Physics Chapter Review Chapter 25- The Eye and Optical Instruments Ethan Blitstein The Human Eye As light enters through the human eye it first passes through the cornea (a thin transparent membrane of a spherical shape) with a refractive index of 1.38. The majority of the refraction of light occurs at the air cornea surface because the change in the index is the greatest there out of all the surfaces inside the eye. After passing through the cornea, the light enters a watery substance with a refractive index of 1.34, called an aqueous humor. o The incoming light encounters an iris, which is the curtain that open and closes to regulate the amount of light entering the eye. The opening is called the pupil. The human eye also refracts again at the crystalline lens, where light entering the pupil passes through. o The crystalline lens is a small, transparent body that has multiple different refractive indices layers. The effective overall index of the lens is 1.42 Both of these refractions form a real inverted image on the retina because your eye is a convex lens and somehow through your nerves and brain, the image appears erect. Depending on the distance an object is from the eye, the eye adjusts through accommodation. o Accommodation is a process by which the shape of the crystalline lens varies and hence the total optical power of the eye varies. Accommodation enables the eye to have clear vision for a variety of object distances, both near and far. For a person with normal vision a point classified as far is and for a person with normal vision a point classified as near is approximately 25 cm. Myopia (nearsightedness) is the condition of the eye where there is too much converging power for the length of the eye. As a result, a distant object is imaged directly in front of the retina and appears blurred. o Myopia can be corrected/treated by placing an appropriate negative lens in front of the eye through glasses, contacts, etc. Hyperopia (farsightedness), is the opposite of myopia. Hyperopia is the condition of the eye in which there is too little converging power for the length of the eye. As a result a close object is imaged behind the retina and appears blurred. o Hyperopia can be corrected/treated by placing an appropriate positive lens in front of the eye through glasses, contacts, etc. Astigmatism occurs because an asymmetrically shaped cornea exists.
o Astigmatism can be corrected/treated through an asymmetric lens. The location of an image point on the retina can be located when a ray is drawn from the object point straight through a point O on the eye s optical axis, 16 mm in front of the retina. o The reduced eye is a drawing (see example 1 below) in which the eye is represented by a circle, with one ray drawn through O to help determine image size. If an object subtends at an angle θ, the height of the image is given by the formula h= (16mm)(tanθ) Example 1: Visual acuity is the eye s capacity to see fine detail. o A measure of acuity is the minimum angle between points that can be resolved by the eye. However, this angle is dependent on the individual s eye. For the most part the angle lies between θmin= 5 x 10-4 rad and θmin= 4 x 10-4. The angular magnification of an optical instrument, M, is defined by the equation: M= (θ /θ) o Where θ is the angle formed at point O in the reduced eye by the retinal image without the instrument, and θ is the angle subtended by the retinal image with the instrument. The Magnifier A magnifier is a positive lens or lens system. The magnifier gives an angular magnification when used by a relaxed normal eye. o Here M = (d/f). D represents the near-point distance, and f is the focal length of the lens. For maximum magnification, the magnifier is held close to the eye, and the object is places so that an image is formed at the eye s near point, which gives a magnification of M= ((d/f)+1) The compound microscope has an objective and an ocular. o The objective is a short focal length, positive lens system, which forms a magnified real image of a small image place outside its focal point. (focal point is determined by the refractive index of the lens material) o The ocular functions as a magnifier, allowing the eye to come very close to the image produced by the objective.
o The overall angular magnification of the microscope is the product of the objective and ocular magnifications. Therefore, M= (M1M2) o The resolution of a microscope may be limited by aberrations or diffraction. A telescope has both an ocular and an objective. o The ocular serves the same function it does in the microscope- to allow the eye to come close to the image formed by the objective. o The objective is a long-focal length, positive lens system or mirror, which is used to form a real image of an object in the distance. The angular magnification of the astronomical telescope equals the ratio of the respective focal lengths of the objective and ocular. M = (f1/f2) F1 is for objective and F2 is for the foc eyepiece magnification (ocular)
Questions: 1. How does the human eye work? 2. What happens to an image inside the eye? 3. What is a crystalline lens? 4. What is the effective overall index of the crystalline lens? 5. What factor, other than the curvature of the surfaces of the lens, determines the location of the focal point? 6. Why does an astigmatism occur? 7. Explain hyperopia 8. Explain myopia 9. What is visual acuity? 10. Where does the average measure of acuity angle lay between? 11. What is a magnifier? 12. What is an objective in a compound microscope? 13. What is an ocular in a compound microscope? 14. What is an objective in a telescope? 15. What is an ocular in a telescope? 16. What is the maximum angular magnification of a lens having a focal point of 9.42 cm. 17. An astronomical telescope has an objective lens with a focal length of 3000 mm. The astronomer is using an eyepiece with a 30 mm focal length. What is the angular magnification power of the telescope? 18. Based on the reduced eye drawing with the 16 mm given, if θ = 60, what is h? 19. If an angle viewed by the reduced eye formed at point O by the retinal image using the microscope is 80, and the angle subtended by the retinal image viewed with the instrument is 75, what is the angular magnification of the optical instrument? Answers: 1. Light passes through the cornea and pupil (lenses). The iris (muscle) controls how much light can pass through. There are rods on the back of the eye that reads the image and then it is sent through the optic nerve. 2. The image is actually inverted because your eye is a convex lens and somehow through your nerves and brain, the image appears erect. 3. The crystalline lens is a small, transparent body that has multiple different refractive indices layers 4. The effective overall index of the crystalline lens is 1.42. 5. The refractive index of the lens material. 6. An astigmatism occurs because an asymmetrically shaped cornea exists.
7. Hyperopia (farsightedness), is the opposite of myopia. Hyperopia is the condition of the eye in which there is too little converging power for the length of the eye. As a result a close object is imaged behind the retina and appears blurred. 8. Myopia (nearsightedness) is the condition of the eye where there is too much converging power for the length of the eye. As a result, a distant object is imaged directly in front of the retina and appears blurred. 9. Visual acuity is the eye s capacity to see fine detail. 10. For the most part the angle lies between θmin= 5 x 10-4 rad and θmin= 4 x 10-4. 11. A magnifier is a positive lens or lens system. The magnifier gives an angular magnification when used by a relaxed normal eye. 12. The objective is a short focal length, positive lens system, which forms a magnified real image of a small image place outside its focal point. (focal point is determined by the refractive index of the lens material) 13. The ocular functions as a magnifier, allowing the eye to come very close to the image produced by the objective. 14. The ocular serves the same function it does in the microscope- to allow the eye to come close to the image formed by the objective. 15. The objective is a long-focal length, positive lens system or mirror, which is used to form a real image of an object in the distance. 16. M= ((25cm/f) +1) M=25/9.42 +1 = 2.65 +1 = 3.65 17. M= (F1/F2) = 3000/30 = 100 18. h= (16mm)(tan60) = 16/1.7321 = 9.24 mm 19. M= (θ /θ) = (75/80) = 0.9375