The distance between the object and the image formed in a lens is 15cm. The ratio between the
object size and the image size is 3:2. An inverted and real image is formed. Find the nature of the lens
and the focal length of the lens.
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(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,
f
1
=
v
1
−
u
1
⟹
10
1
=
v
1
−
−15
1
⟹
v
1
=
10
1
−
15
1
⟹
v
1
=
30
3−2
⟹v=30cm
The image is formed on the other side of the lens at a distance of 30 cm.
Magnification, m=−
u
v
=−
−15
30
=−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,
f
1
=
v
1
−
u
1
⟹
10
1
=
v
1
−
−8
1
⟹
v
1
=
10
1
−
8
1
⟹
v
1
=
40
4−5
⟹v=−40cm
The image is formed on the same side as the object at a distance of -40 cm.
Magnification, m=−
u
v
=−
−8
−40
=+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.