Physics

Laboratory Manual
For
Applied Health Sciences
Students

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 Contents
Exp. (1): Young’s Modulus of Bone-Equivalent Material.

Exp. (2): The viscosity coefficient (η) of a solution.

Exp. (3): Calibration of a Thermocouple.

Exp. (4): Specific heat of water.

Exp. (5): Measurement of short length using vernier calliper.

Exp. (6): Ohm’s law.

Exp. (7): Dielectric constant.

Exp. (8): Determination of the power of a convex lens.

Exp. (9): Determination of the power of a concave mirror.

Exp. (10): X-Ray demonstration.

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 Exp. (1): Young’s Modulus of Bone-Equivalent
Material
Aim:

1- Determine Young‟s modulus of compact bone.

2- Determine the mass of the hook (pan).

Theory

We shall discuss the deformation of solids in terms of the concepts of stress and strain.
Stress is a quantity that is proportional to the force causing a deformation; more specifically,
stress is the external force acting on an object per unit cross sectional area. The result of a stress
is strain, which is a measure of the degree of deformation.
It is found that, for sufficiently small stresses, stress is proportional to strain; the constant
of proportionality depends on the material being deformed and on the nature of the deformation,
we call this proportionality constant the elastic modulus. The elastic modulus is therefore defined
as the ratio of the stress to the resulting strain:
The elastic modulus = (1.1)

The measurement of elasticity of a material is described by the term elastic modulus, also
referred to as modulus of elasticity or Young s modulus, and denoted by the variable E. The
word modulus means ratio. The elastic modulus represents the stiffness of a material within the
elastic range. The elastic modulus can be determined from a stress strain curve, by calculating
the ratio of stress to strain, or the slope of the linear region of the curve.
The elastic quantities of a material represent a fundamental property of the material. The
interatomic or intermolecular forces of the material are responsible for the property of elasticity.
The stronger of the forces, the greater the values of the elastic modulus and the more rigid or stiff
the material. Because this property is related to the forces within the material, it is usually the
same when the material is subjected to either tension or compression.
The property is generally independent of any heat treatment or mechanical treatment that
a metal or alloy has received, but is quite dependent on the composition of the material.

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 Elasticity can be defined as: “It 1s the property of the strained bodies to re-establish their
initial shape and volume after the external forces has ceased to act”
There are two types of deformation:
 Elastic deformation: It‟s the deformation which vanishes with lifting off their external
forces.
 Plastic deformation: It‟s the deformation which remains after the external stresses have
been lifted.
Hooke established a relation between the elastic deformation and the internal forces
acting in a material.

F = kx (1.2)

Young's Modulus and Hooke's Law:

Hooke's law can be restated as follow: “For any elastic material, the stretching stress is
directly proportional to the longitudinal strain”

Where:

: Stress (N'/m2).

Strain (no units).

: Constant called Young's Modulus of elasticity

Young's modulus can be defined as: “It is the ratio of stress to strain, and it has a unit of
N/m2

= (1.3)

Bones as well as some other materials have shown at elastic behavior, it consists of two
quite different materials plus water which are:
Collagen
It is the major organic fraction: it represents 40% of the weight and 60% of its volume.
Collagen is flexible like rubber. Thus it gives the elastic properties of solid bone.

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Bone minerals:
It is the inorganic fraction it represents 60% of the weight and ay 40% of its volume.
Bone minerals are fragile and it is believed to be made up of Calcium hydroxyapatite
[Ca10(PO4),(OH)2]. Which are rod shaped rystals with diameters of 20- 0 and length of 0-
100 .
Bones are of two types:
 Solid or compact: which is of constant density throughout life and ρ=1.9 g/cm3 Its
Young's modulus equals 1.8X1010 N/m2 or 1.8X1011dyne/cm2.
 Spongy or trabecular. Its Young's modulus equals 8X107 N/m2.
As shown from Young's modulus, solid bone has better elastic properties. Bone minerals
are as strong as granite in compression and 25 times stronger than granite under tension. Bones
are not as strong under tension as they are in compression however; bones are stronger under
tension than many other materials such as porcelain and concrete.
In this experiment, a load beam is fixed horizontally at one end (cantilever) is used to
determine Young‟s modulus. If the cantilever has a load (W= mg), suspended at the extreme end,
the depression y of the end is given by the equation:

Y= W Y= mg (1.4)

Where:
B and d: are the length, the breadth, and the thickness of the cantilever respectively.
Y: is the Young s modulus for the material of the beam.
Procedure:

 Take the zero reading of the cantilever without the pan and any loads.
 Apply various loads on the scale pan or hook (50 g) increments, then record the
corresponding reading of the pointer at the extreme end of the cantilever each time.
 Tabulate the observations and draw the relation between the applied mass (m) and the
corresponding depression y from the graph find the slope = And from the relation y=

, slope =

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  We can calculate Young's modulus from the relation:

Y=

 The straight line will intersect the mass axis at a value corresponding to the mass of the
pan (hook), m

Results:

Load (M) mg Depression (y) cm

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 Exp. (2): The viscosity coefficient (η)
of a solution.

Aim:
Determination of the viscosity coefficient (η) of a solution.
Theory:
Any liquid flowing in a tube may be considered as composed of concentric layers,
moving with different velocities. The layers adhering to the sides of the tube may be considered
as stationary, and the velocity of the layers will increase they approach to the middle of the tube
this can be summarized as shown in the following diagram:
 This main reason of the difference in velocity between the different layer is the frictional
force (F).these force are parallel to the moving layer.
 This tangential frictional force (F) is required to maintain constant difference in velocities
of the various layer of the liquid moving in a tube.
 This force (F) is directly proportional to the difference in velocity between the layer of
area of contact, and inversely with the distance between the two layers. This can be
concluding in the following equation:

Fα F= η (2.1)
Where:
A layer of area.
∆V The change in velocities of the moving layers.
d The distance between the moving layers.
η The proportional constant which is called the viscosity coefficient.

It can define as; the frictional resistance offered by a liquid against the displacements of its own
molecules.
Or “It is the frictional force necessary to move layer of liquid of area 1cm with velocity 1cm/sec
over another layer at distance of 1 cm”

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The unit of the coefficient of viscosity is “poise” and its SI unit is dyne sec/cm2
1 poise = 1 dyne. Sec/cm2 =1 gm/sec.cm
The viscosity of a liquid is generally measured by observing the time required for a
define volume of liquid to move or flow through a standard capacity tube.
The law governing the flow of liquids through capacity tube was observed by poiseuille
which is

η= (2.2)
Where:
v: the liquid volume.
L: the capillary tube length.
R: the radius of the capillary.
P: the pressure which affects the liquid flowing
t: the flowing time of the liquid.
The device used for these measurement is called Ostwald' Viscometer as show in the
figure.
The time of flow of equal volumes of two liquids through the same capillary and under
the same conditions is t1 and t2, the ratio of the viscosity coefficient of the two liquids is:

= (2.3)

If solution 2 is water thus:

= (2.4)

Where:
η1 is the viscosity coefficient of solution.1
η2 is the viscosity coefficient of solution2.
ρ1 and ρ2 are their densities.
The viscosity coefficient η is determined by this relation is a relative quantity; thus it has
no unit, which can be defined as it‟s the ratio of the absolute viscosity of any liquid to that of
water.

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 = (2.5)

If the absolute viscosity of water η = 0.01

= (2.6)

Procedure:
(1) The viscometer is cleaned up well
(2) A definite volume of the liquid is introduced to the tube of bulb (C) of the viscometer. Let it
10 ml.
(3) The liquid is forced-up through the capillary tube by suction until reaching mark (a).
(4) Allow the liquid to flow back through the capillary tube with nothing the time of flow till
liquid reaches The mark (b) of the viscometer
(5) Repeat step 3 and 4 two or three times and record the average time flow.

(6) Clean the viscometer by running 2 or 3 times in the viscometer.

(7) Repeat steps 2, 3 and 4 distilled water i. e. 10 ml water. And record the time of flow of water
record the time of flow of water
(8) alculate (η) from the equation:

= .

(9) The absorbed value of viscosity of the solution will be calculated from the equation:

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 Exp. (3): Calibration of a Thermocouple

Aim:

 Determine the room temperature.

Theory:

The thermocouple is the most common type of temperature sensor, primarily because it is
inexpensive and easy to use. In fact, it is used in many places familiar to you: in the home, it is
used to control the temperature of the furnace, water heater, and the kitchen oven; in the
automobile, it is used to monitor coolant and oil temperature, and even to control the air
conditioner.
Thomas Johann Seebeck (1770-1831) discovered that a circuit comprised of dissimilar
metals produces a voltage (and current) when the two dissimilar junctions are exposed to
different temperatures. This phenomenon, called the Seebeck Effect, is depicted in Figure 1. The
voltage produced is proportional to the temperature difference between the junctions. The
voltage produced is small, on the order of millivolts.

Figure (5.1): Thermocouple circuit connected to a voltmeter.

In order to calibrate a thermocouple, one of the junctions must be held at constant

temperature. For the sake of simplicity, let us set the reference temperature, T1, to be 0°C. This
can easily be achieved by submersing the junction in a bath of ice water, commonly called an ice
bath reference. In Part A of this experiment, we will use this technique to calibrate two

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thermocouples. Later in this course, we will develop a simpler method that eliminates the ice
bath.

Procedure:

(1) The thermocouple is connected as shown in the

figure.

(2) Immerse electric heater in the beaker until

boiling point.

(3) Immerse the thermocouple and start to cool

down the water in a step of 10 C record the
potential difference V at each temperature.

(4) Tabulate the results and draw the standard

curve and find the room temperature.

Results:

Emf. Temp.

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 Exp. (4): The Specific Heat of Water

Aim:

 Determine the specific heat of water.

Theory:
Temperature is a measure of the average kinetic energy of the atoms or molecules of a
substance, i.e. the higher the temperature, the faster the molecules are moving around in the
substance. Temperature is difficult to measure directly, so we usually measure it indirectly by
measuring physical properties that change with temperature.
Heat, on the other hand, is defined as the transfer of energy between objects because of a
temperature difference.
Transferring energy into (out of) a system as heat is one way to increase (decrease) the
temperature of a substance, if energy Q transfers to a sample of a substance with mass m and the
temperature of the sample changes by ∆T; hence heat is given by:
Q = mc∆T (4.1)
The value c is the specific heat of the substance. In fact, this equation defines the specific
heat. The specific heat is the amount of heat energy required to raise the temperature of 1 Kg of a
substance by 1 degree Celsius.

c= (4.2)

The unit of specific heat is cal/(kg.co) where, Calorie is the energy needed to raise the
temperature of 1 gram of water through 1°C.
The mechanical equivalent of heat:
1 cal = 4.186 J (4.3)

Specific heat is an important quantity because it can be used to calculate the number of
calories required to heat a known mass of a substance.
The higher the specific heat of a substance, the less its temperature will change when it
absorbs a given amount of heat. Conversely, the lower the specific heat of a substance, the more
its temperature will change when it absorbs a given amount of heat. Among common substances
that are part of our environment, water has a relatively high specific heat 4186 J/(kg.co).

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 When two objects at difference temperatures are placed in contact with each other, the
quantity of heat lost by the hot object is equal to that gained by the cold object provided that no
heat is lost to the surrounding this is because of energy conservation.
Conservation of energy allows us to write the mathematical representation of this energy
statement as:
Qcold = - Q hot (4.4)
The negative sign in the equation is necessary to maintain consistency with our sign convention
for heat.
If we have an amount of water of mass mw and a source of thermal energy like an electric
heater, so we can arrange a system to calculate the specific heat of water as:
Heat lost by the electric heater = heat gained by the water
H = CW mw ∆T (4.5)
H = P. t (4.6)

Where P is electric power, t is the time taken to raise the temperature by amount ∆T.

P. t = Cw mw ∆T (4.7)

P = (4.8)

Tools:
(1) Cylinder beaker.
(2) Stopwatch.
(3) Immersed heater.
(4) Thermometer.
(5) Power supply.
(6)
Method:
1- By using the measuring cylinder take 200 ml of water and put it into a beaker.
2- Immerse the immersion heater in the beaker and it should be covered by the water.
3- Take the zero reading of the water temperature T0.
4- Turn on the power supply of the heater and stir the water continuously.

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 5- Register the temperature, T, every 60 second by using stopwatch.
6- Tabulate the values of time, t, and the difference in the temperature ∆T.
7- Plot the graph between the time, t, and the difference in the temperature ∆T.
8- Calculate the specific heat of water from the relation;

Cw =

Results:
T0 = …………….
P = ………………
mw = ……………..

Time (t) sec. Temperature ( 0C) ∆T (C0)

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 Exp. (5): Measurement of Short Length Using
Vernier Calliper

Aim:

 Study the vernier calliper as a tool to measure small lengths.

Theory:

The purpose of this experiment is to learn how to use the vernier caliper in measuring small
lengths, which cannot be measured by general meter stick. This device can measure length in the
range of tenth of millimeters. Vernier caliper gives accurate reading of fractional part of a scale
division.
The vernier caliper consists of two scales
• Fixed main scale.
• Movable or vernier scale, which is arranged to slide along the fixed scale.
This can be shown diagrammatically as follow;

As shown in Figure 1, the divisions on the main scale are centimeters and further divided into
tenth of centimeters or millimeters.
The vernier or movable scale contains ten divisions; these ten divisions are equal to nine
divisions in the main scale when the two jaws of the caliper are closed as shown in Figure 2.

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How we can take any reading?
1. Take the main reading, by the (0) of the vernier scale.
2. Determine the decimal line of the vernier scale which coincides with a line of the main scale.
This can be shown diagrammatically in Figure 3

The reading by zero of movable scale is 0.4 + something?

Look for the first line of the movable scale which coincides with the main scale, we see that, the
6th line of the movable scale is coincident, thus the final reading will be 0.4 + 0.06 = 0.46 cm.

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Another example,

The main reading = 1.3 +?? cm

We find that the 5th line of the movable scale is coincident, thus the final reading will be 1.3 +
0.05 = 1.35 cm.

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 Exp. (6): Ohm’s law

Aim:

(1) Verify Ohm's law.

(2) Determine the value of unknown Resistance.

Theory:

When beginning to explore the world of electricity and electronics, it is vital to start by
understanding the basics of voltage, current, and resistance. These are the three basic building
blocks required to manipulate and utilize electricity.
These concepts can be difficult to understand because we can‟t see them. We can‟t see by
the naked eye the energy flowing through a wire or the voltage of a battery. In order to detect this
energy transfer, we must use measurement tools such as multimeters to visualize what is
happening with the charge in a system.
Electric current is flow of charges (electrons) in a conductor per unit time. The work
done necessary to transfer one coulomb of charge between two points through a conductor is
called voltage and potential is the other term of voltage. For example, the first element has more
charges, so it has higher potential. On the
other hand, the second element has
charges that are fewer charges so it has
lower potential. The difference between
two points is called potential difference.
Resistance is a material‟s
tendency to resist the flow of charge
(current). All materials are difference in
allowing electrons flow, materials that
allow many electrons to flow freely are
called conductors such as copper, silver,
aluminum. In contrast, materials which
reduce or prevent electrons to flow are
called insulators such as plastic, rubber,

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glass and dry paper. Another type of materials, semiconductors have characteristics of both
conductors and insulators, and examples are carbon, silicon and germanium.
The resistance of a wire of length “L”, cross section area “ ” would be theoretically given as:

R=
Where “ρ” is the “resistivity” of the material.
For example: Copper: 1.72x10-8 Ω•m
Carbon: 3.5x10-5 Ω•m
The most applied relation for a wide range of components used in electrical circuits is
that known as Ohm's Law it states that the current through a conductor between two points is
directly proportional to the potential difference or voltage across the two points, and inversely
proportional to the resistance between them.
The mathematical equation that describes this relationship is:
V=IR
Where I is the current, The SI unit for measuring an electric current is the ampere
(Coulomb/Sec), V is the potential difference in units of volts(joule/ coulomb), and R is the
resistance in units of ohms.
More specifically, Ohm's law states that the R in this relation is constant, independent of
the current.

Tools:
(1) DC Power supply.
(2) Ammeter.
(3) Voltmeter.
(4) Variable resistance.
(5) Wires.

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Method:
(1) We will connect the equipment stated as shown in figure.
(2) Switch on the power supply and record the value of the potential difference V and the value
of electric current I by taking the reading of voltmeter and ammeter respectively.
(3) By means of variable resistance, increase the value of electric current I in regular steps, and
determine in each case the corresponding value of the potentials difference.
(4) Tabulate your results as shown and plot a relation between V and I and calculate the value of
unknown resistance from the slope.

Result:

I (mA)

V( volt)

Slope = = R

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 EXP. (7): Dielectric Constant
Aim
 Determination of the dielectric constant of glass, wood and acrylic sheets using two
isolated parallel plates (as a capacitor) and a capacitance meter.
Theory
A dielectric is a non-conducting material, such as rubber, glass, or waxed paper. When a
dielectric is inserted between the plates of a capacitor, the capacitance increases. If the dielectric
completely fills the space between the plates, the capacitance increases by a dimensionless factor
1, which is called the dielectric constant of the material.
The dielectric constant varies from one material to another. In this experiment, we
analyze this change in capacitance in terms of electrical parameters such as electric charge,
electric field, and potential difference.
We can perform the following experiment to illustrate the effect of a dielectric in a
capacitor. Consider a parallel-plate capacitor that without a dielectric has a charge Q0 and a
capacitance C0. The potential difference across the capacitor is ∆V0 = Q0/C0. Figure (1a)
illustrates this situation. Note that no battery is shown in the figure; also, we must assume that no
charge can flow through an ideal voltmeter. Hence, there is no path by which charge can flow
and alter the charge on the capacitor.
If a dielectric is now inserted between the plates, as in Figure (2b), the voltmeter
indicates that the voltage between the plates decreases to a value ∆V. The voltages with and
without the dielectric are related by the factor 1 as follows:

∆V =

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 Because the charge Q0 on the capacitor does not change, we conclude that the
capacitance must change to the value:

C= = =

C=
That is, the capacitance increases by the factor 1 when the dielectric completely fills the
region between the plates. For a parallel-plate capacitor, where C0 =ε0 A/d, we can express the
capacitance when the capacitor is filled with a dielectric as:
C=

It would appear that we could make the capacitance very large by decreasing d, the
distance between the plates.
Experiment and Requirements
1. Bench of plates.
2. 2 plate capacitor with the same dimensions
3. Dielectric sheets of different materials
4. Capacitance meter.
5. Connecting Cables.
Method
1. Connect the circuit as in the figure.
2. Put a glass sheet for example between the two plates then determine the capacitance of the
capacitor from the capacitance meter.
3. Get the dielectric constant of the different materials by the relation C = where ε0 is the

permittivity of free space.

4. Do the same steps for all dielectric materials.

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 Exp. (8): Determination of the power of a convex lens
Aim:
Determine the power of a convex lens.

Theory:
Snell’s law of refraction:

Refraction is the bending of light rays. Normally, light travels in a straight line, when it
passes from one transparent medium to another, such as from air into glass, its direction and
speed will be change. In a vacuum, the speed of light, denoted as "c," is constant. However,
when light encounters a transparent material, it slows down. The degree to which a material
causes light to slow down is called material's refractive index, denoted as "n."

n= (8.1)

Where c is the speed of light in vacuum and v is the phase velocity of light in the medium. The
approximate values of n for common materials are:

 Vacuum = 1
 Air = 1.0003
 Water = 1.33

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Where θ1 is the angle of incidence, θ2 is the angle of refraction and n1 and n2 are the respective
indices of refraction of the materials.

A lens is a transparent curved device that is used to refract light. It's made from
transparent materials (glass). There are two different shapes for lenses:

Fig. 1: convex and concave lens.

A convex lens is thicker in the middle than at the outer edge. Due to this structure, it will
cause the light rays passing through it to bend and converge at the point called focal point as
shown in Fig.2. .The distance f is called the focal length of the lens. For converging lenses, f is
positive.

A concave lens is just the opposite of a convex lens, thicker at the outer edge than in the
middle. It will cause light rays passing through it to spread out as shown in Fig.2. For diverging
lenses, f is negative.

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 Fig. 2: focal length for convex and concave lens.

The lens equation relates the focal length f of a lens is:

P = L+L′ (8.2)

Where l is initial vergance and l′ is the final vergance

= (8.3)

Where d0, is the distance between object and lens (the object distance) and d1 is the distance
between lens and image (image distance).

The power of a lens is related to the focal length of the lens, f by:-

P = ±n/f (8.4)

where n is the refractive index of the medium that the lens is being used in. In air, the
focal length is just the inverse of the power. A lens has a positive power if it is converging and a
negative power if it is diverging. Powers have units of diopter.

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Method:

1. Put the lens between the source of light and the screen, then move the screen until a

clear image is formed on it.

2. Measure the distance between the object (the source of light) and the lens and then

calculate d0.

3. Measure the distance between the screen and the lens and then calculate d1.

4. Change d0 and d1 several times and then calculate the power.

Results:

d0 d1 1/ d0 1/ d1

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 Exp. (9): Determination of the power of a concave
mirror
Aim:
Determine the focal length and power of a concave mirror.
Theory:
A mirror works by reflecting light that falls on it according to the law of reflection. This law
states that when a ray of light is incident on a surface:

1. The angle of incidence is equal to the angle of reflection ( Ɵi = Ɵr ).

2. The incident ray, the normal and the reflected ray all lie in the same plane.

Reflection off of smooth surfaces such as mirrors or polish metal leads to a type of
reflection known as specular reflection, the reflected rays are parallel to each other. Reflection
off of rough surfaces such as clothing, paper, wood and the asphalt roadway leads to a type of
reflection known as diffuse reflection, light is reflected in random direction.

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 Mirrors can be made into different shapes for different purposes. Two of those shapes are
planes and spheres.

A plane mirror is a flat surface. So it's just a smooth, produce virtual, upright images that
are the same size as the object. A Spherical mirror can be convex or concave, depending on
which side you put the mirrored surface on.

Concave mirror is a spherical mirror where the mirrored surface is on the inside of the
spherical curve. Concave mirrors produce different kinds of images, depending on whether the
object is placed further away from the mirror than the focal point or inside the focal point.

A convex mirror is a spherical mirror where the mirrored surface is on the outside of the
spherical curve. Convex mirrors always produce images that are upright, virtual and smaller than
the object.

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The mirror equation relates the focal length f of a mirror is:

The lens equation relates the focal length f of a lens is:

P = L + L′ (9.1)

Where l is initial vergance and l′ is the final vergance

= (9.2)

Where d0, is the distance between object and mirror (the object distance) and d1 is the distance
between mirror and image (image distance).

Method:

1. Put the mirror in front of the source of light, then move the screen until a clear image is
formed on it.

2. Measure the distance between the object (the source of light) and the mirror and then calculate
d0.

3. Measure the distance between the screen and the mirror and then calculate d1.

4. Change d0 and d1 several times and then calculate power.

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Results:

d0 d1 1/ d0 1/ d1

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 Exp. (10): X-Ray Demonstration
Aim:

 Explained how X-ray image is formed.

Theory

X-rays is a kind of electromagnetic wave, it‟s wave-length ranges from 10‟ 9 mm to 10„
mm, it‟s features include:
(1) Fluorescence effect: When X-rays is shot at fluorescent materials likes ZnS, CdS, Nal,
fluorescence will be produced.
(2) Ionization effect: When X-rays is through gas, gas molecules will be ionized.
(3) No refraction: The refractivity of X-rays is about 1.
(4) Diffraction effect: When X-rays passes a crystal; there will be diffractions, the crystal acts as
a diffraction grating.
(5) Penetration effect: Because of its short wave-length, its photon is of high energy and strong
penetration force.
(6) Biological effect: If X-rays is shot at body, it will harm the cells.
When X-rays penetrates an object, there are absorbing differences which are produced by
the difference in structure and thickness of different part, these differences can be transformed
into image.
The X-rays demonstration instrument is seal constructed; the foregoing characters are
taken into consideration when it is designed, so it is safe to do the experiment.
Structure and principle
The instrument consists of a X-rays tube, a high-voltage power supply and a protective
machine case. The X-rays tube is made up of an anode, a cathode, a tube shell, and an exhaust
pipe, leading-outs.
When a high-voltage is put on the X-rays tube, the X-rays tube emits X-rays shot at the
demonstrating object on the loading glass, there will be an image on the object on the screen, and
one can observe the image on the reflector through the lead-glass window.
Anode: The anode is composed of a target, an anode support and an anode stick, it is sealed and
connected to the glass shell with a stem. It accelerates electrons, stands the shock of high-speed

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electrons, emits X-rays, and transmits most of the heat to outside the tube in order not to burn the
target. The anode is usually made of tungsten or molybdenum, sometimes chromium, nickel or
copper is adopted.
Cathode: The cathode is made up of a cathode head, filament, a cathode case and a stem. The
filament is made of tungsten, it emits thermo-electron.
Shell: The shell is used to guarantee the proper function of X-rays tube; it is vacuumed, and is
usually made of material that allows X-rays to pass, such as glass.
Produce of X-rays
The electrons escaped from the thermic-cathode are accelerated in the electric field to
become high-speed electrons, they run into the anode target, turning kinetic energy into
radioactive rays, one mind of these X-rays is white X-rays, the other is characteristic X-rays.

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