Try the lens and mirrors worksheet on this site to practice more problems like these with different object and lens configurations.
Why Understanding the Derivation Matters
You might wonder why we spent time on the lens formula derivation when you could just memorize 1/f = 1/u + 1/v. Here's why it matters: when you understand that this formula comes from similar triangles in a ray diagram, you naturally understand what each variable represents physically. You'll remember whether to add or subtract the reciprocals (addition, because that's what the geometry demands). You'll understand why the formula has reciprocals rather than direct distances. And when you encounter more complex situations—like diverging lenses, virtual images, or lens combinations—you can reason through them geometrically rather than trying to remember which sign convention applies to which formula variation.
The lens formula isn't arbitrary; it's a direct mathematical consequence of how light rays refract through a lens. The similar triangles we identified in the ray diagram create proportional relationships that, when combined algebraically, inevitably produce this elegant equation. This derivation is also excellent preparation for university physics, where deriving equations from first principles becomes standard practice. Master this derivation now, and you'll find that optics problems become exercises in logical thinking rather than formula recall.