Category: Physics

Interference of light waves is a phenomenon that occurs when two or more light waves overlap as they travel through the same medium. The study of interference provides strong evidence that light behaves as a wave and is clearly demonstrated in experiments such as Young’s double slit experiment, where alternating bright and dark fringes are formed on a screen.

Notes

• Interference occurs when two waves merge.
• According to the principle of superposition, the resultant effects of two waves travelling at a given point in the same medium which is the vector sum of their respective displacements.
• Suppose the amplitudes of the two wave pulses are $A_1$​ and $A_2$​, when the pulses are travelling in the same direction, the amplitude $A$ of the resulting wave is given by:

$A=A_1+A_2$

where $A$ is the vector sum of $A_1$​ and $A_2$

• The amplitude of the resulting pulses is the sum of the individual amplitudes of the initial pulses.
• If the resulting pulse has zero amplitude, then the pulses are said to have undergone complete destructive interference.
• Constructive interference occurs when the amplitude of the resulting pulse is bigger than that of the individual pulses.
• In destructive interference, the amplitude of the resulting pulse is smaller than that of the individual pulses.
• Two waves interfere as shown.

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The Young’s Double Slit Experiment

A single slit $S$ is placed in front of a monochromatic light source.

Because it is narrow, it diffracts light that falls on it, illuminating both slits $S_1$​ and $S_2$​ which are narrow, very close together and parallel to $S$.

$S_1$​ and $S_2$​ diffract the light which once more spreads out, superposing in the shaded area.

A series of alternate bright and dark vertical bands are formed on the screen.

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• The fringes are equally spaced and the light intensity at the bright fringe is maximum while at the dark fringe it is minimum.

How Interference Fringes are Formed

From the descriptions above, interference is a phenomenon which is exhibited by progressive waves and results from the interaction of wavetrains of same frequency and constant phase (coherent wave trains).

In an ordinary light source, light is produced as a result of electron transitions in the atoms of the source. The emitted bursts of waves last within $10^{-9}$10−9 to $10^{-8}$10−8 seconds and are out of phase with each other. Hence, two such light sources cannot be coherent owing to the random emission of light waves. They produce a uniform illumination instead of bright and dark fringes because the interference pattern that forms changes so rapidly.

In the two slits experiment, slits $S_1$ and $S_2$​ are equidistant from $S$. As a wavefront from $S$ reaches $S_1$​ and $S_2$​, each slit is considered as a new light source, such that the two slits form two coherent source as shown:

• $S_1$​ and $S_2$​ are equidistant from $S$.
• As a wavefront from $S$S reaches $S_1$​ and $S_2$​, each slit is irradiated as a new light source, such that the two slits form two coherent sources.
• A central bright fringe forms at $O$ when $S_1O = S_2O$ such that the path difference is zero.
• Moving outward on one side of the central bright fringe, the first bright forms at $P$P where:

$S_2P-S_1P=\lambda$

• For the dark fringe at $R$R:

$S_2R-S_1R=\frac{1}{2}\lambda$

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FORM II MATHEMATICS
CAT 2 – 2019

Intsructions:

• Answer all the questions: 50 marks
• show all your working
• non programmable electronic calculators may be used

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1. The exterior angles of a hexagon are:

Find the value of the smallest angle.(2 marks)

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2.

Simplify:$$\frac{p^4 + 2p^2q^2 + q^4}{p^3 – p^2q + pq^2 – q^3}$$

(2 marks)

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3.

Simplify the expression:$$\frac{4x – 9x^2}{3x^2 – 4x – 4}$$

(3 marks)

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4. Evaluate using logarithms:

$$\sqrt[3]{\frac{14.3 \times 0.0090}{76.54}}$$

(4 marks)

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5.

Three business partners, Atieno, Wambui and Mueni contributed shs. 50,000, 40,000 and 25,000 respectively to start a business. After sometime, they made a profit which they decided to share in the ratio of their contributions. If Mueni’s share was shs. 10,000, by how much was Atieno’s share more than Wambui’s? (3 marks)

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5.

In the figure below, angles ABC and ADC are equal. Angle ACD is a right angle. The ratio of the sides: AC:BC = 4:3

f the area of triangle ABC is 2$\mathrm{4cm2}$ .Find the area of triangle ACD.3 mks

6.

The angle of elevation of the top of a cliff from point P is $45^\circ$. From a point Q, which is 10 m from P towards the foot of the cliff, the angle of elevation is $48^\circ$.Calculate the height of the cliff.(4 mks)

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7.

8. A solid S is made up of a cylindrical part and a conical part. The height of the solid is 4.5 m. The common radius of the cylindrical part and the conical part is 0.9 m. The height of the conical part is 1.5 m.

a) Calculate the volume of solid S, correct to 1 decimal place.4 mks

8.

In the figure below, $\overrightarrow{CA}=k,\ \overrightarrow{AX}=\overrightarrow{AY}=\frac{1}{7}\overrightarrow{AB},\ \overrightarrow{CB}=a$

Express $\overrightarrow{CX}$CX in terms of $a$a and $k$k. $(3 \text{ mks})$(3 mks)

9.

A bus left a petrol station at 9:20 a.m and travelled at an average speed of $75 \text{ km/h}$ to a town $N$N. At 9:40 a.m, a taxi travelling at an average speed of $95 \text{ km/h}$ left the same petrol station and followed the route of the bus. Determine the distance from the petrol station covered by the taxi at the time it caught up with the bus. $(3 \text{ mks})$

10. Solve the simultaneous equations:

11.

The hire purchase (H.P) price of a public address system was Ksh 276,000. A deposit of Ksh 60,000 was paid followed by 18 equal monthly instalments. The cash price of the public address system was 10% less than the H.P price.

Calculate:

(i) the monthly instalment. $(6 \text{mks})$

(ii) the cash price. $(2 \text{mks})$

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11 (b)

The cost of a car outside Kenya is US $5,000. You intend to buy one such car through an agent who deals in Japanese Yen. The agent will charge you 20% commission on the price of the car and a further 80,325 Japanese Yen for shipment of the car.

How many Kenya shillings will you need to send to an agent to obtain the car given that:

• 1 US $ = 65.00 Yen
• 1 US $ = 100.00 shillings $(3 \text{mks})$

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11 (c)

A salesman earns a basic salary of Ksh 9,000 per month. In addition, he is also paid a commission of 5% for sales above Ksh 15,000. In a certain month, he sold goods worth Ksh 120,000 at a discount of $2\frac{1}{2}\%$.Calculate his total earnings that month.(3marks)

12.

The frequency distribution table below represents the number of kilograms of meat sold in a butchery.

Mass in kg	1–5	6–10	11–15	16–20	21–25	26–30	31–35
Frequency	2	3	6	8	3	2	1

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(a) State the modal frequency. $(1 \text{mk})$

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(b) Calculate the mean mass. $(5 \text{mks})$

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(c) Calculate the median mass. $(4 \text{mks})$

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Heat is a form of energy that is transferred from one body to another due to a difference in temperature. The study of quantity of heat and heat capacity helps us understand how substances absorb, store, and transfer thermal energy. Quantity of heat refers to the amount of thermal energy gained or lost by a substance, while heat capacity is the measure of the amount of heat required to raise the temperature of a body by one degree Celsius or one Kelvin. These concepts are important in explaining everyday phenomena such as heating water, cooking food, and the functioning of heating systems. Understanding quantity of heat and capacity provides a foundation for studying thermal physics and helps in analyzing how different materials respond to heating and cooling processes.

Heat is a form of energy that flows from a region of higher temperature to a region of lower temperature.

When heat is supplied to a substance, two things may happen:

• The temperature of the substance increases
• The substance changes its state (for example, melting or boiling)

Understanding how heat affects matter is important in physics, engineering, and everyday life.

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Heat and Temperature Change

The rise in temperature of a substance depends on:

• The mass of the substance
• The type of material
• The amount of heat supplied

This explains why water heats more slowly than metals.

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Heat Capacity

The figure below illustrates comparing of different capacity of water being heated for sometimes by a bunsen burner. The beakers are identical and the graph shows how their temperature changes with time.

Definition:
Heat capacity is the amount of heat required to raise the temperature of a body by 1°C (or 1 K).

Formula:

Q = Cθ

Where:
Q = heat energy (J)
C = heat capacity (J K⁻¹)
θ = temperature change (°C or K)

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Specific Heat Capacity

Definition:
Specific heat capacity is the amount of heat required to raise the temperature of 1 kg of a substance by 1 K.

Formula:

Q = mcθ

Where:
m = mass (kg)
c = specific heat capacity (J kg⁻¹ K⁻¹)
θ = temperature change

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Key Concept

Different materials respond differently to heat:

• Water has a high specific heat capacity → heats slowly
• Metals have low specific heat capacity → heat quickly

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Relationship Between Heat Capacity and Specific Heat Capacity

C = mc

This means total heat capacity depends on both:

• Mass
• Type of material

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Heating Experiment

Aim:
To observe how heat affects temperature change in water.

Procedure:

1. Measure a fixed volume of water
2. Record its initial temperature
3. Heat the water
4. Record time taken to reach a higher temperature
5. Repeat using different volumes

Observation:

• Larger volume → heats more slowly
• Smaller volume → heats faster

Conclusion:
The quantity of heat required depends on the mass of the substance.

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Worked Examples

Example 1

Heat capacity = 460 J K⁻¹
Temperature change = 45°C − 15°C = 30°C

Q = Cθ
Q = 460 × 30
Q = 13,800 J

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Example 2

Power = 50 W
Time = 9 minutes = 540 s

Q = Pt
Q = 50 × 540
Q = 27,000 J

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Example 3

Mass = 60 g = 0.06 kg
Specific heat capacity = 390 J kg⁻¹ K⁻¹
Temperature change = 50°C

Q = mcθ
Q = 0.06 × 390 × 50
Q = 1170 J

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Table of Specific Heat Capacities

Substance	Value (J kg⁻¹ K⁻¹)
Water	4200
Alcohol	2400
Kerosene	2200
Ice	2100
Aluminium	900
Copper	390
Lead	130

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Mass, weight and density are fundamental physical quantities that help us understand how matter behaves, how heavy objects are under gravity, and how compact substances are.

Mass

Mass of an object is the quantity of matter in it. it remains constant regardless of location or the force of gravity acting on it.

The SI unit of mass is Kilogram. The symbol for Kilogram is Kg. A kilogram is the mass of a piece of metal that is stored at Sevres, near Paris. It is used as the standard for measuring masses. Sevres is where the International office of weights and Measurements is located.

Kilogram can be broken into smaller units so that with have a sub-multiples of a kilogram which we can represent with prefixes.

the table below shows some sub-units of a gram, which is the most common unit of measuring masses. 1000 grams = 1 Kilogram.

sub unit and symbol	equivalent in grams
picogram (pg)	10-12
nanogram (ng)	10-9
microgram (μg)	10-6
milligrams (mg)	10-3
centigrams(cg)	10-2
decigram (dg)	10-1

The table belows sub units of kilogram as multiples of gram

sub unit	equivalence in grams	equivalent in kilograms
Decigram(Dg)	10	0.01
Hectogram(Hg)	100	0.1
Kilogram	1000	1.0
tonne	1000000	1000

Revision exercise

1. Convert 0.02 g to milligrams.
2. Convert 0.75 kg to grams.
3. Express 250 cg as grams.
4. what is 0.6 g in centigrams.
5. Convert 8000 ng to micrograms.
6. Convert 5 µg to nanograms.

Weight

Weight of an object is the pull of gravitational force on it. The pull of the earth, sun and moon on an object is called the force of gravity due to the earth, sun and moon respectively. The SI unit of weight is newton.

Weight of an object = mass x gravitational force

in other words: Weight(N) = mass(kg) x g (N/kg)

Density

Density of a substance is the mass of a unit cube of the substance, that is; density is mass per unit volume.

The SI unit of density is kilogram per cubic metre (kgm^-3).

Another common unit of expressing density is grams per cubic centimeters (gcm^-3).

Example problems

1. Find the mass in Kilograms of an ice cube of side 6cm If the density of ice is 0.92gcm^-3 .

solution

volume of the cube = 6cm x 6cm x 6cm = 216cm³

mass = density x volume

mass = 216cm³ x 0.92gcm^-3 = 198.72 g

2. Find the volume of cork in cubic metre of mass 48g given that density of cork is 0.24gcm^-3.

solution

1 cubic metre = 1000000 cm³

That is:

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Uniform Circular Motion refers to the motion of an object traveling at a constant speed along a circular path. The object’s speed remains constant. However, its velocity is constantly changing. This change occurs due to the continuous change in direction as the object moves along the curve.

Key characteristics of uniform circular motion includes:

1. Constant Speed: The object moves with a constant speed along the circle. However, because the direction is always changing, the velocity, which is a vector, changes continuously.
2. Centripetal Force: An object follows a circular path only if a force acts toward the circle’s center. This is called the centripetal force.

Consider a particle moving along a circular path when it moves from point A to point B as in figure below.

The reference point OA makes an angle with the line OB that represents the line joining the new position of the object and the center of the circle. The object has covered a distance S along the circle making an arc with θ.

Angular displacement is the angle swept by a line joining end of an object in a circular path with the center of the path when it moves from one point to another in a circular motion.

Angles in circular motion are usually expressed in radians θ^c.

That is:

θ(radians)=Sr

it therefore follows that S = rꝊ^c

A radian is defined as an angle subtended at the center of a circle by an arc length equal to the radius of the circle.

Therefore angle θ subtended by the circumference at the center of a circle of radius r is therefore given by;

θ=circumferenceradius

and we can write circumference in terms of π such that:

2πr=2π^c

We can relate the degrees from 2π^c = 360^o .

Problem

An object traces an arc of length 10.98 while attached to a cord of length 3.2 m that is fixed on a fixed surface on a flat smooth surface. Determine the angular displacement by the object.

Solution

We visualize the setup as in figure below:

from the equation S = rꝊ^c

θc=Sr=10.983.2=3.43 radians

if we can express answers in degrees, then 3.43 radians = 196.52^o .

Practice Question

An object moves a distance of 80.12 π along a circular path of radius 3.8m. Determine it’s angular displacement.

Angular velocity for a uniform circular motion

Angular velocity is defined as the rate of change of angular displacement with time.

we can shorten the equation by using symbols alone:

ω=ΔθΔt

The SI Units of angular velocity is radians per second (rads-1)

Consider the equation that relates angular displacement in radians with the arc length made by the object:

θ=Sr

to get the rate of change of speed, we divide both sides with t as they both represent displacement of the object.

θt=Srt=1r(St)

but distance s when divided by time gives velocity. That is;

v=St

where v is the linear velocity representing the velocity of the object along the circular path.

ω=θt

Hence the equation above becomes:

hence, the angular velocity can be expressed in terms of linear velocity and radius as show in equation below:

ω=vr

similarly, linear velocity can be expressed in terms of angular velocity as v=ωr.

An object in circular motion has both linear and angular velocity. The time taken to make one complete circle is called the periodic (T) and is given by T= circumstances/speed.

Let us now consider the time taken for a body to make one complete circle in a circular path. At that one circle, it will have covered the circumference of a circle. The time taken to complete such one revolution is called periodic time (T).

from the equation :

time=distancespeed

and that:

T=circumferencespeed

and circumference = 2πr, hence

T=2πrωr=2πω

Hence T in the above equation becomes:

T=2πω

but

1T=ω2π

therefore:

ω=2πT

but since:

1f=T

ω=2π1f=2πf

Example question

A metallic ball is whiled in a horizontal circle making 5 revolution s per second.Determine:

• Period T
• angular velocity and
• linear speed v

Centripetal force

Consider on object m held by a cord om positioned at A. The Object is whirled in a circular motion and after some time Δt, the object is at position B. The velocity of the object in linear direction changes from V_A to V_B. If there was no force acting on the body, the object will not change directions but will go in a straight line. There must be a force that maintain the body at a constant distance distance from the center o.

Centripetal force F_c refers to the force that keeps a body in circular motion. A body in a circular motion is accelerating and from newton’s second law of motion, there must be a force acting on it to cause acceleration. Centripetal force is usually directed towards the center of the circular path. The Centripetal force is the force responsible for the constant change of direction otherwise the body would naturally follow a straight line if there was no force acting to keep the body in circular motion.

The value of the centripetal force is derived from newton’s second law of motion which states that: the rate of change of momentum of a body is directly proportional to the resultant force in the direction of force.

Momentum means mass multiplied by velocity.

Because velocity of a body in circular motion is changing, it’s momentum must also be changing.

The newton’s second law can be described as F=ma, where a = acceleration and m is the mass.

but the acceleration of the body of the body is given by a= v²/r, where v is the linear speed of the object while it is in circular motion. hence

From definition of angular velocity we had shown that, ω is given by v/r, and hence v=ωr.

it follows that F_c = m(ωr)²/r = mω²r²/r = mω²r .

Tension from circular motion

If a body is attached to a string and swung around on a horizontal circle, the centripetal force that keeps the body in the circular orbit is kept as tension in the string. For the body to remain in circular motion, the centripetal force is equal to the tensional force.

From the equation F_c = mω²r , it shows that centripetal force is directly proportional to the angular velocity meaning that a larger force will be required to maintain the body in motion if it is swung faster.

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• Electromagnetic waves
• Centripetal force
• The equation of a circle
• Junior Secondary Examination papers
• Angular Velocity
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Trigonometric questions are math problems that involve angles and ratios using trigonometry. They usually test how well you understand relationships between angles and sides of triangles.

1.Solve the equation 2 cosθ= √3 in the range 0^o≤θ≤360^o

2. Solve for x in the equation -3xsinx=1 in the range 0^o≤θ≤360^o

3. Solve for x in the equation sin²x = 0.25 for 0^o≤θ≤360^o giving your answer in π^c

5. solve the equation 2 sinθ = cosθ in the range 0^o≤θ≤360^o to the nearest whole number

6. Calculate without using log table:

7.Triangle ABC is inscribed in a circle as in figure below. AB = 9.2cm, AC = 7.9 and BC=4.4cm

Find: (a) Angle A (2 marks)

(b) The radius of the circle correct to 1 d.p (2 marks)

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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:

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;

similarly we can subtract equation 2 from 1 as shown:

hence s- d = u + v –v + u

therefore : s- d = 2u and hence;

from the lens formulae:

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

the above equation can be simplified into:

finding the lcm of the denominator, we obtain;

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

and simplifying the above equation in the numerator:

and finding the reciprocal so that we can get f;

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 L₂ that represents different positions of the lens are known as the conjugate points.

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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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