The focal length of a convex lens is 20 cm. If an object of height 2 cm is placed at 30 cm from the lens, find (i) the position and nature of the image (ii) the height of the image (iii) the magnification produced by the lens.

Explanation

`1/f 1/v - 1/u`
` therefore 1/u = 1/f + 1/u = 1/(20 cm) + 1/(- 30 cm) = 1/( 20 cm) - 1/ ( 30 cm)`
` = ( 3-2)/( 60 cm) = 1/( 60 cm)`
v= 60 cm , v is positive.
The image will be formed at 60 cm from the lens and on the other side of the lens with respect to the object. It is a real image.
(ii) ` h_(2)/h_(1) = u/v`
` h_(2)/(2 cm) = (60 cm)/(-30 cm) = -2 `
`h_(20 = -2 xx 2 cm = - 4 cm`
`h_(2) ` is negative. This shows that the image is inverted .
The height of the image = - 4 cm
(iii) ` M = h_(2)/h_(1) = ( - 4 cm)/(2 cm) = -2 `
M is negative, indicating that teh image is inverted.
The magnification produced by the lens = -2