Physics effects

The teflon film test

noreply@blogger.com (Dr Gouri Goutam Borthakur) — Wed, 19 Jun 2013 22:09:00 +0000

This is a test fo purity of water.If water in the drop is not pure, its contact angle over Teflon surface will be less than 110 degree.
This can be a student project: Can we check the purity of milk and muster oil using this non destructive technique ?
Reference :
1. www.uweb.engr.washington.edu/education/pdf/ASHsurfscontact angles05.pdf

Human body resistance

noreply@blogger.com (Dr Gouri Goutam Borthakur) — Sun, 12 May 2013 22:42:00 +0000

Resistance: Positive or negative?
Ohms law states that
Voltage = (+Resistance )x Current
Compare this with the eqn for st line
y= mx+c; if y:voltage, then m:Resistance and x: current, c: intercept which is zero
Meaning that more the voltage, more the current will be.(As in the case with passive devices like resistors, junction diodes)
If Resistance is negative, then the reverse will happen : current will increase, even when voltage decreases implying that some latent power generation mechanism is working behind(as in the case with active devices like Gunn diodes and tunnel diodes)

Resistance of Human body:
There are a lot of factors involved and not every person has the same electrical resistance.But what ever value it may have, the value is always positive, as the human body is a electrically passive system(Dissipates current passing through it, and doesnot generate extra current) .For instance, men tend to have lower resistance than women. Just like for the resistors used in electronics, the resistance of a person's arm depends on the arm's length and diameter. Resistance goes up with length and down with diameter. Since men tend to have thicker arms and legs (more muscle), they usually have lower resistance. (An implication of this is that the lethal current for men is higher than that for women.) A rough value for the internal resistance of the human body is 300-1,000 Ohms. Naturally, the resistance also depends on the path that electricity takes through the body - if the electricity goes in the left hand and out the right foot, then the resistance will be much higher than if it goes in and out of adjacent fingers.

Within the body, the tissues with the greatest resistance are bone and fat - nerves and muscle have the least resistance. That said, the majority of the body's resistance is in the skin - the dead, dry cells of the epidermis (the skin's outer layer) are very poor conductors. Depending on the person, the resistance of dry skin is usually between 1,000-100,000 Ohms. The skin's resistance is much lower if it is wet or burnt/blistered. This means that when a person is electrocuted in real life, the body's resistance drops as the skin is burned. To determine a person's total resistance, just add together the resistance of each part of the body - remember that the electricity must pass through the skin twice (on the way in and on the way out), so the total resistance is:

R_total = R_skin(in) + R_internal + R_skin(out)

Another interesting point to consider is that in addition to acting like a resistor, the epidermis acts like a capacitor if placed in contact with a piece of metal (the underlying tissue is like one plate of a capacitor and the metal surface is like the other plate - the dry epidermis is the less-conductive material or "dielectric" in between) . In cases of electrocution by a DC voltage source, this capacitive property has little importance. But if the electrocution is by an AC source, the epidermis's natural resistance is "shorted out", allowing the current to bypass that part of the body's resistance and making the body's total resistance much lower.

Reference:
1.R. Fish & L. Geddes, Medical and Bioengineering Aspects of Electrical Injuries, c2003 Lawyers & Judges Publishing Company, Inc.
2. http://van.physics.illinois.edu/qa/listing.php?id=6793

Superposition theorm

noreply@blogger.com (Dr Gouri Goutam Borthakur) — Sun, 12 May 2013 07:18:00 +0000

According to this Network theorm,

•To calculate voltage (or current) in any branch

–Replace all other independent voltage sources by short circuits

–Replace other independent current sources by open circuit.

–add all the individual voltages or currents caused by each independent source acting alone

•Note : An ideal voltage source is replaced by a short circuit. An ideal current source is replaced with an open circuit. If the sources are non-ideal ones, their effective resistances orimpedances are placed instead.

Important considerations in applying superposition theorm

• Superposition theorem is not applicable to power.

if V = V1 + V2 I = I1 + I2

VI ¹ V1 I1 + V2 I2

• This theorem is equally applicable to DC and AC.

• This theorem doesn’t hold good when we have just dependent sources.

• While applying, dependent sources will neither be changed to short or open

circuit.

Proof of Superposition theorm

Case 1 :

Case 2:

USing mesh analysis, we will arrive at the same value if current, as shown below. which will be same as calculated by superposition theorem also

Diffraction effect

noreply@blogger.com (Dr Gouri Goutam Borthakur) — Wed, 23 Jan 2013 09:33:00 +0000

Diffration : Sand ball analogy
Consider a sand ball thrown into a open window. If the window size is large in comparison to the dia of the sand ball, the ball will pass undisturbed through the window.If you mount a board so that the ball hits it ofter passing through the window, you will get a single trace on the board.
But, what if the window size is small, comparable to the dia of the ball? Most probably, one or both ends of the ball touch the corners of the window, and eventually those ends of the ball will break apart.So in the board behind the window, you will get a main trace in addition to some small fractions on the board.The pattern of these small traces on the board will depend primarily upon the nature of the window frame.
Now, in the above picture,replace the sand ball by a light particle and the window by a slit, the board by a screen.The light quanta will break and will yeild a main trace along with some small traces and the pattern of distribution of these traces on the screen will depend upon the shape and size of the slit.

Diffraction is actually Interference !

Diffraction can be explained technically in terms of interference of light.So lets start from the youngs DS experiment. S1 and s2 are narrow slits placed close together acting as coherent sources.Let us assume that waves from S1 reaching a particular point P in screen be represented by y1 while Waves reaching same point from S2 be represented by y2
y1 = a1* Sin(wt)
y2 = a2*sin(wt+f)
f is the phase difference at point P.At P , these two waves superpose
y = y1 + y2 = A*Sin(wt +F)
A is the amplitude and F is the phase of the resultant wave at P.
A=√(a1²+a2²+2*a1*a2*cos(f))
F= tan^-1(a2*Sin(f)/(a1+a2*Cos(f)))
A will vary with phase angle f, which in turn will vary with the distance of P from C.At some point A will be maximum(a1+a2), ahile at some nearby points A will be minimum, (a1~a2). Thus as we move along the direction PC, A will keep alternating between Maximim and Minimum values. The points of maximim internsity are called bright fringes while the points of minimum intensity are called dark fringes.here the fringes are of equal width.This pehnomena is known as interference.Here there is a redistribution of light energy on the screen due to two coherent sources placed close together.The width of the dark and bright bands (called fringes) depend on the following factors
1. Separation between the sources d
2. Separation between the source and screen D
3.Wavelength of light L

Now if the sources are brought infinitesimally close together, will there be interference? Yes, there will be.
If more than two sources are placed infinitesimally close together, then ? Ya, of course, there will be interference, but the fringe pattern will be different.What will happen if such infinite number of infinitesimally thin coherent sources are placed close together, infinitesimal distance apart?There will be interference in this case also, but this last situation resembles a beam of light passing through a not-so-narrow slit.And the fringe pattern that will be observed in the last case will be called a fringe pattern due to diffraction of light through the not so narrow slit.
The fringe pattern will depend uponWidth of the slit , say a, and not on the separation between the individual coherent sources. (the latter being infinitesimally small, becomes insignificant) and Wavelength of light. Unlike Interference fringes, The Diffraction fringes are not of equal width, neither of equal internsity and all unequally spaced.

Echelon effect

noreply@blogger.com (Dr Gouri Goutam Borthakur) — Sun, 16 Dec 2012 15:41:00 +0000

The effect is an important factor affecting the acoustics of buildings. This effect is the production of a musical sound of a definite frequency due to reflection of sound from regular structures such as stairs, railings, palings existent inside a auditorium.This occurs due to successive echoes from these regular structures reaching the listener. If the frequency of this musical note is within audible range,. the listner will only hear the sound of this frequency prominently.
The number of equidistant and parallel steps/palings form a grating-like system. When a single
sound pulse falls on the system, it is broken up into a number of pulses by successive reflections from
the steps/palings. Since these steps/palings are equidistant from one another, the portion of the wave
striking each step/paling and getting reflected there travels a little farther and arrives a little later at the
listeners’ ear, than the portion striking the previous step/paling. The succession of pulses at regular
intervals is heard as a short musical note of definite frequency. This is called echelon effect. If d is the

distance between two successive steps/palings, v, the speed of sound and q, the angle of deviation of the reflected pulses, the period of the echo i.e., the time interval between two pulses will be (2dcos q/v) reflected pulses, the period of the echo i.e., the time interval between two pulses will be(v/2dcosq). For normal incidence the frequency will be (v/2d).
This echelon effect can be overcome by spreading a carpet over the steps and/or by making the
spacing of steps irregular. Similarly by making the spacing of palings irregular on ceiling, echelon
effect can be minimised.

Reference :
1. N Subrahmanyam & Brij Lal, Waves and Oscillations (2nd Ed), Vikas Publishing House (2001) , pp348
2. http://www.newagepublishers.com/samplechapter/001627.pdf

A subtle difference between relativistic particle and a non relativistic particle

noreply@blogger.com (Dr Gouri Goutam Borthakur) — Tue, 04 Dec 2012 15:40:00 +0000

How we derive the energy momntum relation of relativistic particle ?
We first use the mathematical relation representing relativistic variation of mass and then use the definition of particle momentum (p=mv) to substitute for m and v/c in einsteins mass energy relation to obtain the expression for the total enegy of a relativistic particle[1].
Momentum of a relativistic particle increases due to increase in its velocity.We expect the energy of the particle to change with change in momentum. But, strikingly enough, it doesnot, as is evident from the graph produced by the following code(plot shown below).The total energy of a relativistic particle remains independent of the momentum.
The reason for this independence is the relativistic variation of mass(the other term in the expression for Energy).

On the other hand, what is the nature of variation of Energy with momentum for a non relativistic particle. This is parabolic , as we all know.

Now compare (p,E )graph of a relativistic particle with that of a non -relativistic one.The former is a straight line parallel to the momentum axis(the red curve), while the latter is a parabola (the blue curve). This is a very subtle difference between a relativistic particle and a non-relativistic particle.
The following matlab code plots total energy of a relativistic particle as well as a non relativistic partivle as a function of its momentum.The graphs shown here are produced with the codes.
*********************************
% Relativistic energy momentum relation
m = 1.0 ; % mass of particle
c =1; % velocity of light
v = 0.1:0.05 :0.9; % velocity of prticle
vs =v.*v ;
p=m*v ; % momentum
ps= p.*p
cs=c*c
vbc =v./c ;
vbcs = vbc.*vbc ;
beta = sqrt(1-vbcs)
mo = m*beta ; %rest mass
mos = mo.*mo
E_rel = sqrt (ps*cs + mos.* cs);
plot(p, E_rel, 'rp:')
E= 0.5*m*vs;
plot (p, E, 'bh-');
*****************************************
Reference :
1. Rajput B S, Singh M P, Physics for Engineers(2nd Ed), Vol 1 ,p 37, Pragati prakashan (2010)

Fibre optics

noreply@blogger.com (Dr Gouri Goutam Borthakur) — Fri, 16 Nov 2012 14:33:00 +0000

Introduction
Since optical frequencies are extremely large in comparison to radio waves and microwaves, signals transmitted through a light wave is capable of carrying more information than conventional radio-waves and microwaves.In the near future, the demand for information will be so high that only a technology based on data transmission vide light waves will be able to meet the demand.Indeed, British Telecom in 1984 stopped laying copper cables and started using Fiber optic cables on its long distance routes.
Advantages of fiber optic cables over metallic cables
1. Higher Bandwidth
2. Less susceptible to interference
3. Much thinner and lighter
4. Allows digital transmission
5. Low attenuation(0.2 dB/km) , as a consequence of which distance between two consecutive repeaters could be as large as 250 km.In comparision , copper cables have repeater stations every few kilometers apart.
Fiber optic cable types :
1. Single mode step index fiber
3. Multimode step index fiber
3. Multimode graded index fiber
Difference between fiber cable types
In single mode step index optical fibers, core dia is ~ 10um, and the refractive index chnages from low RI of the core to high RI of the cladding at the core cladding interface.These cables allow only single mode of the signal to propagate through them.In addition these cables are non dispersive, and offer low loss and higher BW.They are suitable for long distance transmissions, such as in WANs.
On the other hand, multimode cables have five times larger diameter, starting at 50um and they allow over 100 modes to propagate through them. These cables are dispersive, and lower in BW offer high loss. So they are suitable for data sharing in SAN or LANs.The only construction difference between multimode step index fiber and single mode step index fiber is the size of the core diameter.But in the multimode graded index fiber, RI index profile changes in a parabolic manner from the centre to the periphery of the core.This compensates the dispersion effects in MMSI fibers and enhances their BW.
Nature has already used this technology
It is interesting to note that the retina of human eye consists of a large number of rods and cons which have the same kind of structure as the optical fiber. They consists of dielectric cylindrical rods sourrended by another dielectric of slightly lower refractive index. The core diameters are in the range of a few microns. However, the technology is little more advanced, as the light absorbed in these light guides generates electrical signal, which are then transmitted to the brain through various nerves.[1]

reference :
1.Optics(2nd Ed), AjoyGhatak, P655, Tata McGraw Hill (2003)

Appreciation of Arvind Gupta Toys

noreply@blogger.com (Dr Gouri Goutam Borthakur) — Tue, 10 Jul 2012 13:12:00 +0000

Arvind Gupta is an Indian Icon of science popularization. He has designed a house of science demonstration toys (Teaching learning material ) that can be prepared from everyday stuff.These toys can be of much use if a teacher wants to demonstrate the physical phenomena to students.In the table below, I am trying to Tabulate the usefulness of these toys.
I would be glad to know if any one comes with some school project or summer project idea that is inspired by these toys. There fore I have included a further reading tab in the table.
Further, In case, I am able to write a school science project on these toys, I will include them in a new column "My School Projects(MSP)"

Toys	Usefulness	Further reading	MSP
Sudarshan Chakra	To demonstrate Centripetal force	URL	MSP

Mpemba Effect

noreply@blogger.com (Dr Gouri Goutam Borthakur) — Fri, 06 Jul 2012 07:04:00 +0000

Introduction :

Its a counter-intuitive effect exhibited by water, whereby one can observe that, when two equal volumes of water, one slightly warmer than the other, are placed in similar external conditions to freeze, the warmer water is noticed to freeze before the other. This phenomenon is called ‘mpemba effect’ (named after Erasto B. Mpemba, who noticed this phenomenon first, in modern times)

Explanation :

Water usually super-cools at 0°C and only begins freezing below this temperature. The freezing point is governed by impurities in the water that seed ice crystal formation. Impurities such as dust, bacteria, and dissolved salts all have a characteristic nucleation temperature, and when several are present the freezing point is determined by the one with the highest nucleation temperature.

What would happen when two water samples at the same temperature and placed them in a freezer? One sample would usually freeze before the other, presumably because of a slightly different mix of impurities. On removing the samples from the freezer, and then rewarming one sample to room temperature and the other to 80°C and then on freezing them again, it is observed that if the difference in freezing point was at least 5°C, the one with the highest freezing point always froze before the other if it was heated to 80°C and then re-frozen.So, One must take into account the Newtons Law of cooling here .Hot water cools faster because of the bigger difference in temperature between the water and the freezer, and this helps it reach its freezing point before the cold water reaches its natural freezing point, which is at least 5°C lower.

Thus there is a race among the samples to cool faster in a freezer. According to Newtons law of cooling, the hotter sample wins this race.As the two samples reach the vicinity of the freezing point, there again a race among the samples to freeze faster, in which, the sample containing impurities with higher nucleation temperature has the advantage.

Conclusion :
Since it is so difficult to determine the causes of the effect, any applications of the effect are far off in the future. Until a effective experiment can be performed we will never know if the causes of the Mpemba effect are in fact trivial or are instead of a great importance and will help us further to understand our universe.

References :

1. http://phys.org/news188801988.html

2.http://pisci.wikispaces.com/Mpemba+Effect

NGPE 1996 question paper analysis

noreply@blogger.com (Dr Gouri Goutam Borthakur) — Sun, 10 Jun 2012 05:29:00 +0000

I have analysed NGPE 1996 question paper, and from the analysis I saw that following are the topics tested in the question paper
PART -A1
Significant figures
Artificial radioactivity
Atwood Machine
MEchanical

A science toy to demonstrate convection

noreply@blogger.com (Dr Gouri Goutam Borthakur) — Sun, 10 Jun 2012 02:59:00 +0000

What is convection
When heat energy given to the particles of a medium are transferred to the other particles of the same medium by actual movement of the particles, the phenomenon is known as convection.
Examples in nature

Convection occurs on a large scale in atmospheres, oceans, planetary mantles, and it provides the mechanism of heat transfer for a large fraction of the outermost interiors of our sun and all stars. Fluid movement during convection may be invisibly slow, or it may be obvious and rapid, as in a hurricane. On astronomical scales, convection of gas and dust is thought to occur in the accretion disks of black holes, at speeds which may closely approach that of light.

Things required
1. Two equal size glasses.
2. A pot containing hot water
3. A pot containing cold water
4. an Old CD
5. Some edible color or ink.
Procedure
1.Pour some edible color or ink into a glass containing hot water and Mix well.
2.Take snother similer glass and pour ice cold fridge water into it. Cover it with an old CD.
2. cover the CD opening with finger and invert the cold water glass atop the hot water glass.
Observations:
1.convection
2. Density changes with temperature
3. Individual water molecules are not affected/ barely affected by gravitational force.

Reference
1. Arvind gupta, you tube vedio http://www.youtube.com/watch?v=uNIwvDiAVbSY&list=UUT7EcU7rC44DiS3RkfZzZMg&index=1&feature=plcp
2. Examples in nature : http://en.wikipedia.org/wiki/Convection

Editorial of June 2012

noreply@blogger.com (Dr Gouri Goutam Borthakur) — Fri, 01 Jun 2012 14:26:00 +0000

Fate or luck is nothing but the result of your decisions that accumulate you to work hard and smart in that "Practise like a Devil, perform like an Angel"attitude.Wrong decisions may make you trace a path that may end up in a broken bridge or to a land of irrevocable return.But why we take wrong decisions, after all? Few of the factors are
1. speculations
2. Under-estimation of the self
3. Over-estimation of resources
4. Closed narrow view of the future
Here is a exemplary story from corporate world(Doom of Film Photography Giant Kodak Corp."Why KodAK Failed") that will help one to identify the factors that lead us to take wrong decisions. Trying to find the pinning points from the story will be a rewarding experience.

Here is the excerpt of the entire story
"Steve Sasson, the Kodak engineer who invented the first digital camera in 1975, characterized the initial corporate response to his invention this way:

But it was filmless photography, so management’s reaction was, ‘that’s cute—but don’t tell anyone about it.’"

http://www.forbes.com/sites/chunkamui/2012/01/18/how-kodak-failed/

10 Important Physics MCQ/OTQ on MI

noreply@blogger.com (Dr Gouri Goutam Borthakur) — Fri, 11 May 2012 16:47:00 +0000

Objective Type Questions / MCQs

These are some very interesting 1 mark questions from the topic Moment of Inertia. You can post answers as comments if you want explanations and discuss with friends for further insights

For a circular ring, the radius of gyration K (about an axis through its centre and perperdicular its plane) is related to the radius of the ring R as
1. K=R 2. K =R/2 3. K=R/4 4. K=R/3.14
MI of a body is the ratio between the torque and
1. Linear acceleration 2. Centripetal acceleration 3. Centrifugal acceleration 4. Angular acceleration
Mass of a body means
1. Only its gravitational mass 2. Only its inertial mass 3. only its inertia coefficient 4. all above
Moment of inertia ofa body can be defined without any reference to the axis of rotation (True / False)
Torque of a body is equal to its MI when its angular acceleration is _________ (Fill up the blanks)
The plot (Mass, MI) must be a
1. St line passing through the origin 2. St line not passing through the origin 3. St line parallel to the mass axis 4. st line passing throgh the MI axis
What is the minimum no of bodies required to draw a (mass, MI ) graph ?
1. One 2. Two 3. Three 4. Many
The plot (radius of Gyration, MI) must be a
1. parabola symmetric about x axis 2. Parabola symmtric about y axis 3. St line 4. Hyperbola
What is the minimum number of bodies required to draw a (radius of Gyration, MI) graph ?
1. One 2. Two 3. Three 4. Many
We can never experimentally show that the (Mass, MI) graph has actually passed through the origin . Why ?

You can post answers as comments if you want explanations

Diepersion relations with matlab

noreply@blogger.com (Dr Gouri Goutam Borthakur) — Fri, 04 May 2012 02:33:00 +0000

Dispersion relations
This is a topic which vary widely in context. Separate discussions are needed for dispersions relations in plasma, in sound waves, in elastic waves in solid, in water waves, in gravity waves, in electromagnetic waves. etc.
But always a demonstration is required.(it needs to be animated, to represent a real life situation.)I found the following matlab program effective to demonstrate the same
1. Dispersive wavepacket
> with(plots):
animate(sum(.07*(exp(-(.1*k-3)^2)+exp(-(0.1*k+3)^2))*cos(.1*k*x-.1*k/sqrt(1+.1*(.1*k)^2)*t),k=1..100),x=-4..20,t=0..30,frames=60,numpoints=200,color=red);

2. Nondispersive wavepacket
> animate(sum(.07*(exp(-(.1*k-3)^2)+exp(-(0.1*k+3)^2))*cos(.1*k*x-.1*k/sqrt(1+.0*(.1*k)^2)*t),k=1..100),x=-4..20,t=0..30,frames=60,numpoints=200,color=red);

3. Dispersive sinusoidal waves
> animate(sin(x-t)+sin(1.2*x-1.1*t),x=0..50,t=0..63,numpoints=150,frames=100,color=red);

4. Nondispersive sinusoidal waves
> animate(sin(x-t)+sin(1.2*x-1.2*t),x=0..50,t=0..63,numpoints=150,frames=100,color=red);

References
1. http://physics.usask.ca/~hirose/ep225/animation/dispersion/anim-dispersion.html

Microprocessors Vs Microcontrollers

noreply@blogger.com (Dr Gouri Goutam Borthakur) — Tue, 10 Apr 2012 13:41:00 +0000

Microprocessors Vs Microcontrollers

1. Difference in IEEE recognition :
The IEEE defines a microprocessor as an integrated circuit that contains the logic elements for manipulating data and for making decisions. They do not have a definition for microcontroller. In practice a microcontroller is a circuit that provides the control for a system. An example is the chip that controls a microwave oven. Typically it is programmed by the manufacturer of the system and not by the end user. A microprocessor is similar, but is typically programmed by the end user. An example of this is the microprocessor that is the heart of a PC.
2. Difference in Applications
The same chip could be used either as a microcontroller or as a microprocessor although typically microcontrollers are less general purpose and may have smaller data word sizes and support fewer interfaces.
3. Generation Next !
Microprocessors of one generation may be used as the basis for microcontrollers of the next generation. An example of this is the Intel 8080 which was a widely used microprocessor in the early days of the PC and which was later used for microcontrollers in electronic cash registers.
4. Difference in Data and Program Memory Space
In Microprocessors, Data and Program memory is stored in same space. In Microcontrollers, Data and programs are stored in separate memory space.

CDMA or GSM?

noreply@blogger.com (Dr Gouri Goutam Borthakur) — Sun, 01 Apr 2012 04:06:00 +0000

How to know if your Phone is CDMA or GSM?
Locate the sticker inside the battery cover. If the sticker has an ESN (Electronic Serial Numbers) number on it, then the phone is a GSM device. If the sticker has a DEC (Decimal) or HEX (Hexadecimal) number on it, then it's a CDMA phone.

Reference >

1. http://www.ehow.com/how_7334950_tell-cell-phone-gsm-cdma_.html

To compare two low resistances by a potentiometer

noreply@blogger.com (Dr Gouri Goutam Borthakur) — Sat, 31 Mar 2012 02:53:00 +0000

Aim of the Experiment: To compare two low resistances by a potentiometer

Theory :If i1 and i2 be the currents in the potentiometer and the low resistance circuit, and a1, b1, c1 and d1 be the null points corresponding to the potential terminals S3, S4, S5 and S6 respectively, then

i2.R2= i1(b1-c1).p

and i2.R3 = i1(d1-c1).p

Where l1=b1-c1

l2= d1-c1

p is the resistance per unit length of the potentiometer wire

Unknown resistance R2= R3(l1/l2)

Or comparision of two unknown resistances

R2/R3= l1/l2

The image cant be drawn with the softwares I have.In Qucs I can draw, but I cant make it to embed in a text file or convert the drawing to file format accepted by a image editor.On the other hand, if I use a image editor, I cant draw all the circuit elements, so cant complete my circuit.So for the time being, please see the reference below for the circuit diagram.
But finally, I found a tool called xcftools.So here is the algorithm: draw circuit using qucs,--> take a screenshot.--> edit using gimp.--> open terminal and cd to the directory where the .xcf image is located. type xcf2png -o [outputfilname.png] [targetfile.xcf].This produces a png image in the same directory. Here is the circuit diagram for the experiment

Apparatus : Two lowresistances, Potentiometer, Rheostat, A four way key(or two two-way keys) Galvanometer, cells, Resistance boxes.

Circuit connections: R5 is a protective resistance, R1 is Rheostat, To find null points one end of the Galvanometer is connected with one of the four key terminals s3-s6, while the other end is connected to the potentiometer wire through the Jockey J.Care should bbe taken to see that The end of the potentiometer wire that is connected to positive terminal of V2 is connected to positive terminal of V1 also.

Procedure:

Emf of cells V1 and V2 are measured before beginning and at the end of the experiment to see that they have given steady voltages during the experiment.
Connect the key s6 with the galvanometer, and by adjusting the Resistance box R4, bring the null point for to the middle of the last wire of the potentiometer.let this be d1.
Without changing the value of R4, or anything else, obtain null point for key s5.let this be c1.
Repeat step 3 for s4 and s3. Let the readings in these cases be ba and a1.
Calculate R2/R3 for the fixed value of R4.
Repeat steps 2-5 for a different value of R4 and obtain a new set of readings of a1, b1, c1 and d1.Calculate R2/R3 for the fixed value of R4.
Repeat step 6 and obtain still another set of values of a1 , b1 c1 and d1.Calculate R2/R3 for the fixed value of R4.
Interchange the positions of R2 and R3 in the circuit and repeat steps 2-8
Obtain a mean value of R2/R3 from the readings taken in steps 2-8.

Results :

Table 1 : Emf of batteries

Battery	EMF in Volts			Inference
	Before experiment	After experiment	Mean
V1
V2

Table 2 : Determination fo null points

Current in low resistance (indicated by value of R4)

Resistance in

No of observation

Null points corresponding to

Right Gap

Left gap

Wire no

Scale position

Mean

Wire no

Scale position

Mean

Wire no

Scale position

Mean

Wire no

Scale position

Mean

High

moderate

Table 3 : Calculation of the ratio R2/R3

Current in low resistance	Resistance in		No of obeservation	Total balancing length corresponding to				L1= (b1-a1)	L2= (d1-c1)	R2/R3	Mean Ratio
Current in low resistance	Right gap	Left gap	No of obeservation	D1	C1	B1	A1	L1= (b1-a1)	L2= (d1-c1)	R2/R3	Mean Ratio

Conclusion : The mean ratio of the given resistances is …........................ohm

If One of the resistances is known (say R3), the value of the known resistance is R3 =.................ohm

The other unknown resistance is then R2= (l1/l2) .R3 = ….........................ohm.

References :

1. A text book of Practical Physics, Singh S K, New central book agency Pvt Ltd (2008) pp 293-298

Experiment to Calibrate a Polrimeter and to determine the Concentration of an unknown active solution

noreply@blogger.com (Dr Gouri Goutam Borthakur) — Thu, 22 Mar 2012 15:53:00 +0000

Date :
Aim of the Experiment : To Calibrate a Polrimeter and to determine the Concentration of an unknown active solution
Apparatus : 1. Half shade polarimeter (bi-quartz polarimeter )
Theory :
A. To measure the specific rotation
The rotation of the plane of polarisation of plane polarised light by an active solution of length l cm containing m gms of active substance per cc of the solution is given by
q=slm/10= slc/1000
=>c = 1000q/sl ...............................(1)
B. Method of preparing the solution :
The amount of substance required to prepare v cc of solution of c₁% strength is w = c₁v/100 g
From a solution of known concentration c₁, a solution of lwer concentration c₂ can be prepared by adding y cc of distilled water to x cc of the solution of concentration c₁
when
y= (c₁- c_2).x/c₂)   ....................(2)
Procedure :
1.Prepare a solution of concentration c1
1a. Take a clean empty flask and weigh it. let its mass be m1
m1 =
1b . Take some solid w1 g(say ) of the active substance in the flask and weigh it again. Let the new weight of the flask be M1.
Then weight of the active substance is
w= M1-m1
1c. Pour in some distilled water (say 50cc) and shake well to dissolve the solid. Transfer the solution in to measuring cylinder and pour in more water to make the volume 100 cc.This is a soultion of strength c1 % and volume 100 cc.
2. Note the reading of the vernier when the polarimeter tube contains water
2a.Measure the length of the tube T between inner surfaces of its two end plates thrice as l1, l2 and l3.
l1 =           l2 =                              l3 =
Then mean length of the tube l=(l1+l2+l3)/3=
2b. Turn on the source to illuminate the slit.
2c. Fill Tube T with distilled water(with no air bubbles),
2d. Rotate the tube T2 untill the two halves of the field of view are equally bright.Note the readings R1 and R2 of the Verniers V1 and V2 thrice.Note that V1 and V2 are 180 deg apart.
3. To find q For a solution of c1% strength
3a.Empty out the tube T, wash it with a little of the solution of c1 strength for two or three times.Then completely fill the tube with this solution.
3b. Same as 2d. Let R1' and R2' be the new readings of V1 and V2.Then
q₁= (R1~R1')/2 =
and
q₂= (R2~R2')/2 =
Mean optical rotation at this strength is
q= (q₁+q₂)/2
4.To dilute the soultion of c1 strength other lower strengths and to find rotations for those strengths
4a.x cc of stock solution is mixed with y cc of distilled water to have c2 % strength (say 12.5%).
y= (c₁- c_2).x/c₂)
4b. Repeat 3a and 3b for this strength c2 and measure optical rotation at this strength.
4c. By adopting the porcess mentioned in step 4a, solutions of other lower strengths (10%, 7.5%, 5% and 2.5% )are prepared and optical rotations for these strengths are measured as in step 3a and 3b.
5. Drawing the calibration curve and calculate specific rotation
5a. Plot a (concentration, optical rotation)graph.This is known as the calibration curve of the polarimeter.
5b. Take a arbitrary point on the graph and note its coordinates (c', q'), from which s (specific rotation ) is calculated using equation (1)
6. Determine concentration of an unknown active solution
6a.Repeat the step 3a and 3b with the unknown solution.find q for the unknown solution.
6b. From the graph of 5a, find concentration for this q.
Results :

Table 1
Table 2
Table 3
Table 4
Table 5
Plotting of graph :
Precaution and Discussion
Maximum proportional error
To Think :
How to innovate ?
Is there a scope for miniatuarisation in this experiment ?
Is there possibility of automation or sophistication ?
What are the alternative ways to realise the aims of this experiment or to varify the results obtained with this experiment ?
References :
1. Advanced Practical Physics Vol 1 < Mazumdar K G and Ghosh B , Shreedhar Publishers, (2000) ; pp 212-225

Capacitance concepts and review

noreply@blogger.com (Dr Gouri Goutam Borthakur) — Thu, 08 Mar 2012 11:22:00 +0000

What is meant by capacitance or What is the meaning of capacity of a conductor ?
    Capacity or capacitance of an electrical system is the ratio of its charge and its potential.
    OR Capacity or capacitance of a conductor is defined as the charge given to it to raise its potential by unity.
What is the unit of capacitance ?
    The SI unit of capacitance is Farad(F).1 Farad is equal to 1 Coulomb by 1 Volt.
What is the capacitance of an isolated sphere ?
    Capacitance of an isolated sphere is er times the radius of the sphere by 9.0e9
    That is C= er*R/(9.0e9)
What is the capacitance of earth?
    Assuming earth as an isolated sphere of R= 6.4e6 meter, and er~1, C= er*R/(9.0e9)~0.7mF(0.7 milli Farad)
What is the capacitance between two concentric spheres of radii ro > ri ?
       C= (4*pi*eo*er*ro*ri)/(ro-ri) when The outer sphere is earthed, or None of the two is earthed.
And... or Any thing else ? or then ...
       C= (4*pi*eo*er*ro*ro)/(ro-ri) when The inner sphere is earthed, and the outer spehere is charged.
And... or Any thing else ? or then ...
       If the radius of the outer surface is c and its thickness is (b-c), then,
       C= {(4*pi*eo*er*ro*ro)/(ro-ri)}+ {(4*pi*eo*er*ro*ro)/(ro-c)}+4*pi*eo*er*c
What is the capacitance between two concentric spheres of radii ro > ri ?
       C= (4*pi*eo*er*ro*ri)/(ro-ri) when The outer sphere is earthed, or None of the two is earthed.
       C= (4*pi*eo*er*ro*ro)/(ro-ri) when The inner sphere is earthed, and the outer spehere is charged.
       If the radius of the outer surface is c and its thickness is (b-c), then,
       C= {(4*pi*eo*er*ro*ro)/(ro-ri)}+ {(4*pi*eo*er*ro*ro)/(ro-c)}+4*pi*eo*er*c
What is the capacitance of a parallel plate capacitor?
    C= eo*er*(A/d); A is plate area, d is plate separation , er is dielectric constant of the medium between the plates.
What is the capacitance of a parallel plate capacitor of area A and distance d?
    C= eo*er*(A/d); A is plate area, d is plate separation , er is dielectric constant of the medium between the plates.
How the capacitance of a parallel plate capacitor depend on plate area A?
    C increases linearly with A or C is directly proportional to A.
How the capacitance of a parallel plate capacitor depend on plate separation d?
    C increases with decreasing d or C is inversely proportional to d.
How the capacitance of a parallel plate capacitor depend on dielectric constant?
    C increases linearly with er or C is directly proportional to er.
What happens when an insulated uncharged conductor is inserted in between the plates of a parallel plate capacitor ?
    On introducing an insulated uncharged conductor of thickness 't'between the plates of a parallel plate capacitor,     the capacitance changes from eo*er(A/d) to (eo*er*A)/(d-t).As a result capacitance increases.
What happens when an dielectric sheet of thickness 't' is inserted in between the plates of a parallel plate capacitor ?
    On introducing an insulated uncharged conductor of thickness 't'between the plates of a parallel plate capacitor,     the capacitance changes from eo*er(A/d) to (eo*A)/[d-t{1-(1/er)}].As a result capacitance decreases.
what happens When a number of dielectric sheets having same area as the capacitor plate are inserted between the plates ?
    In such a case, When a number of dielectric sheets having same area as the capacitor plate are inserted between     the plates, the capacitance becomes C = eo*A/{(t1/er1)+(t2/er2)+(t3/er3)+....}.
what happens When a number of dielectric sheets having same thickness but different area are inserted between the plates ?
    In such a case, When a number of dielectric sheets having same thickness but different area are inserted between     the plates, the capacitance becomes C = (eo/t)*{(A1*er1)+(A2*er2)+(A3*er3)+....}.

Link to qucs tutorial

noreply@blogger.com (Dr Gouri Goutam Borthakur) — Wed, 07 Mar 2012 03:22:00 +0000

http://qucs.sourceforge.net/examples.html

Analysis of Forced and damped motion

noreply@blogger.com (Dr Gouri Goutam Borthakur) — Tue, 06 Mar 2012 15:58:00 +0000

Analysis of Forced and damped motion by Dr GOURI GOUTAM BORTHAKUR

Get your own Virtual Classroom

Below is an undergraduate level physics test on free forced and damped vibrations aiming to test subject knowledge for IIT-JAM aspirants, CSIRUGC NET aspirants, GATE aspirants in Physics. Please attend the class at http://phy-effects.blogspot.in/2012/03/analysis-of-forced-and-damped-motion.html before appearing for this test. Test your basic knowledge on the topic through a quick mcq type test.Start

For Science writers

noreply@blogger.com (Dr Gouri Goutam Borthakur) — Sat, 03 Mar 2012 11:36:00 +0000

The IMRAD structure is currently the most prominent norm for the structure of a scientific paper. IMRAD is an acronym for Introduction, Methods, Results and Discussion. Scientific papers are typically structured in this basic order:

Introduction - why was the study undertaken? What was the research question, the tested hypothesis or the purpose of the research?
Methods - when, where, and how was the study done? What materials were used or who was included in the study groups (patients, etc.)?
Results - what answer was found to the research question; what did the study find? Was the tested hypothesis true?
Discussion - what might the answer mean and why does it matter? How does it fit in with what other researchers have found? What are the perspectives for future research?

Together with this knowledge of the IMRAD structure, the next important thing iis to know the EASE guidelines in detail.

Design course-lesson 2(Perspectives)

noreply@blogger.com (Dr Gouri Goutam Borthakur) — Tue, 28 Feb 2012 21:32:00 +0000

In the first lesson on product drawing, we learnt practised lines to improve sense of proportion and consistancy.Today, we will learn about perspectives

Understanding perspective is essential to be able to represent three-dimensional objects with correctproportions, which are more realistic and ‘true’ to the eye.Onebasic difference between isometric product drawings and product sketching is the visual effect of depth that is achieved through foreshortening.

What is foreshortening?

Foreshortening is distortion due to perspective when an object appears compressed when seen from any particular viewpoint.

What is Vanishing Point (VP)?

Due to foreshortening, if we were to extend these lines of taper, there would appear a point where these lines seem to emerge from, this point iscalled the vanishing point (VP), as elucidated in Fig: 11

Fig 11: Foreshortening and Vanishing Point.

For more clarity on the vanishing point, hold an object in front of your eyes, and gradually move it away towards your left and then to your right, up and then down and notice that beyond your cone of vision, this object vanishes. This perimeter region, beyond which the object vanishes from your view, is where the vanishing pointslie.

Fig 12: Vanishing Points (VP) with respect to Eye Level or Horizon and the Cone of vision.

One and Two Point Perspectives make use of one and two vanishing points respectively.

What is One Point Perspective?

In One Point Perspective, also known as Parallel Perspective, the object is directly in front of the observer, and not at an angle such that its primary surface is parallel to the imaginary frame of view or picture plane.

Mostly adopted to draw spaces, such as interiors of a room or buildings on a street, one will notice that the objects diminish in size, perpendicularly into the paper towards this single vanishing point.

Fig 13: One Point Perspective.

Exercise 3: One Point Perspective

Assuming a horizon and a single vanishing point, draw a cube in One Point Perspective

What is a Two Point Perspective?

Two Point Perspectives, i.e. perspectives drawn considering two vanishing points are mostly adopted for object drawing. Designers adopt Two Point perspectives for most of their 3D representation.

Fig 14: Cubes in space in Two Point Perspective.

Examples

Exercise 4: Cubes in Two Point Perspective

Assume a horizon and two vanishing points, as discussed above. Extending lines from these points, draw various cubes in space in Two Point Perspective.

We must note that distortions such as circles becoming ellipses for example, occur when represented in perspective.

Fig 15: Circles in Perspective.

Examples

Exercise 5: Circles in Two Point Perspective

Now draw circles onto those various cubes in space drawn by you. Note how the circles get an elliptical profile when in perspective.

Exercise 6: Treasure Box in Two Point Perspective

As you get a hang of the circles and cubes in perspective, try drawing objects that are predominantly combinations of these. Draw these Treasure Boxes to begin with. Notice the ellipsoidal reference lines along the edges of the cube, to correctly draw the open lid.

Fig 16: Treasure Box in Two Point Perspective.

Examples

Reference
1. http://www.dsource.in/course/product-drawing/perspective/perspective.html

Contents of march issue

noreply@blogger.com (Dr Gouri Goutam Borthakur) — Tue, 28 Feb 2012 06:13:00 +0000

Past Issue Forthcoming issue

Present issue

Contents of march issue

Issue 1, Vol 1, March 2012 Published on March 1, 2012.

    Editorial :                                    

 Chaos in the electronics Lab 

Question Hour :                           

 Junction Diode Inventories

Modeling  and Simulations      :    

Hodgekin- Huxley Model

  Physics effects :                            

Photoelectric effect

    Tutorials           :                         

 Finite Element method 

     Devices             :                          

 8051 Simulator

Java tutorial - 8

noreply@blogger.com (Dr Gouri Goutam Borthakur) — Tue, 28 Feb 2012 05:55:00 +0000

In the last tutorial, we learnt to create animations. But there might be lot of flickering. This tutorial demonstrates the problem and offers the solution to avoid flickering by using "double buffering "

[ Home | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 ]

Exercise 7: Backbuffers

Up to this point, the graphics drawn by our applets have been relatively simple. With more complex graphics however, whether in animations or interactive programs, flicker can become a problem. (You may have already noticed subtle flickering in some of the previous applets.)
This example demonstrate the problem. It uses a pseudo-random number generator to produce a big, hairy tangle of lines. The lines follow the mouse cursor.

import java.applet.*;
import java.awt.*;
import java.awt.event.*;
import java.lang.Math;

public class NoBackbuffer1 extends Applet
   implements MouseMotionListener {

   int width, height;
   int mx, my;  // the mouse coordinates
   Point[] points;
   int N = 300;

   public void init() {
      width = getSize().width;
      height = getSize().height;
      setBackground( Color.black );

      mx = width/2;
      my = height/2;

      points = new Point[ N ];
      for ( int i = 0; i < N; ++i ) {
         int x = (int)(( Math.random() - 0.5 ) * width / 1.5);
         int y = (int)(( Math.random() - 0.5 ) * height / 1.5);
         points[i] = new Point( x, y );
      }

      addMouseMotionListener( this );
   }

   public void mouseMoved( MouseEvent e ) {
      mx = e.getX();
      my = e.getY();
      showStatus( "Mouse at (" + mx + "," + my + ")" );
      repaint();
      e.consume();
   }
   public void mouseDragged( MouseEvent e ) { }

   public void paint( Graphics g ) {
      g.setColor( Color.white );
      for ( int j = 1; j < N; ++j ) {
         Point A = points[j-1];
         Point B = points[j];
         g.drawLine( mx+A.x, my+A.y, mx+B.x, my+B.y );
      }
   }
}

The output:

( ) You probably see flickering when you move the mouse over the applet. The lines take a significant amount of time to draw, and since the canvas is cleared before each redraw, the image on the canvas is actually incomplete most of the time.
This second example makes the problem even more pronounced by rendering a bitmap image in the background.

import java.applet.*;
import java.awt.*;
import java.awt.event.*;
import java.lang.Math;

public class NoBackbuffer2 extends Applet
   implements MouseMotionListener {

   int width, height;
   int mx, my;  // the mouse coordinates
   Point[] points;
   int N = 300;
   Image img;

   public void init() {
      width = getSize().width;
      height = getSize().height;
      setBackground( Color.black );

      mx = width/2;
      my = height/2;

      points = new Point[ N ];
      for ( int i = 0; i < N; ++i ) {
         int x = (int)(( Math.random() - 0.5 ) * width / 1.5);
         int y = (int)(( Math.random() - 0.5 ) * height / 1.5);
         points[i] = new Point( x, y );
      }

      // Download the image "fractal.gif" from the
      // same directory that the applet resides in.
      img = getImage( getDocumentBase(), "fractal.gif" );

      addMouseMotionListener( this );
   }

   public void mouseMoved( MouseEvent e ) {
      mx = e.getX();
      my = e.getY();
      showStatus( "Mouse at (" + mx + "," + my + ")" );
      repaint();
      e.consume();
   }
   public void mouseDragged( MouseEvent e ) { }

   public void paint( Graphics g ) {
      g.drawImage( img, 0, 0, this );
      g.setColor( Color.white );
      for ( int j = 1; j < N; ++j ) {
         Point A = points[j-1];
         Point B = points[j];
         g.drawLine( mx+A.x, my+A.y, mx+B.x, my+B.y );
      }
   }
}

The output:

( ) The flickering you see now should be especially bad.
The solution is to use double-buffering : rather than perform drawing operations directly to screen, we draw onto an image buffer (the "backbuffer") in memory, and only after completing this image do we copy it onto the screen. There is no need to erase or clear the contents of the screen before copying (or "swapping", as it's called) the backbuffer onto the screen. During the swap, we simply overwrite the image on the screen. Hence the screen never displays a partial image: even in the middle of swapping, the screen will contain 50 % of the old image and 50 % of the new image. As long as the swap is not too slow, the eye is fooled into seeing a continuous, smooth flow of images.
This example uses a backbuffer.

import java.applet.*;
import java.awt.*;
import java.awt.event.*;
import java.lang.Math;

public class Backbuffer1 extends Applet
   implements MouseMotionListener {

   int width, height;
   int mx, my;  // the mouse coordinates
   Point[] points;
   int N = 300;
   Image img;
   Image backbuffer;
   Graphics backg;

   public void init() {
      width = getSize().width;
      height = getSize().height;
      setBackground( Color.black );

      mx = width/2;
      my = height/2;

      points = new Point[ N ];
      for ( int i = 0; i < N; ++i ) {
         int x = (int)(( Math.random() - 0.5 ) * width / 1.5);
         int y = (int)(( Math.random() - 0.5 ) * height / 1.5);
         points[i] = new Point( x, y );
      }

      img = getImage(getDocumentBase(), "fractal.gif");

      backbuffer = createImage( width, height );
      backg = backbuffer.getGraphics();
      backg.setColor( Color.white );

      addMouseMotionListener( this );
   }

   public void mouseMoved( MouseEvent e ) {
      mx = e.getX();
      my = e.getY();
      showStatus( "Mouse at (" + mx + "," + my + ")" );

      backg.drawImage( img, 0, 0, this );
      for ( int j = 1; j < N; ++j ) {
         Point A = points[j-1];
         Point B = points[j];
         backg.drawLine( mx+A.x, my+A.y, mx+B.x, my+B.y );
      }

      repaint();
      e.consume();
   }
   public void mouseDragged( MouseEvent e ) { }

   public void paint( Graphics g ) {
      g.drawImage( backbuffer, 0, 0, this );
   }
}

The output:

( ) Why do we still see flicker ? Whenever the applet is supposed to redraw itself, the applet's update() function gets called. The java.awt.Component class (which is a base class of Applet) defines a default version of update() which does the following: (1) clears the applet by filling it with the background color, (2) sets the color of the graphics context to be the applet's foreground color, (3) calls the applet's paint() function. We see flickering because the canvas is still cleared before each redraw. To prevent this, we need to define our own update() function, to override the base class' behavior.

import java.applet.*;
import java.awt.*;
import java.awt.event.*;
import java.lang.Math;

public class Backbuffer2 extends Applet
   implements MouseMotionListener {

   int width, height;
   int mx, my;  // the mouse coordinates
   Point[] points;
   int N = 300;
   Image img;
   Image backbuffer;
   Graphics backg;

   public void init() {
      width = getSize().width;
      height = getSize().height;

      mx = width/2;
      my = height/2;

      points = new Point[ N ];
      for ( int i = 0; i < N; ++i ) {
         int x = (int)(( Math.random() - 0.5 ) * width / 1.5);
         int y = (int)(( Math.random() - 0.5 ) * height / 1.5);
         points[i] = new Point( x, y );
      }

      img = getImage(getDocumentBase(), "fractal.gif");

      backbuffer = createImage( width, height );
      backg = backbuffer.getGraphics();
      backg.setColor( Color.white );

      addMouseMotionListener( this );
   }

   public void mouseMoved( MouseEvent e ) {
      mx = e.getX();
      my = e.getY();
      showStatus( "Mouse at (" + mx + "," + my + ")" );

      backg.drawImage( img, 0, 0, this );
      for ( int j = 1; j < N; ++j ) {
         Point A = points[j-1];
         Point B = points[j];
         backg.drawLine( mx+A.x, my+A.y, mx+B.x, my+B.y );
      }

      repaint();
      e.consume();
   }
   public void mouseDragged( MouseEvent e ) { }

   public void update( Graphics g ) {
      g.drawImage( backbuffer, 0, 0, this );
   }

   public void paint( Graphics g ) {
      update( g );
   }
}

Now there should be no apparent flicker:

( )

Update (January 2008): I received the following email message that might be useful to some readers.

Hello,

I recently read your Java Applet Tutorial, it really helped me to understand
basics of java applets, especially backbuffers.
Although i had a problem with smooth animating - quickly moving objects
weren't rendered smoothly.
I thought I've made a mistake implementing backbuffer stuff, but (after a
few hours) i realized that it's everything fine
with this applet when it's ran on Windows.
The strange behaviour was that when I was moving mouse cursor over the
applet, the animation became smooth.
After another few hours i found getToolkit.sync() method which forces
refreshing applet in window system.
The tricky part is that you can't see any difference using Windows, but the
difference is clear in Xorg window system.
maybe you could consider adding a line in your tutorial, maybe someone will
save some time, next time  :)

 public void update( Graphics g ) {
      g.drawImage( backbuffer, 0, 0, this );
>>>>>>getToolkit().sync();
   }


anyways
thank you for your work


Artur Wielogórski
(student @Poznan University of Technology)

References :
http://profs.etsmtl.ca/mmcguffin/learn/java/07-backbuffer/