Thin Lenses: Converging and Diverging Lenses Through Ray Diagrams

What “thin lens” means and why ray diagrams work

A thin lens is modeled as having negligible thickness compared with the object and image distances. In this model, we treat refraction as if it happens at a single plane (the lens plane). That simplification lets us trace a few principal rays to locate the image. Ray diagrams are not just sketches: they encode geometry that matches the lens equations when you use a consistent sign convention.

Converging (convex) vs diverging (concave) lenses

Converging (convex) lens: brings parallel rays to a focus

A converging lens is thicker in the middle than at the edges. Rays that enter parallel to the optical axis bend toward the axis and meet at the focal point on the far side of the lens. Its focal length is positive.

• Key intuition: it can form a real image on a screen when the object is far enough away (beyond the focal length).
• When the object is close: it can produce a virtual, magnified image (magnifying glass behavior).

Diverging (concave) lens: spreads rays as if from a focus

A diverging lens is thinner in the middle than at the edges. Parallel rays leaving the lens spread out; if you extend them backward, they appear to originate from a focal point on the same side as the incoming light. Its focal length is negative.

• Key intuition: by itself, it forms a virtual, upright, reduced image for real objects placed in front of it.
• Everyday feel: it “shrinks” the scene and increases the apparent field of view.

Focal length: what it means physically

The focal length f is the distance from the lens to the focal point for rays that enter parallel to the optical axis. A shorter |f| means stronger bending (more optical power). In practice:

• Large |f|: gentle bending, weaker lens.
• Small |f|: strong bending, stronger lens.

In ray diagrams, the focal points are marked on both sides of the lens at distance |f| from the lens center (with sign handled by convention, not by where you draw the tick marks).

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Principal rays for thin lenses (the “toolkit”)

To locate an image, you typically need only two rays from the top of the object (a third ray is a check).

For a converging lens

• Ray 1 (Parallel → Focus): A ray parallel to the axis refracts through the far focal point.
• Ray 2 (Center ray): A ray through the lens center continues approximately straight (thin lens approximation).
• Ray 3 (Focus → Parallel): A ray aimed at the near focal point exits parallel to the axis.

For a diverging lens

• Ray 1 (Parallel → appears from Focus): A ray parallel to the axis exits diverging as if it came from the near focal point (extend backward to the focus).
• Ray 2 (Center ray): Through the lens center, approximately straight.
• Ray 3 (Toward far focus → Parallel): A ray aimed toward the far focal point exits parallel.

Real vs virtual in the diagram: If the refracted rays actually intersect on the outgoing side, the image is real. If they only intersect when you extend them backward, the image is virtual.

Systematic image construction (step-by-step)

Step-by-step procedure (works for both lens types)

1. Draw the optical axis (a horizontal line) and the lens (a vertical line). Mark the lens center.

2. Mark focal points at distance |f| on both sides of the lens.

3. Place the object (an upright arrow) at distance d_o from the lens on the object side.

4. Trace Ray 1 from the top of the object using the appropriate principal-ray rule (parallel ray rule differs for converging vs diverging).

5. Trace Ray 2 through the lens center (straight line).

6. Find the intersection of the refracted rays. If they diverge, extend them backward with dashed lines to find the virtual intersection.

7. Draw the image arrow at the intersection point: upside down if below the axis (inverted), upright if above (upright). Compare heights to estimate magnification.

How image properties change as you move the object (converging lens)

What a diverging lens does (for any real object)

For a diverging lens with a real object in front:

• Image is always virtual (forms on the object side).
• Image is upright.
• Image is reduced (|m| < 1).

Thin lens equation and magnification (as a check)

Equations

Use these to confirm what your ray diagram suggests:

1/f = 1/d_o + 1/d_i

m = h_i / h_o = - d_i / d_o

Where:

• f = focal length
• d_o = object distance
• d_i = image distance
• h_o, h_i = object and image heights
• m = magnification

Sign convention (use consistently)

• Object distance d_o: positive for a real object placed on the incoming-light side (typical case).
• Image distance d_i: positive for a real image on the outgoing side; negative for a virtual image on the object side.
• Focal length f: positive for converging lenses; negative for diverging lenses.
• Magnification m: negative means inverted; positive means upright. Also, |m| > 1 magnified, |m| < 1 reduced.

Worked examples (ray-diagram prediction + equation check)

Example 1: Converging lens making a real image (projectable)

Given: f = +10 cm, d_o = 30 cm, object height h_o = 2.0 cm.

Ray-diagram expectation: Since d_o > 2f (30 cm > 20 cm), the image should be real, inverted, and reduced, located between f and 2f.

Equation check:

1/f = 1/d_o + 1/d_i  →  1/10 = 1/30 + 1/d_i

1/d_i = 1/10 - 1/30 = (3 - 1)/30 = 2/30 = 1/15  →  d_i = +15 cm

Interpretation: d_i is positive, so the image is real and forms 15 cm on the far side (between 10 and 20 cm, as predicted).

Magnification:

m = -d_i/d_o = -15/30 = -0.50

h_i = m h_o = (-0.50)(2.0 cm) = -1.0 cm

Interpretation: Negative height means inverted; magnitude 1.0 cm means reduced by half.

Example 2: Converging lens as a magnifying glass (virtual image)

Given: f = +10 cm, d_o = 6.0 cm, h_o = 1.5 cm.

Ray-diagram expectation: Since d_o < f, the refracted rays diverge; extending them backward should give a virtual, upright, magnified image on the object side.

Equation check:

1/10 = 1/6 + 1/d_i  →  1/d_i = 1/10 - 1/6 = (3 - 5)/30 = -2/30 = -1/15

d_i = -15 cm

Interpretation: Negative d_i confirms a virtual image 15 cm on the object side.

Magnification:

m = -d_i/d_o = -(-15)/6 = +2.5

h_i = (2.5)(1.5 cm) = 3.75 cm

Interpretation: Positive magnification means upright; |m| > 1 means magnified.

Example 3: Diverging lens (always virtual for a real object)

Given: f = -12 cm, d_o = 24 cm, h_o = 2.0 cm.

Ray-diagram expectation: Image should be virtual, upright, and reduced, on the object side.

Equation check:

1/f = 1/d_o + 1/d_i  →  1/(-12) = 1/24 + 1/d_i

1/d_i = -1/12 - 1/24 = (-2 - 1)/24 = -3/24 = -1/8  →  d_i = -8 cm

Magnification:

m = -d_i/d_o = -(-8)/24 = +0.33

Interpretation: Virtual (negative d_i), upright (positive m), reduced (|m| < 1).

Common interpretation checkpoints (to avoid sign and diagram mistakes)

• If your diagram shows a real image, your calculation should give d_i > 0 and m < 0 (inverted) for a single converging lens with a real object beyond f.
• If your diagram shows a virtual image, your calculation should give d_i < 0 and usually m > 0 (upright).
• Object at the focal point of a converging lens should push the image “to infinity” (equation gives 1/d_i = 0).
• Diverging lens with real object should always yield d_i < 0 and 0 < m < 1.

Everyday connections

Magnifying glass: why it magnifies

A magnifying glass is a converging lens used with the object placed inside the focal length (d_o < f). The lens then forms a virtual image that is upright and larger. Your eye interprets the diverging rays as coming from that larger image location, making fine details easier to see.

Corrective lenses: nearsightedness and farsightedness in lens language

• Nearsightedness (myopia): distant objects focus too strongly in the eye. A diverging corrective lens (negative f) spreads incoming rays slightly so the eye’s own lens can bring them to focus on the retina.
• Farsightedness (hyperopia): the eye doesn’t converge enough for near objects. A converging corrective lens (positive f) pre-converges rays so the eye can focus them properly.

In both cases, the corrective lens is chosen so that the combined system places the final image on the retina for typical viewing distances.

Why a camera lens must move to focus

A camera sensor sits at a fixed position. For a converging lens, the thin lens equation shows that d_i depends on d_o:

• When the subject is far away, d_o is large, so d_i approaches f (image forms near the focal plane).
• When the subject is closer, 1/d_o increases, so 1/d_i must decrease, meaning d_i increases: the image plane shifts farther from the lens.

To keep the image sharp on the sensor, the camera changes the lens-to-sensor separation (or changes effective focal length with internal elements). Ray diagrams reflect this: closer objects require rays to travel farther after the lens before meeting.