determine the focal length of double concave lens that produce an image that is 16 cm behind the lens when the object is 28.5 CM from the lens
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Consider variables u = object distance (in cm), v= image distance (in cm)
Given,
-u + v = 100 ......eq (1)
Magnification = v /u = -3 (real and inverted image)
v = -3u ........eq (2)
Focal length 1/f = 1/v - 1/u ......eq (3)
Substitute eq (2) in eq (1), we get u = -25 cm
eq (2) v = -3 u = - 3 *- 25 = 75 cm
Substitute eq (2) in eq (3), we get f = - 3 u/4 = -3 * -25/4 = 18.75 cm.
Given,
-u + v = 100 ......eq (1)
Magnification = v /u = -3 (real and inverted image)
v = -3u ........eq (2)
Focal length 1/f = 1/v - 1/u ......eq (3)
Substitute eq (2) in eq (1), we get u = -25 cm
eq (2) v = -3 u = - 3 *- 25 = 75 cm
Substitute eq (2) in eq (3), we get f = - 3 u/4 = -3 * -25/4 = 18.75 cm.
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Here is your answer.
Hope it helps.
^_^
Hope it helps.
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