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)

Quantity	Symbol	Sign
Focal length	f	+ for converging, – for diverging
Object distance	u	+ (measured from the lens to the object on the incoming‑light side)
Image distance	v	+ 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 \(m\).

Quantity	Symbol	Value	Sign (IGCSE)
Focal length	f	+10 cm	Positive (converging)
Object distance	u	+6 cm	Positive (object side)
Image distance	v	?	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.