Question

51. A rod of length (f/2) is placed along the axis of a
concave mirror of focal length f. If the near end of
the real image formed by the mirror just touches the
far end of the rod, find its magnification.​

Answer 1

A rod AB of length
f
2
is placed along the axis of a concave mirror of focal length
f
in such a way that the near end of the REAL IMAGE formed by the rod just touches the far end B of the ROD.
The near end of the real image of the rod is the image of far end B of the the rod AB. It is possible only if the far end B of the rod is placed at the center of curvature C of the concave mirror (as shown in the figure).
The image of near end A of the rod is A'
Now PC = the radius of curvature
=
2
f
The object distance for near point of the rod A is
u
So
u
=
P
A
=
P
C
−
A
B
=
2
f
−
f
2
=
3
f
2
The conjugate foci relation of spherical mirror is
1
v
+
1
u
=
1
f
...
...
(
1
)
Where
v
→
Image distance
=
?
u
→
Object distance
=
−
3
f
2
f
→
Focal length
=
−
f
So inserting the values in equation (1)
1
v
−
2
3
f
=
−
1
f
⇒
1
v
=
−
1
f
+
2
3
f
⇒
1
v
=
−
3
+
2
3
f
=
−
1
3
f
⇒
v
=
−
3
f

Negative sign indicates the position of image point A' is at the same side of object point A
So length/size of the image
B
A
'
=
C
A
'
=
P
A
'
−
P
C
=
3
f
−
2
f
=
f
Now Magnification
m
=
Size of image
Size of object
=
f
f
2
=
2
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