# 26.1 Physics of the eye  (Page 3/5)

 Page 3 / 5

The eye can detect an impressive amount of detail, considering how small the image is on the retina. To get some idea of how small the image can be, consider the following example.

## Size of image on retina

What is the size of the image on the retina of a $1\text{.}\text{20}×{\text{10}}^{-2}$ cm diameter human hair, held at arm’s length (60.0 cm) away? Take the lens-to-retina distance to be 2.00 cm.

Strategy

We want to find the height of the image ${h}_{i}$ , given the height of the object is ${h}_{o}=1\text{.}\text{20}×{\text{10}}^{-2}$ cm. We also know that the object is 60.0 cm away, so that ${d}_{o}=60.0 cm$ . For clear vision, the image distance must equal the lens-to-retina distance, and so ${d}_{\text{i}}=2.00 cm$ . The equation $\frac{{h}_{\text{i}}}{{h}_{\text{o}}}=-\frac{{d}_{\text{i}}}{{d}_{\text{o}}}=m$ can be used to find ${h}_{i}$ with the known information.

Solution

The only unknown variable in the equation $\frac{{h}_{\text{i}}}{{h}_{\text{o}}}=-\frac{{d}_{\text{i}}}{{d}_{\text{o}}}=m$ is ${h}_{\text{i}}$ :

$\frac{{h}_{\text{i}}}{{h}_{\text{o}}}=-\frac{{d}_{\text{i}}}{{d}_{\text{o}}}.$

Rearranging to isolate ${h}_{\text{i}}$ yields

${h}_{\text{i}}=-{h}_{\text{o}}\cdot \frac{{d}_{\text{i}}}{{d}_{\text{o}}}.$

Substituting the known values gives

$\begin{array}{lll}{h}_{\text{i}}& =& -\left(1.20×{\text{10}}^{-2}\phantom{\rule{0.25em}{0ex}}\text{cm}\right)\frac{2.00 cm}{\text{60.0 cm}}\\ & =& -4.00×{\text{10}}^{-4}\phantom{\rule{0.25em}{0ex}}\text{cm}.\end{array}$

Discussion

This truly small image is not the smallest discernible—that is, the limit to visual acuity is even smaller than this. Limitations on visual acuity have to do with the wave properties of light and will be discussed in the next chapter. Some limitation is also due to the inherent anatomy of the eye and processing that occurs in our brain.

## Power range of the eye

Calculate the power of the eye when viewing objects at the greatest and smallest distances possible with normal vision, assuming a lens-to-retina distance of 2.00 cm (a typical value).

Strategy

For clear vision, the image must be on the retina, and so ${d}_{\text{i}}=2.00 cm$ here. For distant vision, ${d}_{o}\approx \infty$ , and for close vision, ${d}_{o}=25.0 cm$ , as discussed earlier. The equation $P=\frac{1}{{d}_{\text{o}}}+\frac{1}{{d}_{\text{i}}}$ as written just above, can be used directly to solve for $P$ in both cases, since we know ${d}_{\text{i}}$ and ${d}_{o}$ . Power has units of diopters, where $\text{1 D}=\text{1/m}$ , and so we should express all distances in meters.

Solution

For distant vision,

$P=\frac{1}{{d}_{\text{o}}}+\frac{1}{{d}_{\text{i}}}=\frac{1}{\infty }+\frac{1}{\text{0.0200 m}}\text{.}$

Since $1/\infty =0$ , this gives

$P=0+\text{50}\text{.}0/\text{m}=\text{50.0 D (distant vision).}$

Now, for close vision,

$\begin{array}{lll}P& =& \frac{1}{{d}_{\text{o}}}+\frac{1}{{d}_{\text{i}}}=\frac{1}{\text{0.250 m}}+\frac{1}{\text{0.0200 m}}\\ & =& \frac{\text{4.00}}{\text{m}}+\frac{50.0}{\text{m}}=\text{4.00 D}+\text{50.0 D}\\ & =& \text{54.0 D (close vision).}\end{array}$

Discussion

For an eye with this typical 2.00 cm lens-to-retina distance, the power of the eye ranges from 50.0 D (for distant totally relaxed vision) to 54.0 D (for close fully accommodated vision), which is an 8% increase. This increase in power for close vision is consistent with the preceding discussion and the ray tracing in [link] . An 8% ability to accommodate is considered normal but is typical for people who are about 40 years old. Younger people have greater accommodation ability, whereas older people gradually lose the ability to accommodate. When an optometrist identifies accommodation as a problem in elder people, it is most likely due to stiffening of the lens. The lens of the eye changes with age in ways that tend to preserve the ability to see distant objects clearly but do not allow the eye to accommodate for close vision, a condition called presbyopia    (literally, elder eye). To correct this vision defect, we place a converging, positive power lens in front of the eye, such as found in reading glasses. Commonly available reading glasses are rated by their power in diopters, typically ranging from 1.0 to 3.5 D.

## Section summary

• Image formation by the eye is adequately described by the thin lens equations:
$P=\frac{1}{{d}_{\text{o}}}+\frac{1}{{d}_{\text{i}}}\phantom{\rule{0.25em}{0ex}}\text{and}\phantom{\rule{0.25em}{0ex}}\frac{{h}_{\text{i}}}{{h}_{\text{o}}}=-\frac{{d}_{\text{i}}}{{d}_{\text{o}}}=m.$
• The eye produces a real image on the retina by adjusting its focal length and power in a process called accommodation.
• For close vision, the eye is fully accommodated and has its greatest power, whereas for distant vision, it is totally relaxed and has its smallest power.
• The loss of the ability to accommodate with age is called presbyopia, which is corrected by the use of a converging lens to add power for close vision.

## Conceptual questions

If the lens of a person’s eye is removed because of cataracts (as has been done since ancient times), why would you expect a spectacle lens of about 16 D to be prescribed?

A cataract is cloudiness in the lens of the eye. Is light dispersed or diffused by it?

When laser light is shone into a relaxed normal-vision eye to repair a tear by spot-welding the retina to the back of the eye, the rays entering the eye must be parallel. Why?

How does the power of a dry contact lens compare with its power when resting on the tear layer of the eye? Explain.

Why is your vision so blurry when you open your eyes while swimming under water? How does a face mask enable clear vision?

## Problem exercises

Unless otherwise stated, the lens-to-retina distance is 2.00 cm.

What is the power of the eye when viewing an object 50.0 cm away?

$\text{52.0 D}$

Calculate the power of the eye when viewing an object 3.00 m away.

(a) The print in many books averages 3.50 mm in height. How high is the image of the print on the retina when the book is held 30.0 cm from the eye?

(b) Compare the size of the print to the sizes of rods and cones in the fovea and discuss the possible details observable in the letters. (The eye-brain system can perform better because of interconnections and higher order image processing.)

(a) $-0\text{.}\text{233 mm}$

(b) The size of the rods and the cones is smaller than the image height, so we can distinguish letters on a page.

Suppose a certain person’s visual acuity is such that he can see objects clearly that form an image $4.00 \mu m$ high on his retina. What is the maximum distance at which he can read the 75.0 cm high letters on the side of an airplane?

People who do very detailed work close up, such as jewellers, often can see objects clearly at much closer distance than the normal 25 cm.

(a) What is the power of the eyes of a woman who can see an object clearly at a distance of only 8.00 cm?

(b) What is the size of an image of a 1.00 mm object, such as lettering inside a ring, held at this distance?

(c) What would the size of the image be if the object were held at the normal 25.0 cm distance?

(a) $+62.5 D$

(b) $–0.250 mm$

(c) $–0.0800 mm$

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