04
Determine the focal length and hence the power of a convex lens by
displacement method with the help of an optical bench.
If the object and the image screen be so placed on an optical bench that the distance
between them is greater than four times the focal length (f) of a given convex lens, then there will be
two different positions of the lens which for which an equal sharp image will be obtained on the image
screen. Let the points O and I and L1 and L2 in Fig.
d
u2
B
v2
L1
L2
A
P
Q2
X
u1
v1
Fig: Convex lens on the bench.
The fig represent respectively the positions of the object and the image screen and the two
different positions of the lens for which an equally sharp image is obtained.
Let the distance Ol =D and L1 – L2 = x.
From the lens equation , we have
1
v
−
Or,
1
u
1
1
=
D−u
f
−
1
u
=
1
f
(since u + v =D)
Applying sign convention, u is negative.
1
1
1
+ =
D−u
u
f
Or, u2 – ud = df = 0
Or,
Solving the above equation which is quartic, we have two values of u corresponding to the two
positions of the lens. These are
u1 =
D
2
-√
And u2 =
D2 −4Df
D
2
2
+√
position L1 of the lens
D2 −4Df
2
position L2 of the lens
Then x = L1 ~ L2 = u1 ~ u2 = ± √D2 − 4Df
Or x2 = D2 – 4Df
Or f =
D2 −x2
4D
……………….. (1)
Where D is the distance between the object and the image and must be greater than 4f and x is the
distance between two different positions of the lens.
The power P of the lens s as usual given by the relation,
P=
100
f(in cm)
diopters
Table II
Position of
Lens at
No. of
obs.
Object
(O)
Image
(I)
L1
L2
Displacement
of Lens
X = L 1 – L2
(cm)
Apparent
distance
between
object
and
image
D’ = O~1
Corrected
distance
between
object
and
image
D = D’ + λ
1
2
3
4
5
Table III
No. of
obs.
1
2
3
4
5
Lens
displacement (x)
from Tab. II
Corrected
Distance (d)
from Tab. II
Focal length
D2 − x 2
𝑓=
4D
Mean focal
length (f) cm
Power
P=
100
𝑓