Physics · Medical Physics · A2 Level

Myopia and Its Correction

Myopia, better known as short-sightedness, is probably the most familiar defect of vision — and it's also the cleanest possible worked application of the lens equation and sign convention from two lessons ago. This lesson works through exactly what goes wrong optically, and exactly how a pair of glasses fixes it.

7 min read

What goes wrong in a myopic eye

In a myopic (short-sighted) eye, the total refracting power of the cornea and lens system is too great for the length of the eyeball — or, put the other way round, the eyeball is too long for the eye's own power. Either way, the effect on image formation is the same: light from a distant object, which arrives at the eye as an almost parallel bundle of rays, is brought to a focus too soon, converging to a point in front of the retina rather than exactly on it. By the time the light reaches the retina itself, the rays have already crossed over and begun to spread out again, so the image formed there is blurred.

Near objects are a different story. Because light diverging from a near object needs comparatively more power to bring it to a focus, a myopic eye — which already has excess power — can often focus such an object perfectly well without difficulty. This is exactly why short-sightedness shows up as blurred distance vision with clear near vision, the everyday symptom most people associate with it.

The far point

The far point of an eye is defined as the greatest distance at which an object can be seen clearly without any accommodation being used. For a normal, unaffected eye, the far point is effectively at infinity: a relaxed, unaccommodated eye can focus on the most distant objects without difficulty. For a myopic eye, however, the excess power means the far point is pulled in to some finite distance — objects any further away than this cannot be focused sharply, no matter how much (or little) accommodation is used, because the eye's minimum power is still too great.

Correcting myopia with a diverging lens

The correction for myopia is a diverging (concave) lens placed in front of the eye. Its job is to reduce the total power reaching the eye's own optical system, compensating for the eye having too much power to begin with.

More precisely: the corrective lens is chosen so that parallel rays from a genuinely distant object (effectively at infinity) are made to diverge slightly by the lens, so that after passing through it they appear to be coming from a point at the eye's own far point — a distance the myopic eye can already focus on without help. The eye's own lens system then does the rest, focusing that closer, apparently-diverging virtual image onto the retina exactly as it would for any ordinary near object at that same distance.

In words, the ray picture is: parallel light from a distant object strikes the diverging lens; the lens bends the rays outward very slightly before they enter the eye; instead of parallel rays entering the eye directly, slightly diverging rays enter it, tracing back to a virtual image at the eye's far point; the eye's cornea and lens then focus this virtual image onto the retina as normal.

Worked exampleFinding the correcting lens power for a myopic eye

A myopic eye has a far point of $0.25\text{ m}$. Find the power of the corrective lens needed to allow distant (effectively infinitely far) objects to be seen clearly.

The lens must take parallel rays from infinity and produce a virtual image exactly at the far point.

Object distance: $u = \infty$, so $\dfrac{1}{u} = 0$.

A genuinely distant object is treated as being at infinity in this calculation.

Image distance: $v = -0.25\text{ m}$.

Negative, since the image must be virtual, formed on the same side of the lens as the object, at the eye's far point — following the sign convention set up earlier in this topic.

Lens equation: $\dfrac{1}{u} + \dfrac{1}{v} = \dfrac{1}{f}$, so $\dfrac{1}{f} = 0 + \dfrac{1}{-0.25} = -4.0$.

With $1/u = 0$, the focal length is determined entirely by the image distance.

So $f = -0.25\text{ m}$, and the power is $P = \dfrac{1}{f} = -4.0\text{ D}$.

A negative power confirms this must be a diverging lens, exactly as expected for correcting myopia.

Result: P = −4.0 D (diverging lens)

The pattern to remember

For myopia correction, the object is always taken at infinity ($1/u = 0$), and the required image distance $v$ is simply the eye's own far point, entered as a negative number because it is a virtual image. The focal length — and hence the power — then comes directly from $1/f = 1/v$. The corrective power is always negative, confirming the lens must be diverging.

Key takeaways

Myopia (short-sightedness) occurs when the eye's total power is too great (or the eyeball too long), focusing distant light in front of the retina.
Near objects can usually still be focused normally, since they need more power to begin with — myopia mainly affects distance vision.
The far point is the greatest distance an unaccommodated eye can focus on clearly; for a normal eye it is at infinity, for a myopic eye it is some finite distance.
Correction uses a diverging (concave) lens, which always has negative power under the standard sign convention.
The corrective lens takes parallel rays from a distant object and makes them appear to diverge from the eye's own far point, which the eye can then focus normally.
Calculation pattern: object at infinity ($1/u=0$), image distance equal to the (negative) far point, giving $1/f = 1/v$ directly.

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Sensitivity and Resolution of the Eye

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