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

3.2.3 Thin Lenses

Learning Objective

Describe how converging (convex) and diverging (concave) lenses are used to correct long‑sightedness (hyperopia) and short‑sightedness (myopia).

Key Concepts

• A thin lens forms images according to the lens formula \$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\$ where

  • \$f\$ = focal length of the lens (positive for converging, negative for diverging)
  • \$u\$ = object distance (negative in the sign convention used for the eye)
  • \$v\$ = image distance (positive when the image forms on the opposite side of the incoming light)

• The optical power \$P\$ of a lens is defined as \$P = \frac{1}{f}\$ (with \$f\$ in metres, \$P\$ in dioptres, D). Positive \$P\$ → converging lens, negative \$P\$ → diverging lens.
• The normal (emmetropic) eye forms a clear image of a distant object on the retina when the eye’s focal length \$f_{\text{eye}}\$ equals the axial length of the eye.

How the Eye Becomes Defective

Two common refractive errors are:

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

Corrective Lenses

Corrective lenses are placed in front of the eye to shift the image location so that it falls on the retina.

Why a Converging Lens Corrects Hyperopia

For a hyperopic eye, the image of a distant object would form behind the retina. A converging lens placed in front of the eye creates a virtual image of the distant object at a distance \$v\$ such that the eye now receives light as if it originated from a point closer than infinity. The combined system satisfies:

\$\frac{1}{f{\text{eye}}} = \frac{1}{v{\text{eye}}} - \frac{1}{u_{\text{eye}}}\$

where \$u{\text{eye}} = -v{\text{lens}}\$ (the image distance from the lens becomes the object distance for the eye). By choosing a lens power \$P{\text{lens}}\$ that makes \$v{\text{eye}}\$ equal to the axial length of the eye, the image is formed on the retina.

Why a Diverging Lens Corrects Myopia

For a myopic eye, the image of a distant object forms in front of the retina. A diverging lens creates a virtual image that appears farther away (at infinity) so that the eye’s own strong focusing power now brings the rays to focus exactly on the retina. The required lens power is:

\$P_{\text{lens}} = -\frac{1}{d}\$

where \$d\$ is the distance (in metres) by which the eye’s focal point lies in front of the retina. The negative sign indicates a diverging lens.

Practical Steps for Determining Prescription

1. Measure the far‑point of the eye (the farthest distance at which the eye can see a clear image without correction).
2. For myopia, the far‑point distance \$d\$ is finite; the required lens power is \$P = -1/d\$.
3. For hyperopia, determine the near‑point that the eye can focus on; the required converging power is \$P = +1/f{\text{required}}\$, where \$f{\text{required}}\$ is the focal length that will bring the image onto the retina.
4. Round the calculated power to the nearest 0.25 D, as lenses are manufactured in quarter‑dioptre steps.

Summary

• Converging lenses (positive power) are used for long‑sightedness; they bring parallel rays to a focus sooner, allowing the eye to form a clear image on the retina.
• Diverging lenses (negative power) are used for short‑sightedness; they spread rays so the eye’s excessive focusing power now produces an image at the retina.
• The lens power required is directly related to the eye’s defect distance via \$P = \pm 1/f\$ (in dioptres).