An object is placed 22.0 cm from a screen. (a) At what two points between object and screen may a converging lens with a 3.00 cm focal length be placed to obtain an image on the screen? (b) What is the magnification of the image for each position of the lens?
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An object is placed 22.0 cm from a screen. (a) At what two points between object and screen may a converging lens with a 3.00 cm focal length be placed to obtain an image on the screen? (b) What is the magnification of the image for each position of the lens?
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- The focal length of a diverging lens is negative. If f= -26 cm for a particular diverging lens, where will the image be formed of an object located 53 cm to the left of the lens on the optical axis? cm to the left of the lens What is the magnification of the image?The focal length of a diverging lens is negative. If f = −16 cm for a particular diverging lens, where will the image be formed of an object located 45 cm to the left of the lens on the optical axis? cm to the left of the lensWhat is the magnification of the image?an object of height 5 cm is placed 20 cm in front of a converging lens at focal length 10 cm. Behind the converging lens, and 25cm from it, there is a diverging lens of the focal length of 6 cm. Find the location of the final image, in centimeters, with respect to teh diverging lens. what is the magnification of the final image? what is the height of the final image?
- (a) A 1.00 cm-high object is placed 4.35 cm to the left of a converging lens of focal length 7.85 cm. A diverging lens of focal length -16.00 cm is 6.00 cm to the right of the converging lens. Find the position and height of the final image. cm---Select--- position height Is the image inverted or upright? upright inverted Is the image real or virtual? real virtual cm (b) What If? If an image of opposite characteristic, i.e., virtual if the image in part (a) is real and real if the image in part (a) is virtual, is to be obtained, what is the minimum distance (in cm), and in which direction, that the object must be moved from its original position? cm distance direction ---Select---An object with a height of -0.040 m points below the principal axis (it is inverted) and is 0.140 m in front of a diverging lens. The focal length of the lens is -0.25 m. (Include the sign of the value in your answers.) (a) What is the magnification? (b) What is the image height? m (c) What is the image distance? mA 1.00-cm-high object is placed 3.95 cm to the left of a converging lens of focal length 7.80 cm. A diverging lens of focal length –16.00 cm is 6.00 cm to the right of the converging lens. Find the position and height of the final image. position 7.5 cm in front of the second lens v 0.5466 height Calculate the magnification produced by each lens. Then consider how the magnification relates image size and object size for each lens to find the height of the final image. cm Is the image inverted or upright? upright O inverted Is the image real or virtual? real virtual
- The distance between the lenses in a compound microscope is 18 cm. The focal length of the objective is 1.5 cm. If the microscope is to provide an angular magnification of - 68 when used by a person with a normal near point (25 cm from the eye), what must be the focal length of the eyepiece?In a compound microscope, the objective has a focal length of 1.0 cm, the eyepiece has a focal length of 2.0 cm, and the tube length is 25 cm. What is the magnitude of the overall magnification of the microscope?The image distance for the lens of a person's eye is2.10 cm.Calculate the focal length of the eye's lens system for anobject (a) at infinity, (b) 500 cm from the eye, and (c) 25 cm from the eye.