A magnifying lens of focal length of 20 cm is held 10 cm away from a small object which is 0.25 cm long. The size of the image formed by the lens will be...
Answers
Answer:
According to the question,
Image distance, v=−15 cm (negative due to convex lens)
Focal length, f=−20 cm
Let the object distance be u.
By lens formula:
v
1
−
u
1
=
f
1
[4pt]
⇒
−15 cm
1
−
u
1
=
−20 cm
1
[4pt]
⇒
u
1
=
20 cm
1
+
−15 cm
1
[4pt]
⇒
u
1
=
60 cm
3−4
=
60 cm
−1
[4pt]
⇒u=−60 cm
Therefore, object is placed at 60 cm away from the lens, on the same side as image.
Now,
Height of object, h
1
=5cm
Magnification, m=
h
1
h
2
=
u
v
Putting value of v and u:
Magnification, m=
5 cm
h
2
=
−60 cm
−15 cm
⇒
5 cm
h
2
=
4
1
⇒h
2
=
4
5 cm
=1.25 cm
Thus, the height of the image is 1.25 cm and the positive sign means the image is virtual and erect.
Explanation:
According to the question,
Image distance, v=−15 cm (negative due to convex lens)
Focal length, f=−20 cm
Let the object distance be u.
By lens formula:
v
1
−
u
1
=
f
1
[4pt]
⇒
−15 cm
1
−
u
1
=
−20 cm
1
[4pt]
⇒
u
1
=
20 cm
1
+
−15 cm
1
[4pt]
⇒
u
1
=
60 cm
3−4
=
60 cm
−1
[4pt]
⇒u=−60 cm
Therefore, object is placed at 60 cm away from the lens, on the same side as image.
Now,
Height of object, h
1
=5cm
Magnification, m=
h
1
h
2
=
u
v
Putting value of v and u:
Magnification, m=
5 cm
h
2
=
−60 cm
−15 cm
⇒
5 cm
h
2
=
4
1
⇒h
2
=
4
5 cm
=1.25 cm
Thus, the height of the image is 1.25 cm and the positive sign means the image is virtual and erect.