An object 4 cm high is placed at a distance of 10 cm from a convex lens of focal length 20 cm. Find the position, nature and size of the image.
Object distance, u = -10 cm (It is to the left of the lens.)
Focal length, f = + 20 cm (It is a convex lens.)
Putting these values in the lens formula, we get:
1/v- 1/u = 1/f (v = Image distance)
1/v -1/-10) = 1/20
or, v =-20 cm
Thus, the image is formed at a distance of 20 cm from the convex lens (on its left side).
Only a virtual and erect image is formed on the left side of a convex lens. So, the image formed is virtual and erect.
Magnification, m = v/u
m =-20 / (-10) = 2
Because the value of magnification is more than 1, the image will be larger than the object.
The positive sign for magnification suggests that the image is formed above principal axis.
Height of the object, h = +4 cm
magnification m=h'/h (h=height of object)
Putting these values in the above formula, we get:
2 = h'/4 (h' = Height of the image)
h' = 8 cm
Thus, the height or size of the image is 8 cm.
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