The position of the object is 30 cm infront of a concave mirror of focal length 15cm, if size of the object is 5cm, find the position and size of the image formed. Explain with the help of suitable ray diagram.
Answers
Answer:
Object distance, u=−30 cm
Focal length, f=−15 cm
Let the Image distance be v.
By mirror formula:
v
1
+
u
1
=
f
1
[4pt]
⇒
v
1
+
−30 cm
1
=
−15 cm
1
[4pt]
⇒
v
1
=−
−30 cm
1
+
−15 cm
1
[4pt]
⇒
v
1
=
30 cm
1−2
[4pt]
⇒
v
1
=−
30 cm
1
[4pt]
∴v=−30 cm
Thus, screen should be placed 30 cm in front of the mirror (Centre of curvature) to obtain the real image.
Now,
Height of object, h
1
=2 cm
Magnification, m=
h
1
h
2
=−
u
v
Putting values of v and u:
Magnification m=
2 cm
h
2
=−
−30 cm
−30 cm
⇒
2 cm
h
2
=−1
[4pt];
⇒h
2
=−1×2 cm=−2 cm
Thus, the height of the image is 2 cm and the negative sign means the image is inverted.
Thus real, inverted image of size same as that of object is formed.
The diagram shows image formation.
solution
Explanation:
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