Power of accommodation

Hold a finger an arm's length away, then slowly bring it toward your nose — for a while it stays sharp, and this is not luck. It is the eye's power of accommodation: the ability of the flexible eye lens to change its own curvature, and hence its focal length, so that objects anywhere from a book held close to a mountain far away all land in sharp focus on the retina. No other part of the eye moves to do this — the lens itself reshapes, pulled and released by the ciliary muscles. This article stays inside the same picture Pass 1 built — near point at 25 cm, far point at infinity, power $P = 1/f$ — but pushes it one step further with the lens formula, turning "the lens gets more curved for near objects" into an actual number: how many dioptres of extra power the eye musters between looking at infinity and reading at 25 cm.

The core insight

The power of accommodation is the eye lens's ability to change its focal length (via the ciliary muscles) so that it can form a sharp image on the retina for objects anywhere between the near point (25 cm) and the far point (infinity).

The core idea

The eye lens is not rigid glass — it is a flexible, jelly-like transparent structure whose curvature is controlled entirely by the ciliary muscles attached to it through suspensory ligaments.

Ciliary muscles relaxed → lens pulled thin, flat, long focal length, low power → suited to distant objects, down to the far point (infinity for a normal eye).
Ciliary muscles contracted → tension on the ligaments released, lens bulges thicker and more curved, short focal length, high power → suited to near objects, down to the near point (25 cm for a normal eye).

Power and focal length are linked by

P = \frac{1}{f}

where: $P$ is the power of the lens in dioptres (D), and $f$ is the focal length in metres, taken positive for the converging eye lens. A shorter $f$ means a larger $P$ — more bending power, needed for the more divergent rays reaching the eye from a nearby object. Because the retina sits at a fixed distance behind the lens, the eye cannot simply move its lens back and forth like a camera moving its lens barrel; instead it must change how curved the lens is, image distance held essentially constant. That single constraint — fixed image distance, variable object distance, variable focal length — is exactly what lets accommodation be worked out with the ordinary lens formula, treating the eye as one effective converging lens.

Figure

Ciliary muscles relaxed give a thin flatter lens focusing a distant object while ciliary muscles contracted give a thicker more curved lens focusing a near object, both images landing sharply on the retina

Derivation AThe power of accommodation in dioptres

The goal is to turn the qualitative statement "the eye lens curves more for near objects" into a definite number of dioptres, using nothing but the lens formula already known from the ray-optics chapter, treating the eye as a single effective converging lens with its image always formed at the same fixed distance on the retina — a standard simplification for this level that lumps the cornea's fixed bending together with the lens's adjustable bending into one convenient thin lens.

Step 1. Set up the fixed geometry. Take the distance from the eye's lens to the retina as $v = 2.5\ \text{cm} = 0.025\ \text{m}$ , positive, since the real image always forms on the retina behind the lens. This distance does not change as the object moves — only the lens's focal length adjusts.

Step 2. Apply the lens formula for the far point. Using the Cartesian sign convention, $\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f} = P$ , with $u \to -\infty$ (far point at infinity), so $1/u \to 0$ . Then $P_{\text{far}} = \dfrac{1}{v} = \dfrac{1}{0.025\ \text{m}} = 40\ \text{D}$ . This is the eye's minimum power, with the ciliary muscles fully relaxed.

Step 3. Apply the lens formula for the near point. Now $u = -0.25\ \text{m}$ (near point, object distance measured negative, on the same side as the incoming light). $P_{\text{near}} = \dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{0.025} - \dfrac{1}{-0.25} = 40 + 4 = 44\ \text{D}$ . This is the eye's maximum power, with the ciliary muscles fully contracted.

Step 4. Subtract to get the power of accommodation. The power of accommodation is defined as the extra power the eye lens can supply above its relaxed minimum: $\Delta P = P_{\text{near}} - P_{\text{far}} = 44 - 40 = 4\ \text{D}$ . A normal eye can add up to about $4$ dioptres of power to shift its focus all the way from infinity down to 25 cm — a compact number that quantifies exactly what Pass 1 described only in words.

Worked example

Question: Using the same eye model as above ( $v = 2.5\ \text{cm}$ ), find the power the eye lens must supply to sharply focus on an object placed $1\ \text{m}$ away, and check that this power lies within the eye's accommodation range found above.

Step 1. Identify the known values: $v = 0.025\ \text{m}$ (fixed retina distance), $u = -1\ \text{m}$ (object 1 m away, sign convention).

Step 2. Apply the lens formula: $P = \dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{0.025} - \dfrac{1}{-1} = 40 + 1 = 41\ \text{D}$ .

Step 3. Compare with the range found in the derivation: the eye's power must lie between $P_{\text{far}} = 40\ \text{D}$ (looking at infinity) and $P_{\text{near}} = 44\ \text{D}$ (looking at the near point). Since $40\ \text{D} < 41\ \text{D} < 44\ \text{D}$ , an object at $1\ \text{m}$ is comfortably within the range the ciliary muscles can achieve — only a small, partial contraction is needed, adding just $1\ \text{D}$ over the relaxed state.

Answer: the eye lens must supply a power of 41 D to focus on an object 1 m away, which lies well within its 40 D to 44 D accommodation range.

Common traps

Forgetting that $u$ is taken as negative (object distance) in the sign convention used with the lens formula — dropping the sign flips the answer.
Treating the retina-to-lens distance $v$ as something that changes with the object's position — it is fixed; accommodation works by changing $f$ (via curvature), not $v$ .
Reporting the power of accommodation as $44\ \text{D}$ or $40\ \text{D}$ instead of the difference, $4\ \text{D}$ — it is defined as the range the lens can add, not either endpoint alone.
Forgetting this two-power model (40 D relaxed, 44 D contracted, 2.5 cm retina distance) is a simplified numerical picture for this level; the eye's real total power is shared with the fixed-power cornea, which this model absorbs into the single effective lens for simplicity.

Ask the tutor: want to see how these same near point and far point numbers change when the eye develops myopia or hypermetropia — ask the AI tutor to walk through the lens-power correction for a defective eye.

Why do we fix the retina distance at 2.5 cm — is that a real measured value?

Yes, it is a commonly used approximate value for the distance from the eye lens to the retina in an adult human eye, and it is what makes the lens-formula calculation possible at this level. The exact anatomical distance varies slightly between individuals, but 2.5 cm is the standard reference value used in this kind of numerical, just as 25 cm is the standard near point. Both are treated as fixed, known constants for the calculation, not values you derive.

Why does the near point give the maximum power and not the far point?

Power increases as focal length decreases, $P = 1/f$ . A nearby object sends out rays that diverge more sharply, so the lens must bend them more strongly — shorter focal length, higher power — to still bring them to a focus at the fixed retina distance. A distant object's rays are nearly parallel and need the least bending. So maximum curvature and maximum power happen at the near point, and minimum curvature and minimum power happen at the far point (infinity), matching what Pass 1 already established qualitatively.

Is this 4 D calculation something I need to memorise, or can I always derive it?

You should be able to derive it any time from just three things: the lens formula $1/v - 1/u = 1/f = P$ , the fixed retina distance (usually given, commonly 2.5 cm), and the near and far point values (25 cm and infinity for a normal eye, or different values if the question describes a defective eye). Compute $P_{\text{near}}$ and $P_{\text{far}}$ separately using the given $u$ values, then subtract. The number 4 D is specific to the standard 25 cm and 2.5 cm values; a different retina distance or near point in a question will give a different answer.

Power of accommodation

The core idea

Worked example

Common traps

Before this

Where this leads