Typesof illuminationin slit lamp

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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.

5: 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.

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3: 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?

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Douglas College Physics 1207 Copyright © August 22, 2016 by OpenStax is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.

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1: 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?

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Figure 3 shows the accommodation of the eye for distant and near vision. Since light rays from a nearby object can diverge and still enter the eye, the lens must be more converging (more powerful) for close vision than for distant vision. To be more converging, the lens is made thicker by the action of the ciliary muscle surrounding it. The eye is most relaxed when viewing distant objects, one reason that microscopes and telescopes are designed to produce distant images. Vision of very distant objects is called totally relaxed, while close vision is termed accommodated, with the closest vision being fully accommodated.

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The eye is perhaps the most interesting of all optical instruments. The eye is remarkable in how it forms images and in the richness of detail and colour it can detect. However, our eyes commonly need some correction, to reach what is called “normal” vision, but should be called ideal rather than normal. Image formation by our eyes and common vision correction are easy to analyze with the optics discussed in Geometric Optics.

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3: (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?

As noted, the image must fall precisely on the retina to produce clear vision — that is, the image distance di must equal the lens-to-retina distance. Because the lens-to-retina distance does not change, the image distance di must be the same for objects at all distances. The eye manages this by varying the power (and focal length) of the lens to accommodate for objects at various distances. The process of adjusting the eye’s focal length is called accommodation. A person with normal (ideal) vision can see objects clearly at distances ranging from 25 cm to essentially infinity. However, although the near point (the shortest distance at which a sharp focus can be obtained) increases with age (becoming meters for some older people), we will consider it to be 25 cm in our treatment here.

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We will use the thin lens equations to examine image formation by the eye quantitatively. First, note the power of a lens is given as P=1/f , so we rewrite the thin lens equations as

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What is the size of the image on the retina of a 1.20 x 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.

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

Figure 1 shows the basic anatomy of the eye. The cornea and lens form a system that, to a good approximation, acts as a single thin lens. For clear vision, a real image must be projected onto the light-sensitive retina, which lies at a fixed distance from the lens. The lens of the eye adjusts its power to produce an image on the retina for objects at different distances. The center of the image falls on the fovea, which has the greatest density of light receptors and the greatest acuity (sharpness) in the visual field. The variable opening (or pupil) of the eye along with chemical adaptation allows the eye to detect light intensities from the lowest observable to 1010 times greater (without damage). This is an incredible range of detection. Our eyes perform a vast number of functions, such as sense direction, movement, sophisticated colours, and distance. Processing of visual nerve impulses begins with interconnections in the retina and continues in the brain. The optic nerve conveys signals received by the eye to the brain.

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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).

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.

3: (a) -0.233 mm (b) The size of the rods and the cones is smaller than the image height, so we can distinguish letters on a page.

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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 Figure 3. 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.

For clear vision, the image must be on the retina, and so di = 2.00 cm here. For distant vision, do ≈ ∞ , and for close vision, do = 25.0 cm, as discussed earlier. The equation [latex]   P = \frac{1}{d_o} + \frac {1}{d_i} [/latex] as written just above, can be used directly to solve for P in both cases, since we know di and do. Power has units of diopters, where  1 D = 1/m , and so we should express all distances in meters.

Look at the central transparent area of someone’s eye, the pupil, in normal room light. Estimate the diameter of the pupil. Now turn off the lights and darken the room. After a few minutes turn on the lights and promptly estimate the diameter of the pupil. What happens to the pupil as the eye adjusts to the room light? Explain your observations.

Refractive indices are crucial to image formation using lenses. Table 1 shows refractive indices relevant to the eye. The biggest change in the refractive index, and bending of rays, occurs at the cornea rather than the lens. The ray diagram in Figure 2 shows image formation by the cornea and lens of the eye. The rays bend according to the refractive indices provided in Table 1. The cornea provides about two-thirds of the power of the eye, owing to the fact that speed of light changes considerably while traveling from air into cornea. The lens provides the remaining power needed to produce an image on the retina. The cornea and lens can be treated as a single thin lens, even though the light rays pass through several layers of material (such as cornea, aqueous humour, several layers in the lens, and vitreous humour), changing direction at each interface. The image formed is much like the one produced by a single convex lens. This is a case 1 image. Images formed in the eye are inverted but the brain inverts them once more to make them seem upright.

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(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.)

4: Suppose a certain person’s visual acuity is such that he can see objects clearly that form an image 4.00 μm 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?

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We want to find the height of the image hi given the height of the object is ho = 1.20 x 10-2 cm. We also know that the object is 60.0 cm away, so that do = 60.0 cm. For clear vision, the image distance must equal the lens-to-retina distance, and so di = 2.00 cm . The equation [latex]\boldsymbol{ \frac{h_i}{h_o} = m {-} \frac{d_i}{d_o} = m }[/latex] can be used to find hi with the known information.

We understand that di must equal the lens-to-retina distance to obtain clear vision, and that normal vision is possible for objects at distances do = 25 cm to infinity.