focal length by displacement

Focal length by displacement is a method of estimating focal length of a lens where we change the position of a lens and observe the image formed on a screen while distance between object and the lenses is unchanged.

Ensure you have the following apparatus:

Lens Holder

The screen

cross wires

bulb as a source of light

candle as a source of light

metre rule

Table of Contents

Procedure to estimate focal length by displacement

Estimate the focal length of the lens by focusing a distance object
Set the apparatus as in figure below ensuring that the distance between the object and the screen is more than 4f where f is the focal length estimated above.

Obtain the image of the illuminated object on the screen when the lens is at position L₁
move the lens to position L₂ where another clear but diminished image is formed on the screen as Without changing the position of the object on the screen, shown below.

measure u and v for position L₁ and the new distance u₁ and v₁ for position L₂.
Determine the displacement d .

Deriving the displacement formular

from the diagram above, the distance between the point object and the screen is s. from the diagram, it is shown that the distance s is given by u+v.

i. e. s = u+v ………………………………..(1)

we get the distance between new and original position of the lens by use of expressions:

d=u’- u where u’ is the new object distance and u the original object distance

d can also be obtained from v-v’ which is the original image distance and image distance when the lens is displaced by distance d.

i.e d=u’-u and d = v-v’

but u’=v and v’=u

and therefore:

d=v-u………………………………….(2)

adding (1) and (2);

hence s+ d= u + v + v –u

and so: s + d = 2v and hence;

$$V = \frac{s+d}{2}$$

similarly we can subtract equation 2 from 1 as shown:

hence s- d = u + v –v + u

therefore : s- d = 2u and hence;

$$u = \frac{s-d}{2}$$

from the lens formulae:

$$\frac{1}{f} = \frac{1}{u}+\frac{1}{v}$$

we can substitute values of u and v in terms of s and d as obtained in the expressions above. And hence;

$$\frac{1}{f} = \frac{1}{\frac{s-d}{2}}+ \frac{1}{\frac{s+d}{2}}$$

the above equation can be simplified into:

$$\frac{1}{f} = \frac{2}{s-d}+\frac{2}{s+d}$$

finding the lcm of the denominator, we obtain;

1f=2(s+d)+2(s−d)(s−d)(s+d)=2s+2d+2s−2ds2−d2

$$\frac{1}{f} = \frac{2(s+d)+2(s-d)}{(s-d)(s+d)}$$

and simplifying the above equation in the numerator:

$$\frac{1}{f} = \frac{4s}{s^2 – d^2}$$

and finding the reciprocal so that we can get f;

$$f = \frac{s^2-d^2}{4s}$$

f=s2–d24s

from the above equation: s²-d² = 4fs

a plot of s²-d² against s results to a straight line through the origin with a slope equal to 4f.

different values of s are obtained by changing distance between the object and the screen and then calculating the corresponding distance d.

The two positions L₁ and L₂that represents different positions of the lens are known as the conjugate points.

Procedure to estimate focal length by displacement

Deriving the displacement formular

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