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InClass

High School. PHYSICS. InClass. by SSL Technologies. with S. Lancione. Exercise-53. Lenses. Part 2 /2. Lenses. PART-2 /2. THE EYE The eye is an optical “instrument”. It contains a converging lens

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  1. High School PHYSICS InClass by SSL Technologies with S. Lancione Exercise-53 Lenses Part 2 /2

  2. Lenses PART-2 /2 THE EYE The eye is an optical “instrument”. It contains a converging lens used to focus images on the “retina” (a kind of screen at theback of the eye). Images on the retina are inverted. Click

  3. Lenses PART-2 DEFECTS IN VISION NEARSIGHTEDNESS The ability to see objects clearly which are nearbut objects far away appear blurred. Nearsightedness, known as myopia, is caused by the fact that rays of light focus in front of the retina of the eye. Myopia is corrected by using a diverging lens as illustrated in the diagram below. FARSIGHTEDNESS The ability to see objects clearly which arefarbut objects close to the eye appear blurred. Farsightedness, known as hyperopia, is caused by the fact that raysof light focus behind the retina of the eye. Hyperopia is correctedby using a converging lens. Click

  4. Lenses PART-2 CORRECTING DEFECTS IN VISION Click

  5. Lenses PART-2 Reminder In applying the lens equation, be sure to use the signs carefully: do is the object distance and is always positive f is the focal length and is positive for convex lenses but negative for concave lenses di is the image distance and is positive for real images but negative for virtual images Click

  6. EXERCISES

  7. Question-1 Remember: A negative diindicates a virtual image. 8 A lens forms an image 10 cm high. If the object is 4 cm in heightand situated 8 cm away, what is the focal length of the lens? Note that while the standard unit for length in a physics formulais the meter, since in this case we have lengths on both sidesof the equation, we need not convert to meters as the units automatically cancel out. Click

  8. Question-2 When the magnification is less than one, the image is smaller than the object. Negative image distance means a virtual image. The focal length of a concave lens is 10 cm. An object, whoseheight is 2 cm, is placed 15 cm in front of the lens. Determinethe characteristics of the image. Virtual Type (real or virtual): _______________ Location: _______________ Magnification: _______________ Height: _______________ Attitude (upright/inverted): _______________ 6 cm 0.4 0.8 cm Upright Click

  9. Question-3 When the magnification is greater than one, the image is greater than the object. An object whose height is 4 cm is placed 6 cm in front of aconverginglens. If the focal length of the lens is 8 cm,determine the characteristics of the image. Virtual Type (real or virtual): _______________ Location: _______________ Magnification: _______________ Height: _______________ Attitude (upright/inverted): _______________ 24 cm 4 16 cm Upright Click

  10. Question-4 The focal length of a camera is 10 cm. The lens forms an imagethat is 4 cm high when the negative (film) is 12 cm from the lens. a) What is the object distance? Click

  11. Question-4 The negative sign indicates inversion. The focal length of a camera is 10 cm. The lens forms an imagethat is 4 cm high when the negative (film) is 12 cm from the lens. do = 60 cm (previously calculated) b) What is the object height? Click

  12. Question-4 The negative sign indicates inversion. The focal length of a camera is 10 cm. The lens forms an imagethat is 4 cm high when the negative (film) is 12 cm from the lens. ho = 20 cm (previously calculated) c) What is the magnification factor? Click

  13. Question-5 A 35mm slide (the object) is placed 8.2 cm from a projectionlens whose focal length is 8 cm. Determine: a) The image distance. Click

  14. Question-5 The negative sign indicates inversion. A 35-mm slide (the object) is placed 8.2 cm from a projectionlens whose focal length is 8 cm. Determine: b) The image height. Click

  15. Question-5 The negative sign indicates inversion. A 35-mm slide (the object) is placed 8.2 cm from a projectionlens whose focal length is 8 cm. Determine: c) The magnification factor. Click

  16. Question-6 The focal length of a magnifying glass is 10 cm. The lens is usedto view a stamp that is 2.0 cm in height. If the stamp is placed6.0 cm away from the magnifying glass, calculate the height ofthe image. Click

  17. Question-7 Reminder: concave lenses have a negative focal length. The negative sign indicates a virtual image. An object is 4 cm from a concavelens whose focal lengthis 12 cm. Where will the image be located? Click

  18. Question-8 The negative sign indicates a virtual image. An object 3 cm high is located 30 cm from a concave lens whosefocal length is 15 cm. Determine: a) The image distance. Click

  19. Question-8 An object 3 cm high is located 30 cm from a concave lens whosefocal length is 15 cm. Determine: b) The magnification. Click

  20. Question-8 An object 3 cm high is located 30 cm from a concave lens whosefocal length is 15 cm. Determine: c) The size of the image. Virtual d) The type of image: ___________________ Upright e) The attitude of the image: ___________________ Click

  21. Question-9 Define each of the following terms: a) Myopia: The problem of not seeing far objects clearly. b) Hyperopia: The problem of not seeing near objects clearly. Click

  22. Question-10 Illustrated below are two common eye problems or defects.State the name of each defect and draw the appropriate lensin order to correct the problem. Myopia Hyperopia Click

  23. SSLTechnologies.com/science The end

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