Demonstration Project File PHYSICS

Submitted by: - ROHIT KUMAR HOODA Class: - XII A Roll No.:-

CERTIFICATEy This is to certify that ROHIT KUMAR HOODA of class XII A has successfully completed the given task of demonstration project file for the academic year 2011-2012 and submitted to Mrs. Anjana (PGT.PHYSICS)

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Demonstration Experiment 1:AIM: To study the dependence of diffraction pattern ondifferent factors.

APPARATUS: Screen, iron standing uprights, slits, laserlight, uprights with optical bench.

THEORY:In diffraction of a single of width a the conditions required for maxima & minima are as follows.

y For maxima:A sin = (2n-1) /2 y For minima:A sin = n Fringe width obtained in a single slit for secondary maxima. = D /a y For central minima:=2D /a

METHODSR. NO

ACTIVITY PERFORMED(I)Laser light is selected to be used.

QUES. ASKEDWhy do we use laser lights?

ANSCONCLUSIONIt is a monochromatic light of coherent particles forming beam. For good diffraction Pattern monochromatic& coherent source is required.

1.

2.

Teacher formed the two blades to be used as a Slit.

Why do we use blade?

Since it has Blades joined sharp edges & can be used as Thus, slits. condition of diffraction is fulfilled.

Diffraction pattern is obtained on the screen.

What do we observe?

3.

Diffraction pattern is obtained on screen.

Fringe is obtained on screen with central maxima of maximum intensity.

Slit width changed: (i) width increased:

How does the pattern change?

Fringe width decreases.

Distance between fringe is inversely proportional to a.

(ii)slit width is decreased:

4.

What is the effect on fringe width? What is effect on the brightness of fringe?

Fringe width increases & its intensity decreases.

Fringe is increases proportional to a 1/a I a

(iii) Increasing the distance D.

What on the Fringe width screen pattern decreases. change?

D

RESULTFrom above demonstration experiment we conclude that if slit width is comparable to wavelength of light used diffraction pattern is obtained. According to the formula for fringe width,

= D /d We conclude, (a) (b) (c) 1/d D

Diffraction pattern due to single slit:

PRECAUTIONS:(i) Experiment should be performed in dark room. (ii) Alignment of slit, source of light & screen must be proper.

Demonstration Experiment 2:AIM:-To study the detailed behavior of logic gates.

APPARATUS: (i) ideal p-n junction. (ii) Battery. (iii) Bulb.(v) Wire.

THEORY:Logic gates are digital circuit that produces a single output voltage from a combination of two or more input voltage. The level of that output voltage depends upon the level of the input voltage level of 5v & 0v. In logic gates, the 5v is refereed as 1 and the 0v is refereed as 0 . Various input combination of 1 and 0 produced different output, according to a logic gate truth table. In this way output is displayed.

OR GATE:Represented by + sign.

Operations:y Case 1: When both A & B are connected to earth, both the diodes do not conduct and therefore no voltage develops across resistance. Hence the output y is 0. y Case 2: When A is connected to earth& B is connected to positive terminal of battery 5v, the junction diode does not conduct while other conduct being forward biased. As we have taken therefore, the output y will be 1. y Case 3: When A is connected to positive terminal of battery & B to earth, the first diode will conduct while the other does not, so A will be forward biased. No voltage drop takes place.

Therefore, the output y will be 1. y Case 4: When A and B are connected to positive terminal of battery 5v both the diodes being forward biased, they will conduct .since diode are ideal & connected in parallel, the voltage drop across resistance cannot exceeds 5v. Hence output y will be 1. Circuit diagram:-

Truth Table for OR gate:-

In OR GATE: The output of an OR gate assumes 1-if one or more input assume 1. The truth table of OR gate which combine the input A&B to given output y. Off - > 0 On - > 1 The input is introduced through the switch A & B. The lighting of the bulbs is the output. Here we find that the bulb glows when lither switch A is closed or B is closed or both the switches are closed the bulb remains off only when both the switches A and B are open.

NOT GATE:REPRESENTED BY: ~

OPERATIONS:y CASE 1: When A is earthed the base of the transits or also get earthed. Now, base emitter junction is not forward biased but basecollector junction is reversed biased. As the emitter current is zero, the base current is also zero. Under these conditions the transistor is said to be in cut off mode and voltage will be +5v w.r.t earth due to battery in collector circuit. Hence the output y is 1. y CASE 2: When A is connect to positive terminal of battery 5v the baseemitter junction gets forward biased. There will be emitter

current, base current and collector current. The value of resistance is so adjusted, a lager collector current flows. In this situation, the transistor is said to have gone to situation state. The voltage drop across resistance due to forward biasing of emitter is just equal to 5v. Therefore, the output y is o.

Circuit diagram for NOT gate:

Truth table:

NOT GATE: The output of a NOT gate is 1 if input is 0 and vice versa. Off - > 1 On -> 0 In this gate, off corresponds to 0 and on corresponding to 1. The input is introduced through the switch A. the lighting of the blubs glows only when switch A is open or off and blub does not glow when switch A is close.

AND GATE:REPRESENTED BY .

OPERATIONS:y CASE 1: When both A&B are connected to earth, both the diode get forward biased & hence conduct. The diodes being ideal, no

voltage drop take place across either diode. Therefore a voltage drops of 5v thus the output y is 0. y CASE 2: When A is earthed and B is connected to positive terminal of battery 5v, the first diode will conduct while the second dose not since diode are ideal, no voltage drop of 5v take place across resistance now the output y is 0. y CASE 3: When A is connected to positive terminal of battery 5v, B is connected to earth. The first diode will not conduct while the second will conduct. Since diodes are ideal, no voltage drop of 5v takes place across resistance. Now the output y is 0 (zero).

y Case 4: When A & B both are connected to positive terminal of the battery 5v, none of the diode will conduct. There will no current through resistance. Hence the output is 1.

Circuit diagram:-

Truth table:-

In AND gate:The output of an AND gate assume 1 only if all the inputs assume the inputs assume 1. The truth table of AND gate which combines the inputs A & B to give output y is represented by: Y=A.B OFF-> 0

ON-> 1 The inputs are introduced through the switches A & B. The lighting of the bulb is the output. Here we find that the bulb glows only when both the switches A & B are closed. The bulb remains off when either switch A or switch B or both are open.

NAND GATE:This gate is made by connecting the output of AND gate to the input of a NOT gate. Represented by (A.B)

The truth table of NAND gate can be obtained by logically using the truth table of AND & NOT gates. Truth table for NAND gate:

NOR gate:This gate is made by connecting the output of NOT gate & OR gate. Represented by- (A+B)

The truth table of NOR gate can be obtained by logically using the truth table of OR and NOT gates.

Truth table for NOR gate:-

Result:The output of the logic gates follows the truth table of gates.

Precautions: Connections should be clean & neat. Connections should be tight.