Converging and Diverging Lenses Revision Notes | A Level AQA Physics

1. Why Lenses Matter in Medical Physics

Every time a clinician uses an ophthalmoscope, a microscope, or prescribes corrective lenses, the physics of refraction is doing the work. Understanding how converging and diverging lenses form images — and how to calculate their properties precisely — sits at the heart of the AQA medical physics option.

This lesson builds from the geometry of refraction through to the three key equations you will use in the exam: the power formula, the lens equation, and the magnification formula. Each one connects directly to how the eye forms images and how optical instruments correct or exploit that process.

2. Converging and Diverging Lenses: Core Properties

Picture a beam of parallel rays travelling along the principal axis and hitting a lens. What happens next depends entirely on the lens shape. A converging (convex) lens is thicker at the centre than at the edges; it bends rays inward so they meet at a single point on the far side of the lens. That meeting point is the principal focus, labelled F, and the distance from the optical centre of the lens to F is the focal length, f.

A diverging (concave) lens is thinner at the centre. It spreads rays outward, so they never actually meet on the far side. Instead, if you trace those diverging rays backwards, they appear to come from a point on the same side as the incoming light. That apparent source point is still called the principal focus, but it is a virtual focus — no light actually passes through it.

The sign convention is not arbitrary — it is the key that makes the lens equation work consistently for every situation. Converging lenses have a positive focal length because their principal focus is real (light genuinely passes through it). Diverging lenses have a negative focal length because their principal focus is virtual.

3. Lens Power and the Power Formula

A short focal length means a lens bends light sharply — it is a powerful lens. A long focal length means gentle bending — a weak lens. This inverse relationship is captured in the power formula:

The unit of power is the dioptre (D), defined as one reciprocal metre (m$^{-1}$). A converging lens with f = +0.25 m has power P = +4 D. A diverging lens with f = −0.50 m has power P = −2 D. The sign carries physical meaning: positive power converges light, negative power diverges it.

In clinical optics, lens prescriptions are written directly in dioptres — a prescription of −3.00 D tells you immediately that the corrective lens is diverging with a focal length of −0.33 m. This is why power, not focal length, is the preferred quantity in medical contexts.

4. The Lens Equation and Image Formation

Knowing the focal length tells you about the lens itself. To find out where an image forms for a specific object position, you need the lens equation:

The equation is symmetric in u and v, which reflects a deep optical principle: if you swap object and image positions, the geometry still works. In practice, you are usually given two of the three quantities and asked to find the third — rearrange algebraically before substituting numbers to avoid sign errors.

Once you have calculated v, interpret its sign. A positive v means the image is on the opposite side of the lens from the object — it is a real image that can be projected onto a screen. A negative v means the image is on the same side as the object — it is a virtual image that cannot be projected but can be seen by looking through the lens.

Worked example: Applying the Lens Equation: Converging Lens

An object is placed 0.30 m in front of a converging lens of focal length 0.20 m. Calculate the image distance and state whether the image is real or virtual. [3 marks]

1. $$\frac{1}{u} + \frac{1}{v} = \frac{1}{f}$$ — [M1] Stating the correct formula earns the method mark.
2. $u = +0.30$ m (real object), $f = +0.20$ m (converging lens) — Apply the real-is-positive convention before substituting. Both are positive here.
3. $$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{0.20} - \frac{1}{0.30}$$ — [M1] Correct rearrangement and substitution.
4. $$\frac{1}{0.20} = 5.0 \text{ m}^{-1}, \quad \frac{1}{0.30} = 3.\overline{3} \text{ m}^{-1}$$ $$\frac{1}{v} = 5.0 - 3.\overline{3} = 1.\overline{6} \text{ m}^{-1}$$ — Work with exact fractions where possible to avoid rounding errors mid-calculation.
5. $$v = \frac{1}{1.\overline{6}} = 0.60 \text{ m}$$ — [A1] v = +0.60 m. The positive sign confirms a real image on the far side of the lens.

Answer: v = +0.60 m; the image is real (positive v) and forms 0.60 m from the lens on the opposite side from the object.

Notice that the object was placed beyond the focal length (0.30 m > 0.20 m). This is the condition for a converging lens to produce a real image. When the object is placed inside the focal length (u

5. Linear Magnification

The lens equation tells you where the image forms. The magnification formula tells you how big it is relative to the object:

The magnitude of m tells you the size relationship: |m| > 1 means the image is larger than the object (magnified); |m|

For the worked example above, v = +0.60 m and u = +0.30 m, so m = v/u = 0.60/0.30 = +2.0. However, this calculation uses only the magnitudes of the distances. Under the AQA real-is-positive convention, magnification is properly defined as the ratio of image height to object height: a real image formed by a converging lens is inverted, so the image height is negative relative to the object height, giving m = −2.0. AQA mark schemes award marks for stating that m is negative for an inverted (real) image. Students should be careful not to confuse the ratio of unsigned distances with the signed magnification that indicates orientation.

Under the AQA real-is-positive sign convention, the sign of m = v/u does indicate image orientation: a negative m means the image is inverted, and a positive m means the image is upright. For a real image formed by a converging lens with the object beyond the focal point, v is positive and u is positive, but the image height is negative (the image is inverted), so m is negative. For a virtual image formed when the object is inside the focal length, v is negative and u is positive, giving a negative ratio — but in this case the image is upright, so care must be taken with the sign of the image height. In AQA exam answers, always state: negative m = inverted image; positive m = upright image.

Worked example: Calculating Magnification and Image Height

Using the previous example (u = 0.30 m, v = 0.60 m), calculate the linear magnification. If the object is 0.05 m tall, find the image height. [3 marks]

1. $$m = \frac{v}{u}$$ — [M1] Correct formula stated.
2. $$m = \frac{0.60}{0.30}$$ — [M1] Both values carry their signs from the lens equation result.
3. $$m = 2.0$$ — The image is twice the size of the object.
4. $$\text{image height} = m \times \text{object height} = 2.0 \times 0.05 \text{ m} = 0.10 \text{ m}$$ — [A1] Image height = 0.10 m. The image is magnified (larger than the object).

Answer: m = 2.0; image height = 0.10 m (the image is magnified to twice the object size).

6. Worked Example: Diverging Lens

Diverging lenses always produce virtual, upright, diminished images of real objects — a result that follows directly from the mathematics. Working through a full example cements the sign convention and shows why the image is always on the same side as the object.

Worked example: Full Calculation for a Diverging Lens

An object is placed 0.40 m in front of a diverging lens of power −5.0 D. Calculate (a) the focal length, (b) the image distance, and (c) the magnification. [5 marks]

1. $$f = \frac{1}{P} = \frac{1}{-5.0} = -0.20 \text{ m}$$ — [B1] f = −0.20 m. The negative value confirms a diverging lens.
2. $$\frac{1}{u} + \frac{1}{v} = \frac{1}{f}$$\n$$u = +0.40 \text{ m}, \quad f = -0.20 \text{ m}$$ — [M1] Real object gives positive u; diverging lens gives negative f.
3. $$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{-0.20} - \frac{1}{0.40}$$\n$$= -5.0 - 2.5 = -7.5 \text{ m}^{-1}$$ — [M1] Both terms are evaluated carefully with their signs before combining.
4. $$v = \frac{1}{-7.5} \approx -0.133 \text{ m}$$ — [A1] v is negative — the image is virtual, on the same side as the object.
5. $$m = \frac{v}{u} = \frac{-0.133}{+0.40} \approx -0.33$$ — [A1] The negative sign of m arises because v is negative (virtual image); |m| = 0.33 < 1 confirms the image is diminished (one-third the object size). For a diverging lens with a real object, m is always negative and |m| < 1.

Answer: (a) f = −0.20 m; (b) v ≈ −0.13 m (virtual image); (c) m ≈ −0.33 (diminished, upright image; negative m because v is negative).

This result is characteristic of all diverging lenses with real objects: v is always negative, |m| is always less than 1, and the image is always virtual. These are the properties that make diverging lenses useful for correcting short-sightedness — the virtual image they produce is closer to the eye than the original object, bringing it within the eye's range of focus.

8. Exam Technique and Common Mistakes

The three equations in this topic are straightforward to state but straightforward to misapply under exam conditions. The most reliable approach is to write down the sign convention, assign signs to all given quantities, and only then substitute into the formula. Skipping the sign-assignment step is the single most common source of lost marks.

For extended-response questions asking you to evaluate image properties, structure your answer around the three characteristics — real/virtual, magnified/diminished, upright/inverted — and justify each one with reference to the sign and magnitude of v and m. This approach maps directly onto the marking points examiners use.

Across all three equations — P = 1/f, the lens equation, and m = v/u — the unifying thread is the real-is-positive sign convention. Master that convention and the mathematics becomes a reliable tool for predicting and describing image formation in any optical system, from a simple magnifying glass to the corrective lenses used in clinical practice.