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Question
A convex lens of focal length 0.10 m is used to form a magnified image of an object of height 5 mm placed at a distance of 0.08 m from the lens. Calculate the position, nature and size of the image.
Solution
Given:
Focal length, f = 0.10 m = 10 cm
Object distance, u =-0.08 m =-8 cm
Height of the object, h = 5 mm = 0.5 cm
Applying lens formula, we get:
1/v-1/u = 1/f
or, 1/v = 1/f + 1/u
or, 1/v = 1/10 + 1/(-8)
Image distance, v =-80/2 =-40 cm
or, v =-0.40 m
Thus, the position of image is 0.40 m from the lens on the same side as the object (on the left of lens).
Magnification, m = h'/h = v/u
m = h'/0.5 = (-40)/(-8) = + 5
or, size or height of image, h' = 2.5 cm or 25 mm
Further, the value of magnification is positive; therefore, the image is virtual and erect.
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