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Category: Physics

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

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.

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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.

Related topics
March 20, 2026
Estimating focal length of a lens
Estimating the focal length of a lens is a fundamental experiment in optics. It helps learners understand how lenses form images and bend light. Focal length is the distance between the optical center of a lens and its principal focus. Focal length determines how strongly the lens converges or diverges light rays. In this topic, students explore practical methods of measuring focal length using simple laboratory setups. This includes forming sharp images of distant objects or using object–image distance measurements. Through this investigation, learners develop essential skills in observation, measurement, and application of lens formulas, building a strong foundation for further studies in physics and optical instruments.

We can always determine the focal length of the given lenses by applying various methods of estimating focal length of a lens.

This methods may includes:
- Focusing on a distance object
- lens formula experiments
- Using non-parallax method
- Using an illuminated object
- Displacement method
Page Contents

Focusing a distance object to estimate focal length

We arrange the white screen, convex lens on a lens holder and ruler such that rays of light from a distance object are incident on a lens that is close to the white screen.

See the diagram below

We adjust the lens position to and fro until we obtain a sharp image of a distance object on the screen.

Distance object means an object that is at large distance relative to the focal length of the lens. For instance an object 30 metre from a lens whose focal length is 21cm is a distance object. We note that, object position is many times longer compared to the focal length of the lens.

Distance between the lens and screen where the sharp image of a distance object is formed is considered to be the focal length of the lens. The area occupied by the image is the focal plane of the lens.

The estimated focal length is not exact but can be 2 cm plus or minus the real focal length.

This method of estimating focal length depends on the fact that parallel rays from infinity converges at the focal point on the screen.

Non-parallax method of estimating focal length

Parallax is a displacement or difference in the apparent position of an object viewed along two different lines of sight and is measured by the angle of inclination between those two lines.

Non-parallax is therefore when there is no difference in apparent position when an object is viewed along two different lines of sight.

An optical pin fixed on a cork that is supported by a clamp is placed above a lens that is on the mirror as shown below. The cork is such that it slides up and down the glass rod. see the diagram below.

Adjust height of the pin until it’s image is seen on the mirror.

The position of the pin is adjusted until the image from the mirror and the object pin seems to be moving together when you move your eyes.

Distance between lens and the pin when there is no parallax between image and object is the focal length of the lens.

Non-parallax: Using an illuminated object to estimate focal length

A bulb is placed behind a hole with a cross wire on a cardboard so as shown in figure below. A lens on a lens holder is placed between a mirror and the cardboard.

The cardboard together with the source of light is moved along the metre rule until a sharp image of the cross wire is formed along the cross wire object as shown. The figure shows two rays emerging from the point source towards the mirror through the lens

The lengths f gives the focal length of the lens.

Explanation

The ray striking the mirror are reflected back along the same paths of the incidence so that the image of the source coincides with the source itself. This image can be received on a screen placed at the same position as the source as shown.

If both lens and the mirror are perfectly vertical or parallel , image coincides perfectly with the illuminated crosswire. This makes it hard to see when Estimating focal length.

It is therefore necessary to tilt either the lens or the mirror a little so that the image can be mapped besides the hole.

In the above arrangement, the object pin is moved towards the lens or away until it coincides with it’s inverted image. This occurs when the pinhead is vertically above the center of the lens.

At a point where the object and the image perfectly coincides, there is no relative motion between them. As the eye is moved perpendicular to them, they all move together as one.

The distance between the pin and the lens is then measured as the focal length of the lense.

NB: Focal length increases as thickness of the lens decreases. This is because thick lenses refracts and deviates light more sharply than a thin lenses. Therefore, rays emerging from thick lens tends to converge earlier because because of the sharp bending in the lens.

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March 20, 2026
Focal length from lens formula
Determining Focal length from lens formula involves doing experiments to find different positions of image while adjusting object positions.

Focal length of a lens can be determined by investigating relationship between image distance and object distance by obtaining image distances from varying object distances.

The mirror formula describes the relationship that exists between the focal length, image distance and the object distance. Using the mirror formula derived earlier, We describe the experiment here and explains how to extract the the focal length from the relationship.

The mirror describes the relationship that exists between the focal length, image distance and the object distance.

The unknown Focal length of a lens can be determined experimentally by use of lens formula derived earlier. We describe the experiment here and explains how to extract the the focal length from the relationship.

We draw the graph of the of reciprocal values of image distance against the reciprocal values of object distance. The intercept of the graph is used to estimate the focal length of the lens used.

Apparatus
- Metre rule
- lens and a lens holder
- source of light
- screen
- cardboard with a cross wire
procedure
- set the apparatus as shown
- You place the object at the zero centimeter mark
- set the object distance by placing the lens at a reasonable distance from the object like 80cm from the object.
- Adjust the screen to observe a sharp image on the screen.
- Record a distance between the screen and the lens where you spot a sharp image on the screen is the image distance.
- Record the image and the object distance
- Reduce the object distance u by about 5 cm then adjust the screen until you see another sharp image on the screen.
- reduce the distances distance again by 5 cm and repeat the procedure above.
- Fill the table as shown below
From the data obtain a graph of 1/u against 1/v by plotting.

A typical graph will be as shown:

Graphical analysis of the Lens formula

The lens formula is stated as:

1f=1u+1v

From the above formula, one can see that the sum of reciprocals of the image length from the lens and the reciprocal of the image distances equals reciprocal of the focal length.

At the (1/v) intercept the value of (1/u)= 0. The lens formular becomes:

1f=0+1v

We eliminated 1/u from the formula after it become zero such that:

1f=1v

The value of f^-1 (1/f) is equal to 1/v meaning that we can approximate the reciprocal of f as the reciprocal of the image distance v read at the intercept.

At the (1/u) intercept, the the value of (1/v) =0. The lens formula becomes

1f=1u+0

The formular is reduced to be as follow:

1f=1u

Determining the Focal length from lens formula

as a process of Determining Focal length from lens formula, from the graph, we can deduce that 1/u and 1/v gives reciprocal of the Focal length 1/f at the intercepts.

we can get two values of f from the 1/v and 1/u intercepts such that:

f1=(1V)−1andf2=(1U)−1

The focal length f is the average of f₁ and f₂ such that:

f=f1+f22

Question for practice

The table below shows values of object distance u and corresponding value of image distances a for a convex lens.

object distance u(cm) 10 15 20 25 30 35
image distance v(cm) 40.0 17.1 13.1 11.8 10.9 10.4

a table showing relationship between image distance and object distance

plot a suitable graph and from the graph determine the focal length of the lens.

Related Topics
References:
- Secondary Physics Student’s Book Four. 3rd ed., Kenya Literature Bureau, 2012. pp. 1-42.
- Abbot A. F. (1980), Ordinary Level Physics, 3rd Edition, Heinemann Books International,
  London.
- Nelkon M. and Parker P., (1987), Advanced Level Physics, Heinemann Educational
  Publishers, London.
- Tom D., and Heather K. Cambridge IGCSE Physics. 3rd ed., Hodder Education, 2018, https://doi.org/978 1 4441 76421. pp. 106-142.
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March 12, 2026
A concise Introduction to thin lenses
Table of Contents
Introduction to thin lenses involves general description about properties of thin lenses.

A lens can be defined as a piece of curved glass or plastic that makes things look larger, smaller or clearer when you look through it.

In human eye, one component is a lens and so we can also define lens as the transparent part of the eye, behind the pupil, that focuses light so that you can see clearly.

The idea behind lens operations is that when a light ray passes from air which is more optically denser than the lens material, it is refracted.

When many rays passes through the lens, they all refracted the same way and so they meet at a common point. Sometimes they don’t meet but instead they are scattered after refraction but they are seemed to be spreading from a common point.

Lenses are usually made of glass, transparent plastic or perspex.

common application of lenses includes cameras, spectacles,telescopes, microscopes, film projectors and the human eye.

A thin lens means a lens whose thickness is negligible compared to the radius of curvature of the lens surfaces.

Types of thin lenses

The basic two types of lenses are convex and concave lenses.

Convex lenses are also called converging lenses as they cause the rays that passes through it to meet at a point. Convex lenses are thickest at the middle and they thin in as you move towards their edge.

In this lesson we will be talking about biconvex lenses meaning that it is symmetrical if we cut it long it’s edges. Both sides of it’s services at the center are bulging outwards and the edges are curved inwards uniformly on both sides.see the figure below

Bi convex lens

showing symmetrical in bi-convex lens

Concave lenses are also called diverging lenses as they cause the rays passing through them to be spreading from a common pint. Concave lenses are thinnest at the middle and they they become thicker as you move towards the edges.

Illustrating concave lens

illustrating bi-concave lens

There are variations of convex and concave lenses as illustrated in figures below

plano convex lens

convex meniscus lens

plano concave lens

concave meniscus lens

Effects of lenses on Parallel rays of light

A cardboard with parallel slits is placed between the mirror and a bi-convex lens as in figure below

The mirror is set such that it reflects the sun rays so that the rays passes through the slits before they reach the lens.

After making observations, the bi-convex lens is replaced with a concave lens

Observation

when a convex lens is used, the rays are converged at a point on the paper and then diverge as they continue as shown.

illustrations of parallel rays as they pass through a bi-convex lens

When concave lens is used , the rays diverge as if they were from the focal point in front of the lens as shown.

illustrations of parallel rays as they pass through a biconcave lens

Investigating convergence and divergence of light by thin lenses using a ray box

A ray box acts as a source of parallel beam. A spot light can also be used.

A parallel beam is directed incident to the the lens as shown

Parallel rays of light incident to a convex lens

A white paper is placed on the other side of the lens and it’s position adjusted until a sharp point is observed.

observation

When a convex lens is used, the rays are converged at a point on the paper and then diverges as they continue as shown below

Parallel beam after passing through a converging lens

If convex lens was replaced with concave (diverging) lens, the rays will be observed diverging as if they are coming from a point on the other side of the lens. see the diagram below.

Parallel beam incident to diverging lens

Explanations

Light is usually refracted when it passes through a glass prism. A lens can be considered as an assembly of many tiny prisms where each prism refracts light as in figure below.

Illustrations of bi-convex lens as an assembly of prisms

Please note that, the middle part of the prism is like a rectangular glass prism and a ray that is incident to it at a perpendicular angle passes through without being refracted. As we may see in other lessons, a ray of light that passes normally through the geometrical center of the lens, passes through undeviated.

The figure below shows representation of concave lens as an assembly of prisms.

illustrations of concave lens as an assembly of prisms

conclusions

Rays of light that passes through a lens converges at a fixed point from the lens if the lens is a converging lens or diverge from a common imaginary point if the lens is a diverging lens.

The point at which the rays emerging from the lens converge or seems to diverge from is referred to as the principal focus.

A convex lens has a real principal focus while a concave lens has a virtual (imaginary) principal focus.

Related Topics
References:
- Secondary Physics Student’s Book Four. 3rd ed., Kenya Literature Bureau, 2012. pp. 1-42.
- Abbot A. F. (1980), Ordinary Level Physics, 3rd Edition, Heinemann Books International,
  London.
- Nelkon M. and Parker P., (1987), Advanced Level Physics, Heinemann Educational
  Publishers, London.
- Tom D., and Heather K. Cambridge IGCSE Physics. 3rd ed., Hodder Education, 2018, https://doi.org/978 1 4441 76421. pp. 106-142.
March 10, 2026
Questions involving pressure
Questions involving pressure are physics questions that test your understanding of the concept of pressure, how it is calculated, and how it applies in solids, liquids, and gases.

They may involve:
- Calculating pressure using formulas
- Explaining real-life applications
- Understanding liquid pressure
- Understanding atmospheric pressure
- Applying Blaise Pascal’s Principle
1. The figure below shows Hare’s apparatus used for comparing liquid densities.
Use the information given in the above diagram to calculate the density of liquid x given that density of water is 1000kgm^-3(2mks)

answer

$$ \delta_w g h_w = \delta_wgh_x $$ $$delta_x = \frac{1000 \times 30}{25} = 1200kgm^{-3}$$ $$1000 \times 30 = \delta_x \times 25 \ $$

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March 9, 2026
Perfect squares
A perfect square is a number that is obtained when a whole number is multiplied by itself. In other words, it is the result of squaring an integer. For example, $1 = 1 \times 1$ $1 = 1 \times 1$ 1=1×1, $4 = 2 \times 2$ $4 = 2 \times 2$ 4=2×2, $9 = 3 \times 3$ $9 = 3 \times 3$ 9=3×3, and $16 = 4 \times 4$ $16 = 4 \times 4$ 16=4×4. These numbers are called perfect squares because they can be arranged to form a perfect square shape.

Perfect squares are important in mathematics because they help us understand square roots, area, and patterns in numbers. The square root of a perfect square is always a whole number. For instance, the square root of 25 is 5, since $5 \times 5 = 25$ $5 \times 5 = 25$ 5×5=25.

in quadratic expressions, a perfect square occurs when the expression can be written as the square of a binomial. This means the quadratic can be expressed in the form: $(a + b)^2 \quad \text{or} \quad (a – b)^2$ $(a + b)^2 \quad \text{or} \quad (a - b)^2$

When expanded, these forms follow special patterns: $(a + b)^2 = a^2 + 2ab + b^2$ $(a – b)^2 = a^2 – 2ab + b^2$

A quadratic expression is called a perfect square trinomial if:
1. The first term is a perfect square.
2. The last term is a perfect square.
3. The middle term is twice the product of the square roots of the first and last terms.
Examples

x²+6x+9 is a perfect square

x² = (x) x (x) =(x)²

9 = 3 x 3 = (3)²

middle term : 6x = 2(x)(3)

Therefore:

x²+6x+9 = (x+3)²

Example 2

assume x²−10x+25 is a perfect square

x²= (x) (x) = (x)²

25 = (5)²

middle term:-10x = -2(x)(5)

therefore x²−10x+25 = (x-5)²

Importance in Solving Quadratics

Perfect square expressions are useful because:
- They make factorization easier.
- They help in completing the square, a method used to solve quadratic equations.
- They simplify graphing, since expressions like $(x – a)^2$ clearly show the vertex of a parabola.
Example problem

factorize x² – 8x +16

solution

The constant term is a perfect square

middle term is twice the constant term

hence 16 = (4)(4)

-8 = -4-4

hence x² – 8x +16 = (x-4)(x-4) = (x-4)²

example

factorize:

$$x^2+\frac{4}{3}x+\frac{4}{9}$$

solution

$$\frac{4}{9} = \frac{2}{3} \ or \ (\frac{-2}{3})$$

$$\frac{4}{3}= \frac{2}{3}+\frac{2}{3}$$

$$x^2 + \frac{4}{3}x + \frac{4}{9} = (x+\frac{2}{3})^2$$

Example

factorize 4x²+20x+25

solution

4x²= (2x)(2x)

25=(5)(5)

hence 4x²+20x+25 = (2x+5)(2x+5) = (2x+5)²

completing the square

Completing the square is a method used to solve quadratic equations and rewrite quadratic expressions in a special form. It helps us change a quadratic expression into a perfect square trinomial, which can then be factored easily.

consider the expression x² +bx

we can add a constant term such that the expression becomes a perfect square.

To complete the square:

Take half of ,square it and then add it to the expression.

that is: add (b/2)² to the expression

Example problem on completing the square

what must be added to x² -18x to make it a perfect square?

$$\text{we must add}: (\frac{1}{2} \times (-18)^2 =(-9)^2 =81$$

so x² -18x is transformed to x² -18x+81 to be a perfect square.

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February 23, 2026
Upthrust in gases
Upthrust in gases is the upward force exerted on an object immersed in a gas, which opposes the object’s weight. Upthrust in gases acts as an upward force on an object immersed in a gas and opposes the object’s weight. It arises from the pressure difference between the top and bottom of the object due to the gas’s density. This concept is similar to upthrust in liquids but involves the behavior of gases, which are much less dense than liquids.

When an object is placed in a gas, the pressure at the bottom of the object is greater than the pressure at the top. This difference in pressure results in an upward force (upthrust). This is what that results to upthrust in gases. The size of this force depends on the volume of the displaced gas and the density of the gas.

Just like liquids, gases exerts upthrust on objects that are in them. Air is the most common gas of interest and so it will be used extensively to illustrate upthrust in gases. We float objects on air. Sometimes, as human beings, we use parachutes to float in air.

The upthrust in gases is small because air has lower density compared to most of substances.

The density of air is about 1.3kgm^-3 or 0.0013gcm^-3.

A balloon can float in air if it contains a gas with a lower density than air. For example hydrogen has a density of 0.09 kgm^-3 whereas helium has a density of 0.18kg^m-3. Therefore a balloon filled with helium or hydrogen will rise on air provided density of the balloon fabric and air will be less than density of air.

Problems involving upthrust in gases

Consider the figure below that illustrates a balloon filled with air .

If we consider the balloon filled with air to a certain volume, the weight of air in the balloon plus its fabric is greater than the weight of air displaced. This is because the volume of air in the balloon is nearly equal to the volume of air displaced.

The upthrust force on the balloon due to the air is thus less than the weight. The balloon thus stay grounded because the it’s weight is less than the upthrust force that could set it up to float on air.

That is W-U > resultant downward forces.

If the balloon is filled with a gas that has a lower density than air, the gas and balloon fabric weigh less than the displaced air. The upthrust force U exerted by the air on the balloon is greater than the weight W of the inflated balloon. This results in the upthrust force being larger. The resultant upward force is greater than W-U and hence the balloon sets to accelerate upward.see the illustrations below.

Example Question

A meteorological balloon has a volume of of 55m³ and is filled with a helium gas of density 0.18kgm^-3. If the weight of the balloon fabric is 170N, calculate the maximum load the balloon can lift given that density of air is 1.3kgm^-3

solution

Volume of air displaced by the balloon is 55m³ which is volume occupied by the balloon.

mass of air displaced by the balloon = 55m³ x 1.3kgm^-3 = 71.5kg

weight of air displaced =71.5kg x 10 Nkg^-1 = 715N

mass of helium in the balloon = 55m³ x 0.18kgm^-3 = 9.9Kg

weight of helium in the ballon = 9.9kg x 10 Nkg^-1 = 99N

Total weight of the inflated balloon = 170N + 99N = 269 N

From the law of floatation, upthrust is the weight of air displaced which should also be weight of the balloon plus the weight of load it will carry. hence,

upthrust = weight of the balloon + load in the ballon = weight of air displaced

269N + load in the ballon = 715N

load in the balloon = (715-269)N=446N

So the total mass of the goods to be included in the balloon should never exceeded 44.6kg.

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February 6, 2026
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Sources of water are the natural places from which humans, animals, and plants obtain water for their daily needs. The main source of water is rain, which supplies rivers, lakes, and underground water. Surface water comes from rivers, lakes, streams, and dams, while groundwater is found below the earth’s surface in wells, boreholes, and springs. Oceans and seas are also sources of water, although their water is salty and must be treated before it can be used for drinking. These sources of water are very important because they support life, agriculture, and economic activities.

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