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Example. A 2.0 cm high object is placed 5 cm in front of a +10 cm focal length lens. Where is the object located? Is it real or virtual? Find the height of the image. Image. F. P.A. Object. F. The image is upright, virtual, and enlarged!. Example.
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Example A 2.0 cm high object is placed 5 cm in front of a +10 cm focal length lens. Where is the object located? Is it real or virtual? Find the height of the image. Image F P.A. Object F The image is upright, virtual, and enlarged!
Example A 4.0 cm high object is placed 18 cm in front of a +8 cm focal length lens. Where is the object located? Is it real or virtual? Find the height of the image. F Object F
Diverging Lens Principal Rays Image F P.A. Object F 1) Rays parallel to principal axis bend and seem to come from focal point. 2) Rays through center of lens are not refracted. 3) Rays toward F emerge parallel to principal axis. Image isvirtual, upright and reduced.
Diverging Lens (Example) An object is placed 4 cm in front of a diverging lens. The resulting image is 1 cm high and is located 2 cm in front of the lens. Find the focal length of the lens, the magnification, and the height of the object.
Camera (far object out of focus) Film The image of a distant object is very near the focus. In order for our picture to be “sharp”, we must move the lens such that the image (point where light rays intersect) is on the film.
Camera (far object in focus) Now the image is located on the film. This will produce a picture that is “in focus”. But, what happens if we want to photograph something nearby?
Camera (near object out of focus) Now the image would be located behind the film. We must move the lens again so that the image falls on the film.
Camera (near object in focus) Now the image would be located behind the film. We must move the lens with respect to the film again so the image falls on the film.
Camera (Example) A 300 mm focal length camera can be adjusted from 300 mm to 400 mm from the film. For what range of object distances will this camera be useful?
The Human Eye Light is refracted by the cornea and lens. A real, inverted image is formed on the retina. The eye automatically adjusts its focal length to focus on objects at different distances.
Eye (Relaxed) 25 mm Determine the focal length of your eye when looking at an object far away.
Eye (Tensed) 250 mm 25 mm Determine the focal length of your eye when looking at an object up close (25 cm).
Near Point, Far Point Eye’s lens changes shape (changes f ) Object at any so can have image be at retina (si = approx. 25 mm) Can only change shape so much “Near Point” Closest so where image can be at retina Normally, ~25 cm (if far-sighted then further) “Far Point” Furthest so where image can be at retina Normally, infinity (if near-sighted then closer)
Nearsighted If you are nearsighted, your “far point” is too close (less than infinity). To correct your vision, you can use a diverging lens that will create an image of the distant object at your far point.
Correcting Nearsightedness do=∞ di Eye can focus on this image. x Far Point = 50 cm (from eye) and X = 2 cm = 48 cm We want di = Far Point – X f = −48 cm Need glasses with this f in order to see distant objects.
Refractive Power of Lens Diopter = 1/f where f is focal length of lens in meters. In our previous example the person needed glasses with f = − 48 cm. Doctor’s prescription reads: 1/(−0.48m) = –2.1 Diopters (D)
Farsighted If you are farsighted, your “near point” is too far away (more than ~25cm). To correct your vision, you can use a converging lens that will create an image of the nearby object at your near point.
Correcting Farsightedness di do x Eye can focus on this image. Near Point = 80 cm (from eye) and X = 2 cm We want to see objects as close as 25 cm. do = 25 cm – X = 23 cm di = Near Point – X = 78 cm f = 32.6 cm
As you look at this question, the image of this question on the retina of your eye is __________. A. real and upright B. real and inverted C. virtual and upright D. virtual and inverted cornea retina lens
You are working at a cash register and you are watching a customer walk away after paying you. As you watch the customer, the lenses of your eyes bulge ________ and the radii of curvature of the lens surfaces __________. cornea A. less . . . decrease B. less . . . increase C. more . . . decrease D. more . . . increase retina lens
Example A person can just barely read a book when it is held out at arm’s length (50 cm). What power of reading glasses should be prescribed for him if he wants to read at 25 cm, assuming they will be placed 2 cm in front of the eye?
Angular Size Angular size tells you how large the image is on your retina, and how big it appears to be. q q q q Both are same size, but nearer one looks bigger.
Unaided Eye Bring object as close as possible (to near point N) object h0 q N How big the object looks with unaided eye. **If q is small and expressed in radians.
Magnifying Glass magnifying glass virtual image object hi ’ ho so si Magnifying glass produces virtual image behind object, allowing you to bring the object to a closer so: and larger q Compare tounaided eye Ratio of the two angles is themagnitfying powerMP:
Angular Magnification (Example) A magnifying glass with a focal length of 6 cm is placed a distance of 5 cm from a newspaper. What are the linear and angular magnifications?
Example A magnifying glass with a focal length of 9 cm produces an angular magnification of 7. What is the linear magnification of the lens?