1.An object of height 0.05 meters is located 0.6 meters in front of a thin conve
ID: 1462086 • Letter: 1
Question
1.An object of height 0.05 meters is located 0.6 meters in front of a thin converging lens. The lens forms an image of the object at point A, which is 0.3 meters beyond the lens. a) What is the focal length of the lens? b) What is the magnification of the image? c) What is the height of the image formed at position A? d) When the object is moved slightly closer to the lens. the image will - increase in size and move closer to the lens. - increase in size and move farther away from the lens. - decrease in size and move closer to the lens. - decrease in size and move farther away from the lens. - stay in the same place.Explanation / Answer
a) 1/f = 1/v - 1/u ; v = 0.3m = image distance, u = - 0.6m = object distance.
1/f = 1/0.3 - 1/(- 0.6 m) = 3/0.6
f = 0.6/3 = 0.2 m.
b) magnification, m = v/u = 0.3/-0.6 = - 0.5 ; therefore, image is half the siz of the object and inverted (negative sign
of magnification tells that image is inverted.).
c) since, magnification, m = 0.5
therefore height of the image = m* height of object = 0.5 *0.05 m = 0.025 m.
d) increase in size and move further away from the lens. When the object the at 2f distance, the image will be formed at 2f distance on the other side of same size.
from infinity of 2f distance of a convex lens image is inverted and smaller than equal to object size. equality hold at 2f distance.
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