Prove that a convex lens produces an 'n' times magnified image when the object distances from the lens have magnitude (f ± f/n). Here f is the magnitude of the focal length of the lens. Hence
find the two values of object distance for which a convex lens of power 2.5 D will produce an image that is four times as large as the object.
Magnification produced by any lens,
m = v/u = f / (f + u) ............... (1)
given m = ± N
± N = f / f + u .......... (from - 1)
or, f + u = ± f / N
or, u = - f ± f / N
Hence magnitude of object distances,
|u| = f ± f / N
Given P = 1/f = + 2.5 D
f = 1/2.5 = 0.4 m = 40 cm
Also N = 4
|u| = 40 ± 40/4
|u| = 40 ± 10
|u| = 50 cm or 30 cm.
(a) For a convex lens, f >> 0 and for an object on left, u < 0. When the object is placed within the focus of a convex lens,
0 < |u| < f or 0 < 1 / |u| > 1/f
1/v = 1/f + 1/u
1/v = 1/f - 1/|u| <0
i.e. v < 0 so a virtual image is formed on left.
Now as u < 0 and v < 0, so 1/v = 1/f + 1/u
– 1/ |v| = 1/f – 1/|u|
or, 1/|u| – 1/|v| = 1/f
As f > 0
1/|u| – 1/|v| > 0
or, 1/|u| > 1/|v|
|m| = |v/u| > 1
Hence image is enlarged.
(b) For a concave lens f < 0 and for an object on left, u < 0
1/v = 1/f +1/u
1/v = 1/|f| – 1/|u|
1/v = - [1/|f| + 1/|u|] < 0 for all u.
i.e. v < 0 for all values of u. Hence a virtual image is formed on the left.
Also 1/|v| = 1/|f| + 1/|u|
1/|v| > 1/|u|
|m| = |v/u| < 1
i.e. the image is diminished in size.