Describe the use of converging and diverging lenses to correct long-sightedness and short-sightedness

3.2 Thin Lenses – Correcting Long‑sightedness and Short‑sightedness

Learning objective

Explain how converging (convex) and diverging (concave) lenses are used to correct hyperopia (long‑sightedness) and myopia (short‑sightedness). Include the action of each lens on a parallel beam, the required image characteristics, the relevant sign conventions, and the quantitative use of the lens formula and optical power as required by the Cambridge IGCSE 0625 syllabus.

Key terminology

• Principal axis – the straight line that passes through the centre of a thin lens and is perpendicular to its surfaces.
• Principal focus (focal point) – the point on the principal axis where rays that were originally parallel to the axis converge (convex lens) or appear to diverge from (concave lens) after passing through the lens.
• Focal length, \(f\) – distance between the lens centre and its principal focus. Positive for converging lenses, negative for diverging lenses.
• Optical power, \(P\) – \(P=\dfrac{1}{f}\) (with \(f\) in metres, \(P\) in dioptres, D). Positive power → converging lens; negative power → diverging lens.

Sign conventions (IGCSE)

Lens formula

The thin‑lens equation is

\[

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

\]

Apply the sign rules above when substituting values. The formula is used to:

• Find the image distance for a given object distance and lens power.
• Calculate the required lens power from a measured far‑point (myopia) or near‑point (hyperopia).

Image characteristics of thin lenses

Ray‑diagram construction (textual)

Converging lens – real, inverted image

1. Ray 1: parallel to the principal axis; after refraction it passes through the principal focus on the far side.
2. Ray 2: passes through the centre of the lens; it continues undeviated.
3. The intersection of the two refracted rays on the far side gives the image point. For a distant object (object at infinity) the image forms at the focal plane, a real, inverted image at distance \(f\) from the lens.

Diverging lens – virtual, upright image

1. Ray 1: parallel to the principal axis; after refraction it diverges as if it originated from the principal focus on the same side as the object.
2. Ray 2: aimed toward the centre of the lens; it passes straight through.
3. Extend the two refracted rays backwards; their apparent intersection on the object side is the virtual image. The image is upright, reduced, and appears at a distance \(|f|\) from the lens.

Figure placeholders (to be drawn in class):

• Figure 1 – Converging lens forming a real image on a screen.
• Figure 2 – Diverging lens forming a virtual image seen by looking into the lens.

How the eye becomes defective

1. Long‑sightedness (hyperopia) – The eye is too short or its refractive power is too weak. Parallel rays from a distant object would focus behind the retina.
2. Short‑sightedness (myopia) – The eye is too long or its refractive power is too strong. Parallel rays focus in front of the retina.

Corrective lenses – action on a parallel beam

Why a converging lens corrects hyperopia

For a hyperopic eye the uncorrected focal point lies behind the retina. Placing a converging lens a short distance in front of the eye creates a virtual image of the distant object at a finite distance \(v_{\text{lens}}\). This virtual image becomes the object for the eye:

\[

\frac{1}{f{\text{eye}}}= \frac{1}{v{\text{eye}}}-\frac{1}{u{\text{eye}}},\qquad u{\text{eye}}=-v_{\text{lens}}

\]

Choosing a lens power \(P{\text{lens}}=+1/f{\text{lens}}\) such that the resulting object distance \(v_{\text{lens}}\) makes the eye’s image distance equal to the axial length of the eye brings the image onto the retina.

Why a diverging lens corrects myopia

For a myopic eye the focal point lies in front of the retina. A diverging lens forms a virtual image at the eye’s far‑point \(d\) (the farthest distance the eye can see clearly without correction). The required power is therefore

\[

P_{\text{lens}}=-\frac{1}{d}\;\text{dioptres}

\]

The negative sign indicates a diverging lens. By moving the apparent object to the far‑point, the eye’s own (excessive) power now focuses the rays exactly on the retina.

Practical steps for determining a prescription

1. Determine the eye’s far‑point (myopia) or the extra convergence needed for the near‑point (hyperopia).

   • Myopia: Measure the far‑point distance \(d\) (e.g., 0.30 m). Use \(P=-1/d\).
   • Hyperopia: Measure the near‑point that the eye can focus on without correction, or calculate the extra power needed to bring the image from infinity to the retina.

2. Calculate the lens power with \(P = \pm 1/f\) (use the sign conventions above).
3. Round the result to the nearest 0.25 D, because commercial lenses are supplied in quarter‑dioptre steps.

Worked examples (quantitative)

Example 1 – Myopia

John’s far‑point is 0.28 m. What is the prescription?

1. Use the formula \(P = -\dfrac{1}{d}\):
   

   \(P = -\dfrac{1}{0.28\ \text{m}} = -3.57\ \text{D}\).

2. Round to the nearest 0.25 D → –3.50 D.
3. Lens focal length: \(f = \dfrac{1}{P} = -0.286\ \text{m}\) (negative, confirming a diverging lens).

Example 2 – Hyperopia

Emma’s eye needs an additional converging power of +2.75 D to bring the image onto the retina. What focal length and lens type are required?

1. Positive power → converging (convex) lens.
2. Focal length \(f = \dfrac{1}{P} = \dfrac{1}{+2.75}=+0.364\ \text{m}\).
3. Prescription: +2.75 D (already a standard quarter‑dioptre value).

Optional: Dispersion of light (link to 3.2.3)

Thin lenses also exhibit chromatic dispersion because the refractive index of glass varies with wavelength. Short‑wave (blue) light is bent more than long‑wave (red) light, giving the lens a slightly different focal length for each colour. In the IGCSE syllabus this is an optional point, often illustrated with a prism or a simple “rainbow” produced by a glass prism.

Summary

• Converging (convex) lenses have \(f>0\) and \(P>0\); they bend parallel rays inward, producing a real image at the focal plane. They are used to correct hyperopia by moving the image forward onto the retina.
• Diverging (concave) lenses have \(f<0\) and \(P<0\); they bend parallel rays outward, producing a virtual upright image at the focal length. They are used to correct myopia by moving the apparent object to the eye’s far‑point.
• Apply the IGCSE sign conventions in the lens formula \(\dfrac{1}{f}= \dfrac{1}{v}-\dfrac{1}{u}\) to calculate the required power from a measured far‑point (myopia) or extra convergence (hyperopia).
• Round the calculated power to the nearest 0.25 D for a realistic prescription.
• Understanding image type (real/virtual), orientation (upright/inverted), and size (reduced/enlarged) is essential for constructing accurate ray diagrams and for answering exam questions.