Question

(i) An object is placed at a distance of 20 cm in front of a concave lens of focal length 20 cm. find: (a) the position of the image, and (b) the magnification of the image.

(ii) A concave lens forms an erect image of 1/3rdsize of the object which is placed at a distance 30 cm in front of the lens. Find:
(a) The position of image, and
(b) The focal length of the lens.

Answer 1

(i)Given,
Object distance, u = – 20 cm
Focal length, f = – 20 cm (concave lens)

(a) Lens formula is,
1 / v – 1 / u = 1 / f
∴ 1 / v = 1 / f + 1 / u
∴ 1 / v = 1 / – 20 + 1 / – 20
∴ 1 / v = – 2 / 20
∴ v = -10 cm
Hence the image is 10 cm in front of the lens on the same side as the object.

(b) For a lens, magnification is
m = v / u
∴ m = -10 / – 20
∴ m = + 0.5


(ii) Given,
Distance of the object u = – 30 cm
Magnification m = h’/ h
= 1 / 3
We know that
The magnification is
m = h’ / h
m = v / u
1 / 3 = v / -30
v = -10 cm
Therefore the image is formed at 10 cm from the lens.
Len’s formula is
1 / f = 1 / v – 1 / u
1 / f = 1 / – 15
f = -15 cm
Hence, the focal length is 15 cm and the image is formed at 10 cm from the lens.