Difficulty: Intermediate | Prerequisites: Algebra, basic ray optics (reflection, refraction), familiarity with converging and diverging lenses.
The thin lens equation is one of the most-used tools in introductory optics. It relates the focal length of a lens to the distances of the object and its image. Once you can handle a single lens, the next step is placing two lenses on the same optical axis, which is how many real instruments (cameras, microscopes, telescopes) work. This topic builds directly on the refractive power and human-eye material in Section 19.2 and prepares you for compound optical systems later in the course.
The thin lens equation (1/s + 1/s' = 1/f) lets you find where an image forms for any single thin lens. For two-lens systems, you solve the equation for the first lens, then treat its image as the object for the second lens and solve again. Whether the intermediate image lands to the left or right of the second lens determines whether it acts as a real or virtual object.
Thin lens equation
The relationship 1/s + 1/s' = 1/f, connecting object distance (s), image distance (s'), and focal length (f) for a thin lens. In simple terms, it is the master equation for locating images.
Object distance (s)
The distance from the object to the lens, measured along the optical axis. Positive when the object is on the incoming-light side of the lens.
Image distance (s')
The distance from the lens to the image. Positive when the image forms on the opposite side of the lens from the object (a real image); negative when it forms on the same side (a virtual image).
Focal length (f)
The distance from the lens to the point where parallel incoming rays converge (converging lens, f > 0) or appear to diverge from (diverging lens, f < 0).
Real image
An image formed where light rays actually converge. It can be projected onto a screen. s' is positive.
Virtual image
An image formed where light rays appear to diverge from but do not actually meet. It cannot be projected onto a screen. s' is negative.
Virtual object (in multi-lens systems)
When the image from the first lens would form on the far side of the second lens (beyond it), that image becomes a virtual object for the second lens. The object distance for the second lens is then taken as negative.
1/s + 1/s' = 1/f
Rearranged to solve for the image distance: s' = (1/f - 1/s)^(-1)
s is positive when the object is on the incoming-light side (the usual case).
s' is positive for a real image (opposite side from the object) and negative for a virtual image (same side as the object).
f is positive for a converging lens and negative for a diverging lens.
Identify s (object distance) and f (focal length), including signs.
Plug into s' = (1/f - 1/s)^(-1).
If s' is positive, the image is real and on the far side of the lens. If s' is negative, the image is virtual and on the same side as the object.
The image of the first lens becomes the object for the second lens. Solve the thin lens equation for Lens A first. Then use the result to work out the object distance for Lens B.
If the image from Lens A falls to the left of Lens B (between the two lenses, or further left), it is a real object for Lens B, and its object distance is positive.
If the image from Lens A would form to the right of Lens B (past it), it is a virtual object for Lens B, and its object distance is negative.
Lens A has f = 15 cm. Lens B has f = 20 cm. They are 80 cm apart. An object is 40 cm to the left of Lens A.
Step 1, solve for Lens A: s' = (1/0.15 - 1/0.40)^(-1) = 0.24 m = 24 cm.
The image forms 24 cm to the right of Lens A. Since the lenses are 80 cm apart, this image is 80 - 24 = 56 cm to the left of Lens B.
The image is real and sits between the lenses, so it is a real object for Lens B with s = 56 cm.
Step 2, solve for Lens B: s' = (1/0.20 - 1/0.56)^(-1). This gives the final image location relative to Lens B.
Lens A has f = 43 cm. Lens B has f = 2.5 cm. They are 2.0 cm apart. An object is 25 cm to the left of Lens A.
Step 1, solve for Lens A: s' = (1/0.43 - 1/0.25)^(-1) = -60 cm.
The negative sign means the image would form 60 cm to the left of Lens A (a virtual image, on the same side as the object).
Since that image is to the left of Lens A, it is also to the left of Lens B. It is a real object for Lens B.
The object distance for Lens B: the virtual image is 60 cm to the left of Lens A, and Lens B is 2.0 cm to the right of Lens A, so s for Lens B = 60 + 2.0 = 62 cm.
Step 2, solve for Lens B with s = 62 cm and f = 2.5 cm.
\frac{1}{s} + \frac{1}{s'} = \frac{1}{f}The thin lens equation. s = object distance, s' = image distance, f = focal length.
s' = \left(\frac{1}{f} - \frac{1}{s}\right)^{-1}Rearranged to isolate the image distance. This is the form you will use most often in problems.
P_{\text{total}} = P_1 + P_2Power addition for two thin lenses in contact (separation approximately zero).
Multi-lens systems are the basis of every compound optical instrument. A camera lens is a stack of several elements designed to correct aberrations. A compound microscope uses an objective lens and an eyepiece, each contributing to the total magnification, and the image-of-first-becomes-object-of-second rule applies at every stage.
Students often forget to convert centimetres to metres before plugging into the equation. The dioptre is defined using metres, so mixing units is a common source of wrong answers.
A frequent error in two-lens problems is using the wrong object distance for the second lens. The object distance for Lens B is not the image distance from Lens A directly; you must account for the separation between the lenses.
Students sometimes assume a virtual image from Lens A cannot serve as a real object for Lens B. It can, so long as the virtual image is on the incoming-light side of Lens B.
Forgetting that a negative s' means a virtual image, not an error in the calculation. If s' comes out negative, the image is on the same side as the object.
⚠️ Expect at least one two-lens problem on the exam. Practise the full workflow: solve for Lens A, find the object distance for Lens B (accounting for separation), solve for Lens B.
⚠️ Know the rearranged form s' = (1/f - 1/s)^(-1) by heart. It saves time and reduces algebraic mistakes.
⚠️ Watch your signs. A negative s' is not a mistake; it means a virtual image. A negative object distance for the second lens means a virtual object.
⚠️ Unit consistency: if f and s are in cm, s' comes out in cm. If you need dioptres, convert to metres first.
True or False: A converging lens always produces a real image. (False. If the object is inside the focal point, a converging lens produces a virtual image.)
Fill in the blank: When s' is negative, the image is ____. (virtual)
True or False: In a two-lens system, the object distance for the second lens equals the image distance from the first lens. (False. You must subtract the image distance from the lens separation to find the object distance for the second lens.)
Fill in the blank: A diverging lens has a ____ focal length. (negative)
True or False: Two thin lenses in contact have a combined power equal to the product of their individual powers. (False. Their powers add, not multiply.)
Q: An object is placed 30 cm to the left of a converging lens with f = 20 cm. Where does the image form, and is it real or virtual?
A: s' = (1/0.20 - 1/0.30)^(-1) = (5.0 - 3.33)^(-1) = 1/1.67 = 0.60 m = 60 cm to the right of the lens. s' is positive, so the image is real.
Q: An object is placed 10 cm from a converging lens of focal length 20 cm. Describe the image.
A: s' = (1/0.20 - 1/0.10)^(-1) = (5.0 - 10.0)^(-1) = -0.20 m = -20 cm. The image is virtual (negative s') and appears 20 cm to the left of the lens, on the same side as the object.
Q: Lens A (f = 15 cm) and Lens B (f = 20 cm) are 80 cm apart. An object is 40 cm left of Lens A. Where is the final image relative to Lens B?
A: Lens A gives s' = 24 cm (to its right). Object distance for Lens B = 80 - 24 = 56 cm. Lens B gives s' = (1/0.20 - 1/0.56)^(-1) = (5.0 - 1.79)^(-1) = 1/3.21 = 0.31 m, roughly 31 cm to the right of Lens B. The final image is real.
Q: Why does the problem say "treat the image as a real object" when the first lens produces a virtual image to its left?
A: A virtual image from Lens A that falls on the incoming-light side of Lens B is a real object for Lens B, because light is actually diverging from that point as it reaches Lens B. "Real object" means the object is where the light is coming from, on the correct side.
This material connects directly to the human eye and vision correction notes: the eye is a multi-element optical system (cornea + lens) and corrective glasses add yet another lens to the chain. The same thin lens equation underpins the analysis of microscopes (objective + eyepiece) and telescopes, which appear later in the course. Magnification, covered alongside these instruments, relies on the image and object distances you calculate here.
Thin lens equation, 1/s + 1/s' = 1/f, object distance, image distance, focal length, real image, virtual image, virtual object, converging lens, diverging lens, two-lens system, multi-lens, compound lens, image becomes object, refractive power, dioptre, PHYS 116, general physics, University of Washington, Chapter 19, optics