Question

i) Find the nature, position and magnification of the image formed by a convex lens of focal length 10cm, If the object is placed at a distance of a) 15cm b) 8cm
ii) Which of the above represents the use of convex lens in a) A film projector b) The magnifying glass used by palm reader

Answer 1

(i) Given: convex lens of focal length 10cm.

To find the nature, position and magnification of the image formed for object placed at (a) 15cm (b) 8cm

Solution:

(a) According to the given criteria,

Focal length, $f=10cm$

object distance, $u=-15cm$ i.e., between F and 2F

Applying lens formula,

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}⟹\frac{1}{10}=\frac{1}{v}-\frac{1}{-15}⟹\frac{1}{v}=\frac{1}{10}-\frac{1}{15}⟹\frac{1}{v}=\frac{3-2}{30}⟹v=30cm$

The image is formed on the other side of the lens at a distance of 30 cm.

Magnification, $m=-\frac{v}{u}=-\frac{30}{-15}=-2$

As magnification is negative. So the image is Real, inverted and enlarged in size.

(b) According to the given criteria,

Focal length, $f=10cm$

object distance, $u=-8cm$, i.e., between center of lens and F.

Applying lens formula,

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}⟹\frac{1}{10}=\frac{1}{v}-\frac{1}{-8}⟹\frac{1}{v}=\frac{1}{10}-\frac{1}{8}⟹\frac{1}{v}=\frac{4-5}{40}⟹v=-40cm$

The image is formed on the same side as the object at a distance of -40 cm.

Magnification, $m=-\frac{v}{u}=-\frac{-40}{-8}=+5$

As magnification is positive. So the image is virtual, erect and enlarged in size.

(ii) The convex lens in which object placed at 15 cm is used in a film projector as the image should be produced on other side of the lens.

The convex lens in which object is placed at 8cm is used in the magnifying glass of the palm reader as the image should be in same side as that of lens and should be enlarged.