LECTURE 18 HUMAN EYE

Instructor: Kazumi Tolich Lecture 18

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¨ 19.2 The human eye ¤ Focusing and accommodation ¤ Vision defects and their correction 19.2 The human eye

¨ Incoming light rays are refracted (mostly at the air-cornea Vitreous boundary) to produce a real inverted image on the retina. humor ¤ For a cataract, the lens is removed, and the cornea alone � = 1.34 provides a marginal level of vision. � = 1.40 � = 1.33 ¨ Rods and cones on the retina convert the light into � = 1.00 electrical impulses, which travel down the optic nerve to the brain.

¨ The brain adjusts the image in order for it to appear upright. � = 1.38

¨ The iris determines how much light enters the eye through the pupil. 19.2 Focusing and accommodation

¨ Accommodation is the process of changing the lens shape as the eye focuses at different distances.

¨ The near point (NP) is the closest point on which the eye can focus.

¨ The far point (FP) is the most distant point on which the relaxed eye can focus. Quiz: 19.2-1

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¨ We will approximate the eye as a single thin lens 2.30 cm from the retina.

¨ What is the smallest distance an object can be from the lens such that an image is focused on the retina if the focal length of the lens is 2.10 cm?

� �

� = 2.30 cm � = 2.30 cm Quiz: 19.2-1 answer

¨ The thin-lens equation: + = ¨ � = 2.30 cm and � = 2.10 cm ¨ � = − = − = 0.242 m . .

¨ This is close to the near point of a normal eye, approximately 25 cm.

� �

� = 2.30 cm � = 2.30 cm Quiz: 19.2-2

¨ If an object is moved just inside the near point, where is the image formed? A. Behind the retina. B. In front of the retina. Quiz: 19.2-2 answer

¨ If an object is moved just inside the near point, where is the image formed?

¨ Behind the retina. ¨ The thin-lens equation: + = ¨ � = − = = ¨ When � decreases, � increases. Quiz: 19.2-3

¨ Determine the new near point in meters, if the shortest focal length increases to 2.15 cm due to loss of flexibility of the lens. Assume the lens-to-retina distance is still normal value of 2.30 cm. Quiz: 19.2-3 answer

¨ The thin-lens equation: + = ¨ � = 2.30 cm and � = 2.15 cm ¨ � = − = − = 0.330 m . . ¨ Presbyopia is a refractive error in the eye due to the loss of accommodation.

¨ For children, the NP can be as little as 10 cm, for young adults the average is 25 cm. At age 40–45, the NP moves out and can reach 200 cm by age 60. Quiz: 19.2-4

¨ Let’s now consider the case where the width of the eye decreases to 2.25 cm, but the minimum focal length of the lens is normal value of 2.10 cm. Determine the new near point in meters. Quiz: 19.2-4 answer

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¨ The thin-lens equation: + = ¨ � = 2.25 cm and � = 2.10 cm ¨ � = − = − = 0.315 m . .

¨ A person with shortened eyeball has farsightedness or hyperopia, and cannot see nearby objects.

¨ In both cases, presbyopia and hyperopia, the near point is longer than normal. Quiz: 19.2-5

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¨ Now suppose that the normal eye with its width of 2.30 cm is focused on an object at infinity.

¨ What is the focal length of the lens in cm? Quiz: 19.2-5 answer

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¨ The thin-lens equation: + = ¨ As � approaches infinity, + = or = ¨ � approaches � = 2.30 cm. Quiz: 19.2-6

¨ Let’s now consider the case where the width of the eye increases to 2.33 cm from the normal value of 2.30 cm while the maximum focal length of the lens is still the normal value of 2.30 cm. For an object at infinity, where is the image formed? A. In front of the retina. B. Behind the retina. Quiz: 19.2-6 answer

¨ In front of the retina.

¨ The image is formed 2.30 cm from the lens, but the retina is farther away.

¨ A person with nearsightedness, or myopia, has an elongated eyeball, and cannot see distant objects. Quiz: 19.2-7

¨ Continue to consider the case where the width of the eye increases to 2.33 cm while the maximum focal length of the lens is still 2.30 cm.

¨ What is the far-point distance in meters when this change occurs in the eyeball? Quiz: 19.2-7 answer

18 ¨ The thin-lens equation: + = ¨ � = 2.33 cm and � = 2.30 cm ¨ � = − = − = 1.79 m . . ¨ The small change in the width of the eyeball (3 tenths of a millimeter!!) results in a reduction of the far-point from an infinite distance to a mere 1.79 m. 19.2 Vision defects and their correction

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¨ Presbyopia, hyperopia and myopia can be corrected with corrective lenses.

¨ When corrective lens is used in combination with the eye’s lens, the image produced by the corrective lens acts as the object for the eye’s lens.

¨ Refractive power in diopter D:

1 � = � Quiz: 19.2-8

¨ Jamie’s far point is 2.25 m from her eye. To correct her vision, Jamie needs glasses that create an image of distant objects that is

A. at her near point. B. at her far point. C. at infinity. D. 25 cm from her eye. Quiz: 19.2-8 answer

¨ Jamie’s far point is 2.25 m from her eye. To correct her vision, Jamie needs glasses that create an image of distant objects that is at her far point.

¨ The corrective lenses need to bring the image of a far away object close enough for her to focus.

¨ Yet, they should not bring it too close otherwise every object closer appear too close to her. Quiz: 19.2-9

¨ Jamie’s far point is 2.25 m from her eye. What is the refractive power of her corrective lenses in diopters? Quiz: 19.2-9 answer

23 ¨ The thin-lens equation: + = ¨ The image created by the corrective lens must be at her far point, � = −2.25 m. ¨ The object is far away, so � → ∞, + = or = ¨ � = � = −2.25 m. ¨ � = = = −0.444 D . ¨ The focal length must be in meters. Quiz: 19.2-10

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¨ In the famous novel Lord of the Flies, one of the boys is nearsighted, and his glasses are used to start a fire. Is this a feasible story? A. Yes B. No Quiz: 19.2-10 answer

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¨ No Hyperopia

¨ Concave (diverging) lenses would have not been able to focus the rays of the sun.

¨ Nearsightedness (Myopia) is corrected with a diverging lens. Myopia ¨ Farsightedness (Hyperopia) is corrected with a converging lens.