PHYSICS LABORATORY - Marmara Üniversitesidosya.marmara.edu.tr/fef/fzk/PHYS_LAB_III_Experiment... · 2017-02-27 · 8 8 which is called as “final speed”. We may find this speed

T.C. MARMARA UNIVERSITY

FACULTY OF ARTS AND SCIENCES PHYSICS DEPARTMENT

PHYSICS LABORATORY – III

DEPARTMENT: NAME: SURNAME: NUMBER:

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T.C.MARMARA UNIVERSITY PHYSICS DEPARTMENT

PHYSICS LABORATORY –III MANUAL

EXPERIMENT NO:

EXPERIMENT NAME:

THE DATE:

GROUP NO:

NAME:

NUMBER:

DELIVERY TIME:

REPORT NOTE:

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EXPERIMENT.1

DETERMINATION OF SPECIFIC HEAT OF METALS

Goal: Measuring the specific heat of metals.

Theory:

If energy is given to a matter without any work, temperature of the matter generally increases. The energy that needed to increase the temperature of matters with equal amounts changes with matter to matter. For example energy that needed to increase temperature of 1 kg of water to 1 °C is 4186 J. While energy that needed to increase temperature of 1 kg of copper to 1 °C is only 387 J. Heat capacity of a matter, C, is heat energy that needed to increase temperature of the matter to 1 °C. According to this definition, when heat unit of Q is given to an object, temperature of the object changes as ΔT. So:

Q=C ΔT

Heat capacity (c) of any object is defined as heat capacity per unit volume. So if energy (Q) transfers to m massed object via heat, temperature of the object changes as ΔT. So specific heat of the object is:

Q

cm T

Specific heat is a measurement of heat susceptibility of given energy. If specific heat of a matter is bigger, amount of energy that can change temperature difference is bigger too. According to this definition, m is mass of matter, Q is transferred energy and ΔT is temperature difference:

Q=mc ΔT

For example, heat energy that increase temperature of 0.5 kg water as 3°C is:

(0.5 kg)(4186 J/kg°C)(3°C)=6,28 x 310 J. When heat transfers to an object, both of Q and ΔT

are positive signed. Because both of energy of system and temperature of object increase. When temperature decreases, Q and ΔT are negative and system releases energy. Specific heat changes with temperature but if temperature range isn’t so wide, the difference is negligible and c is assumed as constant. For example specific heat of water changes %1 between 0-100°C at atmospheric pressure.

Heat capacity of water is the highest one among other matters. The reason of very low or high temperature of huge massed water is high heat capacitiy of water. When temperature of water decreases in winter, heat is transferred from water to air. So this cause winds that blow from waters to lands.

If a piece of iron is heated as much as 100°C in boiling water and then the iron is cooled in a water filled calorimeter, heat that released from iron is equal to heat that taken by calorimeter and water.

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2 2 1 2 1(100 ) ( ) ( )Fe Fe aqua aqua Cu Cum c T m c T T m c T T (1)

Fem : Mass of iron, kg

aquam : Mass of water in calorimeter, kg

Cum : Mass of internal container(copper)of calorimeter, kg

aquac : Heat capacity of water, 4200 J/kg

Cuc : Heat capacity of copper, 386 J/kg

Fec : Heat capacitiy of iron, ?

1T and 2T : Initial and final temperatures, °C

Experimental Setup:

Apparatus: 1.) Iron 2.) Calorimeter 3.) Thermometer 4.) Stirrer 5.) Weight

EXPERIMENTAL SETUP AND MEASUREMENTS:

Measurements: 1.) Measure mass of iron and heat it up to 100°C with water. 2.) Weigh internal container of calorimeter and then fill 2/3 of the container by water. Weigh it again and record results to Table.1. 3.) Place calorimeter as Figure.1 and measure initial temperature of water. 4.) Put heated iron to calorimeter and stir water. Measure final temperature of water after reaching thermal balance. (Be careful about thermometer mustn’t touch to iron.)

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Figure.1.1 : Calorimeter Calculations: 1.) Record your measurements and calculations to table below. Table.1 : Measurements

Mass of Iron Fem

kg

Mass of internal container(copper)of calorimeter Cum

kg

Mass of internal container of calorimeter + water Cum + aquam

kg

Mass of water in calorimeter aquam

kg

Initial temperature of calorimeter 1T

°C

Final temperature of calorimeter+iron 2T

°C

2.) Find specific heat of iron using formula(1)

Specific heat of iron Fec

J/kg cal/g

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Error Calculation: 1.) Derive dependent error formula using formula (1) and find maximum absolute error of specific heat of iron. 2.) Show the theorical table that include real specific heats. Compare theorical and experimental value of specific heat and calculate percentage error. 3.) Write probable error sources. Conclusions and comments: Write your conclusions and comments about the experiment.

EXPERIMENT 1. DETERMINATION OF SPECIFIC HEAT OF METALS:

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EXPERIMENT.2

VISCOSITY COEFFICIENT OF GLYCERIN

Goal: Measuring the viscosity coefficient of any liquid by Stoke’s Law.

Theory:

Stoke’s Law: If viscous liquid flows around a globe or a globe moves in a still viscous liquid, a

friction force acts on the globe. We will call this friction force. Because of the viscosity of

liquid, the globe moves with a constant speed, .

According to Stoke’s Law, flow speed is in direct proportion to the friction force and

inversely proportional to radius of globe .

where is viscosity coefficient.

Now we consider a viscous liquid with a falling globe inside. If initial speed of globe is zero

then the friction force is also zero at the beginning. There are only gravity and ascending force

acting on the globe.

Gravity:

Ascending Force:

where is mass density of liquid and is mass density of globe. According to the Newton’s

Second Law, we find the initial acceleration of globe.

Net Force:

In addition to gain velocity due to this acceleration, the globe is acted by an additional force

which is expressed by Stoke’s Law. This additional force is nothing but the friction force

which is mentioned above. Since speed increases, the friction force increases proportionally.

Finally, globe gains a limit speed which gives a friction force equal to the net force acting on

the globe. After that, the acceleration wanishes and the globe moves with a constant speed

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which is called as “final speed”. We may find this speed by making the net force equal to the

friction force.

where is viscosity coefficient of liquid and is the thermal speed of the globe.

Apparatus:

1. Glycerin

2. 100ml graduated cylinder

3. Thermometer

4. 5 steal balls in different radius

5. Stopwatch

Experimental Setup:

1. Fill the graduated cylinder with glycerin. Wait for the bubbles wanish.

2. Clean the balls from grease and oil by sopping in sodium hydroxide. (After that hold the

balls with tweezers!)

3. Measure the diameters of balls with your micrometer and calculate each radius. Write

your results in Table 1.

4. Mark the specified levels A and B where balls fall with constant speed.

5. Let balls go one by one from the surface of glycerin. Keep the falling time from A to B.

(three times for each ball)

6. Use your thermometer to take the temperature of glycerin.

Calculations and Analyses:

Quantities:

Mass density of steal ball:

Mass density of glycerin:

Gravitational acceleration:

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Lenght of A-B:

Temperature of glycerin:

Table 1: Falling Time,

Thermal

Speed,

Steal Balls

Viscosity

Coefficients

Diameter,

Radius,

1. Calculate the thermal speeds and write these down into related columb of Table 1

2. Calculate the viscosity coefficients etc. and get the aritmetic mean of these

quantities.

3. Draw the graph according to Table 1 and calculate the slop of the graph.

4. Calculate the viscosity coefficient of glycerin at the temperature T with this formula

5. Calculate the viscosity coefficient of glyserin at

6. Compare your results with the accepted values in other literatures.

Error Calculation:

1. Calculate your percentage error for .

2. Derive the relative error formula from .

3. Calculate the maximum absolute error for each measure.

4. Calculate the average standart deviation with

5. Itemize the possible causes of error.

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Results and Comments:

Write down your results and interpret your conclusions in accordance with your information you gain in this experiment

EXPERIMENT 2. VISCOSITY COEFFICIENT OF GLYCERIN

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EXPERIMENT.3

THIN EDGES LENS

Goal: Finding focus of thin edges lens

Theory:

Transparent media at least one surface of it is spherical is called lens. If edges according to the middle of the lens are thin, it is called thin edges lens. Otherwise edges according to the middle of the lens are thick, it is called thick edges lens. Some definitions on lens are described in the figure 1.

Figure 3.1

f; focus of lens, do and di; the distance of image and object from the lens, ho and hi heights of image and object, respectively. Between these quantities there are two relations;

Several do-di values pair can be obtained by changing the location of object. If graph 1/do against to 1/di is drawn, points intersect the horizontal and vertical axes of the graph give the values of f.

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Figure 3.2. Graph 1/do against to 1/di

Experimental Setup:

Apparatus: 1.) Light source 2.) Thin edges lens 3.) Screen 4.) Ruler 5.) Paper tape EXPERIMENTAL SETUP AND MEASUREMENTS:

1.) A long strip of paper is placed on table.

2.) Keep away candle from the lens as it could be. And mark the location of the clear image and the object. Then write them to table 1.

3.) Change the location of the thin edges lens and write do-di values to table 1 by getting a clear image again on the screen.

4.) Repeat this process at least ten times.

Measurements: n do(cm) di(cm) 1/ do (1/cm) 1/ di (1/cm) (di/do)=m f (cm)

1 2 3 4 5 6 7 8 9

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Calculations: 1.) By using formula (1), calculate the focus of thin edges lens mathematically that used in this experiment. 2.) Use ten different measuring results, plot 1/do against to 1/di and find the focus of thin edges lens by using graph intersects the axes. 3.) Do the error calculation by comparing theoretical and experimental value of focus. 4.) Calculate the maximum relative error in any measurement from formula (1). 5.) Find the average standard deviation by using calculated f values. 6.) Write error causes in order. 7.) Write the results and comments about experiment via obtained data. 8.) Interpret your conclusions.

EXPERIMENT 3. THIN EDGES LENS

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EXPERIMENT.4

REFRACTION OF LIGHT

Goal: Observing of refraction of light in different medias and calculate the index of refraction.

Theory:

Figure 4.1.The energy spectrum and visible light region

The light beam which is called as “white light” includes different wavelenght of light beams and in the visible spectrum people can observe between 400 and 700 nm wavelenght. This is because the radiation from sun included photons with a large variety of energy values.

Figure 4.2. Beams of sun light

Light travels in straight lines until it encounters another material where it is partially reflected and partially transmitted. The light travels until it meets with a surface and then it is reflected, transmited or absorbed.

After light hits to the surface it has two components : reflected and transmitted. The angle of incidence is equal to the angle of reflection and the angles do not depend on the nature of the material. In refraction the angle of the ray when transmitted through the material changes and depends on the speed of light in the two materials.

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The ratio of how we can illuminate a material depends of the photons that are hit the surface. If the surface absorbs the light totally that means we may not see the material because there will be no reflected light.

Sometimes material can only absorbes some waveleghts so that waveleght will not be reflected. Think about the Sun light which shines the “grass”. After light beam hits the leafs of the grass only green light is reflected and all others are absorbed.

Some materials are transparent. These ones let the light beam transfer inside and light beam will continue its way without losing its energy (maybe just a little energy lose as heat). Thats how the window pane works.

Nature of Light:

Light velocity in vacuum is 300; 000 km/s or 3.108

m/s. Nothing can move faster than this value (nonrelativistically).

Figure 4.3. incident ,refracted and critical angle

n1sin (θ1) = n2 sin(θ2)

θ1 is the angle between the incoming (incident) light and the Normal and θ2 is the angle between

the outgoing (refracted) light and the Normal. When a light beam comes from a medium with a

higher index of refraction and goes to a medium with smaller index of refraction, the beam will be

move away from the Normal and the refracted angle will be bigger than incident angle. But the

refracted angle can not be bigger than 900

. If it is, the light will not transfer to second medium and

will move inside the first one . There will be a maximum limit on the incident angle so as to observe

refraction which is called critical angle θc and above this value the light will not transfer to second

medium . That is called total internal reflection.

Experimental Setup:

Apparatus: 1.) Light source 2.) Materials with different index of refraction 3.) graded angle sheet (attached on the lab. sheet) 4.) Ruler, protractor 5.) Calculator

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EXPERIMENTAL SETUP AND MEASUREMENTS:

1. Place the material on the sheet and use point source and focus it on the origin.

Figure.4.1

2. Use the incident angle values which are given in the table and adjust your point source to these angles. Then one by one measure outgoing angle and fill the table.

3. Draw a graph between the incoming and outgoing angles. What it tells you ?

4. Draw a graph between the sin values of incoming and outgoing angles. What it tells you

5. Incident index of refraction belongs to air. Use the formulation and calculate the index of refraction of unknown material.

6. Do the error calculation by comparing theoretical and experimental value of index of

refractions.

7. Calculate the maximum relative error in any measurements of refractive index.

8. Find the average standard deviation by using calculated n values.

9. Write error causes in order.

10. Write the results and comments about experiment via obtained data.

11. ) Interpret your conclusions.

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EXPERIMENT.4 REFRACTION OF LIGHT

Angle

(θ0) sin(θ0) sin(θi) n0 ni

1 10

2 20 3 30

4 40 5 50 6 60

7 70 8 80

9 90

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EXPERIMENT.5

LIGHT THROUGH A PARALLEL SIDED BLOCK

Goal: Observing of refraction of light in parallel sided block and measuring amount of shift.

Theory:

If a parallel sided block which has 2n refraction index in an ambient that has 1n refraction

index, incident ray that has 1i angle between normal, refracts two times as when it enters and

exits from matter.

Figure 5.1 Snell law for incoming and outgoing rays:

For incoming: 1 1 2 2sin sinn i n i

For outgoing: 2 3 1 4sin sinn i n i

Using these two equations above, we can easily find that 1 4sin sini i and so 1 4i i . This equation

shows that incoming and outgoing rays are parallel to each other. If we want to find amount of parallel shift(d) using trigonometric equations;

2cosL

iAB

, and 1 2sin( )d

i iAB

.

1 2 1 2

2

sin( ) sin( )cos

Ld AB i i i i

iand finally the amount of parallel shift is:

1 1

1

sin( )

cos

i rd L

r (1)

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Experimental Setup:

Apparatus: 1.) Light source 2.) Parallel sided glass block 3.) Millimetric paper 4.) Goniometer EXPERIMENTAL SETUP AND MEASUREMENTS:

Measurements: 1.) Draw three figures that illustrate shape of parallel sided glass block on millimetric paper. 2.) Set incoming angles as 30,50 and 70. 3.) Draw way of outgoing ray. 4.) Measure refraction angles using goniometer and check that your values obtain snell law. 5.) Calculate amount of parallel shift (d) using formula (1).

1 ......L cm

N

1i 2i 3i 4i expd (cm) theod (cm)

1 30 ° 2 50 ° 3 70 °

2 ......L cm

N


4 30 ° 5 50 ° 6 70 °

3 ......L cm

N


7 30 ° 8 50 ° 9 70 °

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Error Calculation: 1.) Compare theorical and experimental value of parallel shift and calculate percentage error. 2.) Find maximum absolute error of any of your measurement. 3.) Write probable error sources. Conclusions and comments: Write your conclusions and comments about the experiment.

EXPERIMENT.5 LIGHT THROUGH A PARALLEL SIDED BLOCK:

Rentech Experimental Set of Fundamental Optics

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EXPERIMENT 5: Experiment of Polarization

Part 1: Polarizer and Screen

1. Light source, polarizer and screen are placed

on optics track.

1.1. Distance between light source and

polarizer is adjusted to;

.

Distance between screen and polarizer is

adjusted to;

.

1.2. White pointer on polarizer is adjusted to;

.

1.3. Light source is opened and image is

formed on screen (Figure-41a).

1.4. Polarizer is removed from optics track and

whether intensity of illumination of the

image changes or not is observed. If it

changes, its reason is explained.

2. Now, second polarizer (analyzer) is placed

behind polarizer (Figure-41b).

2.1. Distance between analyzer and polarizer is

adjusted to;

.

2.2. Firstly, white pointer on analyzer is

adjusted to;

.

2.3. Next, the angle of analyzer is set as

in condition that the angle of

polarizer is kept unchanged ( )

(Figure-41c).

2.4. After this operation, the angle between

axes of first polarizer and second polarizer

(analyzer) will be;

.

2.5. Change of intensity of illuminance of the

image on screen is monitored.

3. Same steps are repeated, changing the angle

of analyzer as increase of 30° between

.

3.1. According to each change of angle, how

intensity of illuminance of the image on

screen changes is observed.

(a)

(b)

(c)

Figure 41: Intensity of light of the image on screen of light

passing from polarizer (a), intensity of light of the image

on screen of light passing from analyzer (b), and the

angle of analyzer with respect to the axis of polarizer (c).


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3.2. When difference of angle between polarizer and

analyzer is , whether image is formed on

screen is determined. If there is no any image on

screen, its reason is explained on experimental

report.

Two polarizers are used in experimental set-up. Light

passes from these polarizers through plane marked

0° and 180° on scale of polarizer. As a result of this,

light passing from first polarizer is linearly polarized.

If direction of second polarizer is same as direction of

vibration of linearly polarized light, that light passes is

observed. If the angle between these two directions

is 90°, that light does not pass is seen.

4. What is polarization?

If components of an electromagnetic wave vibrate in

different directions, phenomenon of reduction of this

vibration to one direction is called “polarization”.

5. What is direction of polarization?

Direction of vibration and direction of propagation of

electric field component of an electromagnetic wave

form plane of polarization, direction of vibration of this

electric field component is called direction of

polarization. Direction of vibration of electric field

vector is chosen for direction of polarization of light

wave.

6. The angle between axes of polarizer and

analyzer is . How does change intensity of

light passing from analyzer with change of

angle ?.

When the axes of polarizer and analyzer are parallel

with each other ( or ), intensity of light

passing from analyzer becomes maximum, in the

case of that the axes are perpendicular to each other

( or ), intensity of light becomes

minimum (zero).


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There are two types of waves, depending on

direction of vibration (shape of propagation). They

are transverse and longitudinal waves. Waves of

which direction of propagation and direction of

vibration are perpendicular to each other are

transverse waves.

Electromagnetic waves are transverse waves.

Waves of which direction of propagation and

direction of vibration are parallel with each other are

longitudinal waves. Sound waves are longitudinal

waves.

Displacement per time of waves is velocity of

waves. Velocity of a wave depends on frequency of

wave and wavelength. It is given by the following

relation:

Here,

In light waves, electric and magnetic field vector are

perpendicular to each other and direction of

propagation. Electric field (E) and magnetic field (B)

of electromagnetic wave change as perpendicular to

each other. In addition, electric and magnetic field

vibrate in same phase. In other words, electric field

and magnetic field increase or decrease together.

Velocity of propagation of electromagnetic waves is

perpendicular to these two fields.

Electromagnetic waves have a feature of

polarization. If components of a wave vibrate in

different directions, phenomenon of reduction to

only one direction is called polarization. Direction of

polarization of an electromagnetic wave is taken at

direction of electric field. If electric field vibrates at

only one direction, it is expressed as linearly

polarized. Direction of vibration of electromagnetic

wave’s electric field is direction of polarization

(Figure-28a).

6. Polarization

(a)

(b)

(c)

Figure-28: Vibration of electric field vector in non-polarized

light and in polarized light (a), polarized light that has angle

θ between its axis and vertical axis is polarized after it

passes from polarizer (b), when axis of polarizer is

perpendicular to axis of analyzer, analyzer does not pass

light (c).

fv (Experimental) (48)

)/( smv : Velocity of wave.

)(m : Wavelength.

)(Hzf : Frequency.


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If the angle between linearly polarized wave and

polarizer is (Figure-28b), magnitude of electric

field vector of wave is given below:

cos0EE (49)

If intensity of light of polarized light is , intensity of

light passing from polarizer will be as below:

2

0 cosII (50)

Here,

:0I Intensity of light.

: The angle between linearly polarized wave

(incident light) and axis of filter.

When non-polarized light passing from a polarizer of

which polarized axis is vertical, light becomes

polarized as vertical. Therefore, polarizer allows

passing only vertical component of incident light. In

the case of use of analyzer, if polarized axis of this

analyzer and axis of polarizer are perpendicular to

each other, light do not pass (Figure-28c).

Two methods are used for polarization of visible

light.

1. To pass light from polarizer,

2. Polarization by means of reflection.


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6.1. Polarizer

Material which selects an electromagnetic wave

vibrating only certain direction from non-polarized

light and obtains polarized light is called polarizer.

When polarizer is used for polarization of visible

light, polarizer allows passing a major part of

incident light in parallel with axis of polarization. In

other words, it allows passing a part of light in

direction of electric field. As a result of this, intensity

of light reduces in half.

When non-polarized light is passed from polarizer,

second polarize is needed to understand whether

light beam is polarized.

When we turn second polarizer by a value between

0° and 180°, light is polarized if intensity of light

changes between maximum and minimum (zero).

Second polarizer used for this purpose is called

analyzer. If intensity of light transmitted to analyzer

is zero at certain position, light is polarized with

%100. Moreover, if intensity of light reaches to

minimum value, light is partially polarized.

As an example, in system of two polarizers, axis of

the first polarizer is vertical and the angle between

axis of the other polarizer and vertical line is 60°. To

find intensity of light passing from analyzer, firstly it

is needed to determine intensity of light passing

from polarizer. Intensity of light passing from

polarizer is a half of intensity of incident light.

Therefore, after non-polarized pass from first

polarizer, its intensity reduces to half (Figure-29a).

012

1II

After light polarized as vertical pass from second

polarizer (analyzer), its intensity decrease:

20

12 )60(cosII

124

1II

As a result, it is found that;

028

1II

As seen, intensity of light passing from analyzer ( )

is equal to 1/8 of intensity of non-polarized light and

polarization occurs at (Figure-29b).

(a)

(b)

Figure-29: Non-polarized light has two equal intensity of

light as horizontal and vertical components. Intensity of

light reduces in half after passing from polarizer (a). A part

of polarized light passes from analyzer (b).


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EXPERIMENT 8: Single Slit Diffraction

Figure-45: Experimental set-up for single slit diffraction (a)

and position of central bright fringe at diffraction pattern

(b).

1. Laser source, single slit plate and scaled

screen are placed on optics track (Figure-

45a).

1.1. In experiment of single slit diffraction, slit of

which width is smallest is chosen.

1.2. Wavelength of laser source used in

experiment is .

1.3. Rectangular slit is put on its holder horizontally

(or vertically) and it is placed in front of laser

source by means of its holder so that they are

close to each other.

1.4. Distance between slit and screen is adjusted

to;

.

This value is noted as distance of screen.

2. Laser source is opened.

2.1. Diffraction pattern formed by laser source, i.e.

bright and dark fringes, is observed on scaled

screen.

2.2. Central bright fringe’s width and brightness is

observed. (Scaled screen used in

experiment has latitudinal and longitudinal

metric ruler.)

If a monochromatic light beam of which wavelength

is λ pass from a single slit of which width is D,

central bright fringe is seen on screen. Intensity of

light of bright fringes on screen declines as it is

approached to sides of screen. So, it is seen that

intensity of light is maximum at central bright fringe

and decreases at sides of screen.

3. At single slit diffraction pattern, distance (x)

between center of central bright fringe and

center of first dark fringe (m=1) is measured

(Figure-45b).

(a)

(b)

Please, don’t look at laser beam directly.


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3.1. Slit width is calculated, using wavelength of

laser ( ), distance between slit plate

and screen ( and measured

distance (x):

4. Experiment is repeated for different slit

width and different screen distance.

5. Why minimum and maximum intensity of

light form at diffraction pattern is explained.

6. Which condition should be satisfied for

occurring diffraction at light wave?.

When slit width is equal to or smaller than

wavelength of light passing from slit, diffraction

phenomenon occurs.

7. How does fringe width ( ) change when

slit width is decreased at single slit

diffraction?.

Fringe width on screen increases.

8. What is relation of fringe width ( )?.

9. If wavelength of light (λ) used at single slit

diffraction is increased, How does fringe

width on screen ( ) change?.

If wavelength of light used at single slit

diffraction is increased, fringe width on diffraction

pattern increases.

D

mLxm


31

Figure-32: Single slit diffraction (a), formation of diffraction

pattern (b), and change of intensity of light at diffraction

pattern (c).

After waves pass from side of an obstacle or a

narrow gap, it changes its direction. This

phenomenon is called diffraction. While

monochromatic light beam incident on single slit

pass from slit gap, it propagates through every

direction as light waves. When these waves

interfere with each other, some waves destruct with

each other by causing that region is dark whereas

some waves construct with each other and bright

regions emerge.

In Figure-(32a), a monochromatic light beam with a

wavelength “λ” incident on a slit with width “D” is

shown. A monochromatic light beam from gap of

which width is “D” meets at point N on the screen at

the distance L. difference between beam passing

from lower side of the slit and beam passing from

the center will be λ/2. Since distance difference to

point N is λ/2, waves from slit interfere destructively

at point N and they emerge dark (minima) fringe.

For dark fringe, the following relation is found:

2/

2/sin

D

(61)

D

sin (62)

Intensity of light is maximum at θ = 0° and minimum

at given angle θ.

Bright (maxima) fringe at point N;

Single slit diffraction pattern formed by light passing

from slit and incident on screen is shown in Figure -

32b. Bright fringes formed at diffraction pattern

weaken through sides of screen. Figure-32c

indicates that distribution of intensity of light as a

function of value of sinθ.

7. Diffraction of Light Wave

(a)

(b)

(c)

mD sin m=1, 2, 3, … (63)

)2

1(sin mD m=1, 2, 3, … (64)


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Here, angle θ determines position of a point on the

screen.

If single slit diffraction is investigated;

1. If << D, then diffraction does not occur.

While a wave travels on linear path, bright

point is seen on screen.

2. If D, it is started to observe diffraction.

Light propagates at all directions after it

passes from slit.

3. If D << , diffraction is observed clearly.

Light beam incident on single slit brings

about many point wave sources at slit

separation. Therefore, light beam from

different points on slit separation interfere

constructively and destructively and so

diffraction pattern is formed.

Intensity of light of bright fringes at diffraction pattern

decreases as sides of screen is gone (Figure-32b).

Figure-(32c) gives distribution of light intensity of

bright fringes as a function of sinθ. Here, angle θ is

angle that determines position of a point on screen.

Figure-33: Distance between a dark fringe formed at

single slit diffraction pattern and central bright fringe.

If at single slit diffraction (Figure-33), distance

between dark fringe and central bright fringe

is , the following relation is written:

Here;

In this equality, distance between central bright fringe

and first dark fringe ( ) is wanted to find, m will be 1.

Fringe width (Δx) is distance between two sequent

bright or two sequent dark fringe. Wavelength is

found in terms of fringe width as below;

L

xD (66)

D

mLxm

(Experimental) (65)

)(mxm: Distance between dark fringe and central

bright fringe.

)(mL : Distance between slit plane and screen.

)(m : Wavelength of light source.

)(mD Width of slit


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As an example, a laser beam with λ=633nm used in

the experiment of single slit and diffraction pattern

emerges on screen at distance L=6 m. On screen,

distance between centers of first dark fringes except

for central bright fringe is calculated as 32 mm (see

Figure-34).

If slit width is calculated from these data;

m=1 is taken for distance between central

bright fringe and first dark fringe.

Distance between central bright fringe and

first dark fringe is;

.

From here, it is found that;

D

mLxm

DLx

1

m

mm

xLD

3

9

1 1016

)10633)(6(

mmmD 24.0104.2 4

Figure 34: Diffraction pattern of beams passing from single

slit on screen.

Single Slit

Wavelength Distance of

screen

Distance of

first dark

fringe

Slit width

(nm) L (m) x1 (m) D(mm)

633 6 16x10-3

. . . . .

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PHYSICS LABORATORY - Marmara Üniversitesidosya.marmara.edu.tr/fef/fzk/PHYS_LAB_III_Experiment... · 2017-02-27 · 8 8 which is called as “final speed”. We may find this speed