3
A 2.0 cm tall Object is placed perpendicular to the principal
axis of a concave mirror of focal length osm The distance
of the object fuom the misure is som Rod. The nature,
position and size of the image found. Represent the
situation with the help of
ray diagram
Answers
Answer:
According to the question:
Image distance, v=−10 cm
Focal length, f=−15 cm (since, concave lens)
Let the object distance be u.
By lens formula:
v
1
−
u
1
=
f
1
[4pt]
⇒
−10 cm
1
−
u
1
=
−15 cm
1
[4pt]
⇒
u
1
=−
−15 cm
1
−
10 cm
1
[4pt]
⇒
u
1
=
30 cm
2−3
=
30 cm
−1
[4pt]
⇒
u
1
=
30 cm
−1
∴u=−30 cm.
Thus, object should be placed at 30 cm in front of lens, in the same side as image.
Now,
Height of object, h
1
=2 cm
Magnification, m=
h
1
h
2
=
u
v
Putting value of v and u:
Magnification, m=
2 cm
h
2
=
−30 cm
−10 cm
⇒
2 cm
h
2
=
3
1
⇒h
2
=
3
2
cm=0.67 cm
Height of image is 0.67 cm
Thus, the image is virtual, diminished, and erect and one-third the size of object.
Explanation:
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