Correcting Eye Defects Revision Notes | A Level AQA Physics

1. Why Eyes Go Wrong — and How Physics Fixes Them

A healthy eye focuses light precisely onto the retina. When the eyeball is the wrong shape, or the lens cannot accommodate sufficiently, that focus shifts — and the image blurs. The physics of correcting this is elegant: you add a lens in front of the eye that shifts the object's apparent position to somewhere the eye can handle.

This section of the AQA A Level Medical Physics specification asks you to work quantitatively with correcting lenses — drawing ray diagrams, calculating lens powers in dioptres, and reading astigmatism prescriptions. Each of those tasks follows directly from the thin lens equation you already know.

2. The Lens Equation and Power — The Core Tool

Before correcting any defect, you need the thin lens equation and the definition of lens power firmly in place. The thin lens equation relates object distance $u$, image distance $v$, and focal length $f$:

Lens power $P$ is the reciprocal of focal length measured in metres. A short focal length means a strongly curved lens that bends light a lot — hence high power.

When two thin lenses are placed in contact, their powers add: $P_{\text{total}} = P_1 + P_2$. This is the principle behind combining the eye's own optical power with a correcting lens — the correcting lens adjusts the total power so that the combined system focuses correctly on the retina.

3. Myopia — Correcting Short-Sightedness

In a myopic eye, the eyeball is too long (or the cornea too curved), so parallel rays from a distant object converge to a focus in front of the retina. The image that lands on the retina is therefore a blurred circle of light rather than a sharp point.

The myopic eye does have a far point — the most distant point it can see clearly without accommodation. For a myopic person, this far point is closer than infinity; it might be 2 m, 1 m, or even less. The correcting lens must take a distant object (effectively at infinity) and produce a virtual image at the patient's far point, so the eye can then focus it onto the retina.

The correcting lens for myopia is a diverging (concave) lens. It spreads the incoming parallel rays so that they appear to diverge from the patient's far point. The eye then treats those diverging rays as if they came from a real object at the far point — which it can focus perfectly.

On a ray diagram: draw parallel rays from a distant object hitting the diverging lens. The refracted rays diverge; trace them back (dashed lines) to find where they appear to originate — that virtual image sits at the far point. The eye, positioned to the right of the lens, receives diverging rays and converges them onto the retina.

Worked example: Calculating the Power of a Myopia-Correcting Lens

A patient's far point is 0.50 m from the eye. Calculate the power of the correcting lens needed so that the patient can see distant objects clearly. [3 marks]

1. Object at infinity: $u = -\infty$ Required image (virtual, same side as object): $v = -0.50$ m — Using the real-is-positive convention, a real object is at negative infinity. The image must be virtual and on the same side as the object, so $v$ is negative. The magnitude equals the far point distance. [M1]
2. $\frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{-0.50} - \frac{1}{-\infty}$ — $\frac{1}{-\infty} = 0$, so the second term vanishes.
3. $\frac{1}{f} = -2.0 - 0 = -2.0 \text{ m}^{-1}$ So $f = -0.50$ m — The negative focal length confirms a diverging lens, as expected for myopia. [M1]
4. $P = \frac{1}{f} = \frac{1}{-0.50} = -2.0$ D — Power is negative, confirming a diverging lens. Always include the unit (D) and the sign. [A1]

Answer: $P = -2.0$ D (diverging lens)

4. Hypermetropia — Correcting Long-Sightedness

Hypermetropia arises when the eyeball is too short or the cornea too flat. Rays from a near object would converge to a focus behind the retina — the eye lacks sufficient converging power to bring them to focus in time. The result is blurred near vision.

The key reference point here is the near point — the closest distance at which the eye can focus clearly with maximum accommodation. For a normal adult eye this is approximately 25 cm. A hypermetropic eye has a near point further away than 25 cm, meaning the person cannot focus on objects that are close.

The correcting lens for hypermetropia is a converging (convex) lens. It takes an object at the standard near point (25 cm) and produces a virtual image at the patient's actual near point — further away, where the eye can focus. The eye then converges those rays onto the retina.

On a ray diagram: draw diverging rays from an object at 25 cm hitting the converging lens. The lens refracts them so they diverge less steeply (or converge), and when traced back, the virtual image appears at the patient's near point. The eye receives rays that appear to come from that more distant point and can now focus them.

Worked example: Calculating the Power of a Hypermetropia-Correcting Lens

A hypermetropic patient has a near point of 1.0 m. Calculate the power of the correcting lens that allows the patient to read a book held at 25 cm. [3 marks]

1. Object distance: $u = -0.25$ m (real object, 25 cm from lens) Image distance: $v = -1.0$ m (virtual image at near point, same side as object) — The object is real so $u$ is negative. The image must be virtual (on the same side as the object) at the near point, so $v = -1.0$ m. [M1]
2. $\frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{-1.0} - \frac{1}{-0.25}$ — Substitute both values carefully, keeping the signs.
3. $\frac{1}{-1.0} = -1.0 \text{ m}^{-1}$ $\frac{1}{-0.25} = -4.0 \text{ m}^{-1}$ $\frac{1}{f} = -1.0 - (-4.0) = -1.0 + 4.0 = +3.0 \text{ m}^{-1}$ — Subtracting a negative is the same as adding. The positive result confirms a converging lens. [M1]
4. $P = +3.0$ D — Positive power confirms a converging lens, as expected for hypermetropia. State the unit and sign. [A1]

Answer: $P = +3.0$ D (converging lens)

5. Comparing Myopia and Hypermetropia — Side by Side

Examiners frequently ask you to compare the two defects. The table below consolidates the key distinctions in a format that maps directly onto 'compare' and 'describe' command words.

Notice the consistent logic: the correcting lens always produces a virtual image at the eye's natural reference point (far point for myopia, near point for hypermetropia). The eye then does the final focusing onto the retina as normal.

6. Astigmatism — Prescriptions and the Cylindrical Lens

Myopia and hypermetropia both assume the eye has the same refracting power in every meridian (plane through the optical axis). Astigmatism breaks that assumption: the cornea or lens is more curved in one meridian than another, like the surface of a rugby ball rather than a football.

The result is that horizontal and vertical lines cannot both be in focus simultaneously. A person with astigmatism might see vertical lines sharply but horizontal lines blurred, or vice versa. The defect is corrected using a cylindrical lens — a lens that has refracting power in one meridian only, leaving the perpendicular meridian unaffected.

In practice, most patients have a combination of spherical defect (myopia or hypermetropia) and astigmatism. The prescription therefore contains three components, written in a standardised format that you need to be able to read and interpret.

Here is an example prescription: −1.50 / −0.75 × 180. The SPH value of −1.50 D corrects the underlying myopia. The CYL value of −0.75 D is the additional diverging power applied in one specific meridian. The axis of 180° tells the optician to orient the cylindrical correction horizontally.

A prescription with CYL = 0 (or no CYL entry) means no astigmatism correction is needed — the defect is purely spherical. When CYL is present, the total power experienced by light passing through the lens depends on the meridian: in the axis direction the power equals SPH alone; perpendicular to the axis the power equals SPH + CYL.

7. Exam Technique — Getting Every Mark

Calculation questions on correcting lenses follow a predictable mark scheme structure. The method mark is almost always awarded for correctly identifying $u$ and $v$ with the right signs before substituting. Losing the sign on $v$ is the single most common reason students drop marks here.

Ray diagram questions reward precision. Draw at least two rays: one parallel to the principal axis (refracted through the focal point for a converging lens, or appearing to diverge from the focal point for a diverging lens) and one through the optical centre (undeviated). Extend refracted rays back with dashed lines to locate a virtual image. Label the object, image, and focal points.

Worked example: Exam-Style Multi-Step: Finding Correcting Lens Power from Near and Far Point Data

A patient's far point is 2.0 m and near point is 0.40 m. (a) Calculate the power of the lens needed to correct their distance vision. (b) With the distance correction in place, determine whether the patient can read a book at 25 cm, given that the corrected eye's near point remains at 0.40 m. [4 marks]

1. Far point = 2.0 m (finite, not infinity) → patient is myopic — A finite far point always indicates myopia. The correcting lens must image infinity to the far point.
2. $u = -\infty$, so $\frac{1}{u} = 0$ $v = -2.0$ m (virtual image at far point) $P = \frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{-2.0} - 0$ — The image is virtual and on the same side as the object, so $v$ is negative. [M1]
3. $P = \frac{1}{-2.0} = -0.50$ D — Negative power confirms a diverging lens. [A1]
4. With the −0.50 D lens, an object at 25 cm ($u = -0.25$ m) forms an image at: $\frac{1}{v} = P + \frac{1}{u} = -0.50 + \frac{1}{-0.25} = -0.50 + (-4.0) = -4.5 \text{ m}^{-1}$ $v = \frac{1}{-4.5} \approx -0.22$ m — The image of the book falls at approximately 0.22 m from the lens. The patient's near point is 0.40 m, so they can only focus on objects at 0.40 m or beyond. Since 0.22 m < 0.40 m, the image is closer than the near point. [M1]
5. Image distance ≈ 0.22 m; near point = 0.40 m 0.22 m < 0.40 m → image is inside the near point — The patient cannot focus on the image because it falls closer than their near point. They would need a separate reading correction (or bifocal lenses). [A1]

Answer: (a) $P = -0.50$ D (diverging lens). (b) The correcting lens images the book to approximately 0.22 m, which is closer than the patient's near point of 0.40 m, so they cannot read at 25 cm with the distance correction alone.

This multi-step problem illustrates a common exam extension: correcting one defect does not automatically fix the other. A patient with both myopia and a reduced near point may need separate distance and reading prescriptions — or a combined prescription with different powers in different zones of the lens.

Across both spec requirements — lens power calculations and astigmatism prescription format — the underlying logic is the same: identify where the eye's focus falls relative to the retina, determine what virtual image the correcting lens must produce, and apply $P = 1/f$ with consistent sign convention throughout.