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Question

A rod of length 10 cm lies along the principal axis of a concave mirror of focal length 10 cm in such a way that its end close to the pole is 20 cm away from the mirror. The length of the image is

A
10 cm
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B
15 cm
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C
2.5 cm
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D
5 cm
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Solution

The correct option is D 5 cm

The rod is lying axially along the principal axis, its two ends will acts as two point objects O1 and O2 for the mirror.
The Object distance for the end close to the pole is 20 cm
R=2f=20 cm
Thus, image I1 for object O1 will form at object itself, because O1 lies at centre of curvature.
|PI1|=20 cm
For object O2, let image is formed at v2 (Taking positive along the incident ray)
u=30 cm
Using mirror formula,
1v2+1(30)=1(10)
(f=10 cm)
1v2=130110=230
v2=15 cm
Thus image distance from pole is 15 cm
|PI2|=15 cm
length of image will be,
li=|PI1||PI2|=2015=5 cm

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