Define and use the terms focal length, principal axis and principal focus (focal point)

3.2.3 Thin Lenses

Learning Objective

Define and use the terms focal length, principal axis and principal focus (focal point) for thin lenses, and apply these concepts to predict image characteristics, draw ray diagrams (including virtual images formed by a converging lens) and explain the use of lenses in vision correction.

Key Definitions (Cambridge terminology)

• Principal Axis: The straight line that passes through the optical centre of a thin lens and is perpendicular to the lens surfaces.
• Optical Centre (O): The geometric centre of a thin lens. A ray that passes through O is not deviated.
• Principal Focus (F) – also called the focal point: The point on the principal axis where rays that were parallel to the axis converge (convex lens) or appear to diverge from (concave lens) after refraction.
• Focal Length (f): The distance between the principal focus F and the optical centre O.

   • Positive (+f) for a converging (convex) lens.

   • Negative (‑f) for a diverging (concave) lens.

Thin‑Lens Approximation

The lens is treated as an infinitely thin plane located at O. This means:

• All refraction is assumed to occur at a single point O.
• Ray‑diagram rules (parallel, central and focal rays) are valid only when the lens thickness is small compared with the object‑lens and lens‑image distances.
• For real lenses the approximation is excellent for the focal lengths and object distances used in the IGCSE syllabus.

Types of Thin Lens

Ray‑Diagram Conventions (Thin‑Lens Approximation)

To locate the image, draw any two of the three standard rays; the third is optional.

1. Parallel Ray: Starts from the top of the object, travels parallel to the principal axis, then after the lens passes through (convex) or appears to come from (concave) the principal focus.
2. Central Ray: Passes straight through the optical centre O and continues undeviated.
3. Focal Ray: Directed toward (convex) or away from (concave) the principal focus before the lens; after the lens it emerges parallel to the principal axis.

Virtual Image formed by a Converging Lens

When the object is placed between the focal point F and the lens, the three rays diverge after the lens. Extending them backwards (i.e. extrapolating the diverging rays) they meet at a point on the same side of the lens as the object – this is a virtual, upright, magnified image.

Object at the Focal Point

If an object is placed exactly at the principal focus (u = f) of a converging lens, the parallel ray and the focal ray emerge parallel after the lens. The rays never meet, so the image is formed at infinity. This principle is used in projectors and in the eye’s accommodation for distant objects.

Mathematical Relationships

Lens Formula (sign‑convention version)

\[

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

\]

Magnification

\[

m = \frac{v}{u}= \frac{\text{image height }(h')}{\text{object height }(h)}

\]

• If m is positive the image is upright; if negative the image is inverted.
• \(|m|>1\) → magnified; \(|m|<1\) → reduced; \(|m|=1\) → same size.

IGCSE Sign Conventions

Image Characteristics Summary

Worked Examples

Example 1 – Convex Lens (Real Image)

A convex lens has a focal length of +8 cm. An object is placed 24 cm in front of the lens. Find the image distance, magnification and describe the image.

1. Lens formula: \(\displaystyle \frac{1}{8}= \frac{1}{v}+ \frac{1}{24}\)
2. \(\displaystyle \frac{1}{v}= \frac{1}{8}-\frac{1}{24}= \frac{3-1}{24}= \frac{2}{24}= \frac{1}{12}\)
3. \(v = +12\ \text{cm}\) → real image on the opposite side.
4. Magnification: \(m = \frac{v}{u}= \frac{12}{24}= -0.5\) (negative sign shows the image is inverted).
5. Result: Real, inverted, reduced (half the object height) image 12 cm behind the lens.

Example 2 – Convex Lens (Virtual Image)

Object distance \(u = +5\ \text{cm}\) (i.e. the object is between the lens and its focal point) for a convex lens with \(f = +8\ \text{cm}\). Find the image distance and magnification.

1. \(\displaystyle \frac{1}{8}= \frac{1}{v}+ \frac{1}{5}\)
2. \(\displaystyle \frac{1}{v}= \frac{1}{8}-\frac{1}{5}= \frac{5-8}{40}= -\frac{3}{40}\)
3. \(v = -13.3\ \text{cm}\) → virtual image on the same side as the object.
4. \(m = \frac{v}{u}= \frac{-13.3}{5}= -2.66\). The image is upright (positive height) and magnified (\(|m|>1\)).
5. Result: Virtual, upright, magnified image appearing 13.3 cm behind the lens.

Example 3 – Concave Lens (Virtual Image)

A diverging lens has a focal length of ‑5 cm. An object is placed 15 cm** in front of the lens. Find the image distance, magnification and describe the image.

1. \(\displaystyle \frac{1}{-5}= \frac{1}{v}+ \frac{1}{15}\)
2. \(\displaystyle \frac{1}{v}= -\frac{1}{5}-\frac{1}{15}= -\frac{4}{15}\)
3. \(v = -3.75\ \text{cm}\) → virtual image on the same side as the object.
4. \(m = \frac{v}{u}= \frac{-3.75}{15}= -0.25\). The image is upright and reduced.
5. Result: Virtual, upright, reduced image 3.75 cm behind the lens.

Vision‑Correction Applications (IGCSE requirement)

• Myopia (short‑sightedness): The eye’s focal point lies in front of the retina. A diverging (concave) lens with negative focal length is placed in front of the eye; it spreads the incoming rays so that the image is formed further back, on the retina.
• Hyperopia (long‑sightedness): The eye’s focal point lies behind the retina. A converging (convex) lens with positive focal length is used; it brings the rays together earlier so the image falls on the retina.
• Magnifying glass: A convex lens used at a distance slightly less than its focal length, producing a large upright virtual image.
• Projector: Object placed at the focal point of a convex lens; the emerging rays are parallel and the image is formed at infinity on a screen placed far away.

Sample IGCSE‑style Question

Q: A converging lens has a focal length of +10 cm. An object 3 cm high is placed 6 cm in front of the lens.

1. State, with justification, whether the image formed is real or virtual.
2. Calculate the image distance and the magnification.
3. Draw a labelled ray diagram showing the formation of the image.

Answer outline:

• Since the object is inside the focal length (u = +6 cm < f), the image will be virtual, upright and magnified.
• Lens formula: \(\frac{1}{10}= \frac{1}{v}+ \frac{1}{6}\) → \(\frac{1}{v}= -\frac{1}{15}\) → \(v = -15\) cm (negative ⇒ virtual, same side as object).
• Magnification: \(m = \frac{v}{u}= \frac{-15}{6}= -2.5\). The image height \(h' = m\,h = -2.5 \times 3 = -7.5\) cm; the negative sign indicates the image is upright.
• Ray diagram: draw the parallel ray through F, the focal ray through the focal point on the object side, and the central ray through O; extend the diverging rays backwards to meet at the virtual image point 15 cm in front of the lens.

Quick Revision Checklist

• Identify the principal axis and locate the optical centre (O) on any lens diagram.
• Mark the principal focus (F) on the axis; remember its sign (+ for convex, – for concave).
• Measure or calculate the focal length (f) as the distance OF.
• Apply the IGCSE sign conventions for u, v, f when using the lens formula.
• Use the three standard rays (parallel, central, focal) to construct ray diagrams – include the case of a virtual image produced by a converging lens.
• Recall that an object at the focal point gives an image at infinity.
• Determine image type (real/virtual), orientation (upright/inverted) and size (magnified/reduced) using the magnification formula.
• Know one everyday use for each lens type and the role of lenses in correcting myopia and hyperopia.
• Practice a full IGCSE‑style question that asks for description, calculation and a labelled ray diagram.