Draw and use ray diagrams for the formation of a virtual image by a converging lens

3.2.3 Thin Lenses – Virtual Image Formation by a Converging Lens

Learning Objective

Draw and use ray diagrams to predict the position, nature and size of a virtual image formed by a converging (convex) lens, and verify the result with the thin‑lens formula.

Key Concepts

  • Converging (convex) lens: thicker at the centre than at the edges; parallel incident rays are brought to a focus on the opposite side of the lens.

  • Diverging (concave) lens: thinner at the centre; parallel incident rays diverge as if they originated from a focal point on the same side as the object. Consequently a diverging lens always produces a virtual, upright, reduced image on the object side.

  • Principal axis: straight line passing through the optical centre and both focal points.

  • Optical centre (O): geometric centre of the lens; a ray that passes through O continues undeviated.

  • Focal points (F1, F2): points on the principal axis where rays that are initially parallel to the axis appear to converge (F2) or diverge from (F1).

  • Virtual image: formed when the refracted rays diverge; the image cannot be projected on a screen and is located on the same side of the lens as the object.

Sign‑Convention (Cambridge IGCSE)

QuantitySymbolSign
Focal lengthf+ for converging, – for diverging
Object distanceu+ (measured from the lens to the object on the incoming‑light side)
Image distancev+ for real images (opposite side), – for virtual images (same side as the object)
Height (object, image)h, h′+ if measured upwards from the principal axis

Lens‑Maker’s Equation & Optical Power (optional)

The focal length of a thin lens is related to its radii of curvature (R1, R2) and refractive index n by

\[

\frac{1}{f}= (n-1)\left(\frac{1}{R{1}}-\frac{1}{R{2}}\right)

\]

The optical power \(P\) is

\[

P = \frac{1}{f}\qquad\text{(dioptres, D)}

\]

Ray‑Diagram Construction

1. Virtual Image (object placed between the lens and its focal point, \(0

  1. Draw a horizontal line – the principal axis.

  2. Mark the optical centre O of the lens on the axis.

  3. Mark the two focal points: F1 on the object side and F2 on the image side, each a distance \(|f|\) from O.

  4. Place the upright object (arrow) between O and F1 so that \(0

  5. From the top of the object draw the three principal rays:

    • Parallel ray: travels parallel to the principal axis, strikes the lens and appears to diverge from F1 after refraction. (Extend the refracted ray backwards; it meets the other ray at the virtual image.)

    • Focal ray: aimed toward the focal point on the image side (F2); after refraction it emerges parallel to the principal axis.

    • Central ray: passes straight through the optical centre O without deviation.

  6. Extend the refracted rays backwards (i.e. behind the lens on the object side) until they intersect. The intersection point is the virtual image.

  7. Label the image:

    • Upright (same orientation as the object).

    • Magnified if the object is close to the lens; the magnification decreases as the object approaches the focal point.

    • Located on the same side of the lens as the object, between the lens and F1.

2. Real Image (object placed beyond the focal point, \(u>f\)) – quick reference

  1. Object is to the right of F1 (i.e. \(u>f\)).

  2. Ray behaviour:

    • Parallel ray: after refraction passes through the near focal point F1.

    • Focal ray: aimed toward F1 before the lens; after refraction it emerges parallel to the principal axis.

    • Central ray: passes undeviated through O.

  3. The refracted rays actually meet on the opposite side of the lens – this point is a real, inverted image.

Worked Example (Virtual Image)

Given a converging lens with \(f = +10\;\text{cm}\) and an object placed \(u = 6\;\text{cm}\) from the lens (i.e. \(u < f\)), find the image distance \(v\) and the magnification \(m\).

QuantitySymbolValueSign (IGCSE)
Focal lengthf+10 cmPositive (converging)
Object distanceu+6 cmPositive (object side)
Image distancev?Negative for a virtual image

Thin‑lens equation (IGCSE sign convention)

\[

\frac{1}{f}= \frac{1}{v}+ \frac{1}{u}

\qquad\Longrightarrow\qquad

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

\]

Substituting the numbers

\[

\frac{1}{v}= \frac{1}{10}-\frac{1}{6}= \frac{3-5}{30}= -\frac{2}{30}= -\frac{1}{15}

\]

\[

v = -15\;\text{cm}

\]

Thus the virtual image forms 15 cm in front of the lens (same side as the object).

Magnification (IGCSE formula)

\[

m = -\frac{v}{u}= -\frac{-15}{6}= +2.5

\]

  • A positive magnification indicates an upright image.

  • \(|m| = 2.5\) shows the image is 2.5 times taller than the object (magnified).

Everyday Applications of Virtual Images

  • Magnifying glass: a convex lens held close to an object creates a large, upright virtual image that the eye can focus on.

  • Microscope eyepiece: the eyepiece forms a virtual image of the real image produced by the objective, providing the final magnification.

  • Reading glasses: a weak convex lens produces a virtual image at the near point of the eye, reducing the accommodation effort for presbyopic readers.

Practical Activity (AO3 – Skills)

  1. Secure a thin converging lens on a sheet of paper and draw a straight principal axis.

  2. Mark the optical centre O and the focal points F1 and F2 (measure the focal length using a distant light source).

  3. Place an upright object (e.g. a pencil) at three different positions between O and F1 (e.g. 2 cm, 4 cm, 6 cm).

  4. Using a ray‑tracing kit or a laser pointer, draw the three principal rays for each object position.

  5. Extend the refracted rays backwards to locate the virtual image; measure its distance from the lens.

  6. Calculate the theoretical image distance and magnification with the thin‑lens equation; compare with your measurements.

  7. Record all data in a table, comment on any discrepancies, and discuss possible sources of error (lens thickness, parallax, non‑thin‑lens effects, etc.).

Common Mistakes to Avoid

  • Confusing the near focal point F1 with the far focal point F2.

  • Drawing the parallel ray through the wrong focal point – for a virtual‑image case it must appear to diverge from F1 after refraction.

  • Forgetting to extend the refracted rays backwards when locating a virtual image.

  • Applying the sign convention incorrectly: remember that \(u\) is always positive, \(v\) is negative for virtual images, and \(f\) is positive for converging lenses.

  • Using the magnitude of \(v\) directly in the magnification formula; the sign of \(v\) determines the sign of \(m\) and therefore the image orientation.

  • Interpreting a negative magnification as “upright”. In the IGCSE convention, positive magnification → upright; negative → inverted.

Summary Checklist

  • Identify lens type and assign the correct sign to the focal length.

  • Confirm the object is placed between the lens and its focal point (\(0

  • Draw the three principal rays accurately, remembering the correct behaviour of the parallel ray for a virtual image.

  • Extend the refracted rays backwards to locate the virtual image.

  • Use the thin‑lens equation with the IGCSE sign convention to verify the image distance.

  • Calculate magnification with \(m = -v/u\); a positive value confirms an upright image.

  • Relate the result to a real‑world example (e.g., magnifying glass) and, if possible, test it experimentally.