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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|

or, |u|<|v|

i.e. |v|>|u|

|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|

or, |v|<|u|

|m| = |v/u| < 1

i.e. the image is diminished in size.

Regards

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