When an object of height 2 cm is placed in front of a concave mirror at a distance of 15 cm
away from it, a real image is obtained 30 cm away on the same side.
a) Calculate the focal length of the mirror.
b) Find out the magnification.
c) Find out the height of the image
Answers
Answer:
According to the question:
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.
Explanation:
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