732
15 Optics
Fig. 15.20 Real images
formed by a convex lens in
two positions
By symmetry v 2 = u 1
(3)
v 1 = u 2
(4)
Then (2) becomes
u 1 − v 1 = d
(5)
size of image
size of object
=
I 1
O
=
v 1
u 1
(6)
Similarly,
I 2
O
=
v 2
u 2
(7)
∴
I 1
I 2
=
v 1
v 2
u 2
u 1
=
v 2
1
v 2
2
(8)
From (1) and (5), v 2 =
D + d
2
and v 1 =
D − d
2
Therefore,
I 1
I 2
=
(D − d) 2
(D + d) 2
(b) Multiplying (6) and (7)
I 1
O
I 2
O
=
v 1
u 1
v 2
u 2
=
u 2
u 1
u 1
u 2
= 1
∴ O =
I 1 I 2
(c) 1
u 1
+
1
v 1
=
1
f
∴
2
D + d
+
2
D − d
=
1
f
where we have used (1), (2) and (4).
∴
f =
D 2 − d 2
4D