Monday, 29 June 2015

Wave Length of Monochromatic Light By Using Michelson Interferometer

Find the wave length of monochromatic light by using Michelson Interferometer

Objectives:
To find the wave length of Helium Neon LASER by using Michelson interferometer up to complete accuracy
Apparatus:
·         Helium Neon LASER
·         Michelson instrumental setup
·         Focusing lens
·         Meter rod
Michelson Interferometer:
The Michelson interferometer was invented in1893 by Albert Michelson, to measure in SI units of the wavelength of the red line of the cadmium spectrum. With an optical interferometer, he measured distance directly in terms of wavelength of light used by counting the interference fringes. In the Michelson interferometer, coherent beams are obtained by splitting a beam of light that originates from a single source with a partially reflecting mirror called a beam splitter. The resulting reflected and transmitted waves are then reflected back by ordinary mirrors to a screen where they interfere to create fringes. This is known as interference by division of amplitude. Fig. 3.1
Schematic Diagram:

 Fig. 3.1

Theory:
Interferometers are used to precisely measure the wavelength of optical beams through the creation of interference patterns.
Light is a transverse wave. When two waves of same wavelength and amplitude travel through same medium, their amplitudes combine. A wave of greater or lesser amplitude than the original will be the result. The addition of amplitudes due to superposition of two waves is called interference. If the crest of one wave meets with the trough of the other, the resultant intensity will be zero and the waves are said to interfere destructively. Alternatively, if the crest of one wave meets with the crest of the other, the resultant will be maximum intensity and the waves are said to interfere constructively.
In constructive interference, a bright fringe is obtained on the screen. For constructive interference to occur, the path difference between two beams must be an integral multiple of  of the wavelength λ,
Where m is the order, with
m =0, 1, 2, 3...
If the path difference between two waves is   the interference between them is destructive, and a dark fringe appears on the screen.

Procedure:
Light from a monochromatic source S is divided by a beam splitter (P), (Fig.3.1) which is oriented at an angle approximately 45° to 50° to the beam, producing two beams of equal intensity. The transmitted beam (L) travels to mirror M1 and it is reflected back to P. 50% of the returning beam is then reflected by the beam splitter and strikes the screen, D. The reflected beam (R) travels to mirror M2, where it is reflected. 50% of this beam passes straight through beam splitter and reaches the screen D.
Since the reflecting surface of the beam splitter Pis partially reflecting surface, the light ray starting from the source S and undergoing reflection at the mirror M2 passes through the beam splitter three times, while the ray reflected at M1 travels through Ponly once. The optical path length through the glass plate depends on its index of refraction, which causes an optical path difference between the two beams. The recombined beams interfere and produce fringes at the screen D. The relative phase of the two beams determines whether the interference will be constructive or destructive. By adjusting the inclination of M1 and M2, one can produce circular fringes, straight-line fringes, or curved fringes.

Conclusion:
The results shows that the actual wavelength of He-Ne LASER is greater than the calculated value, this error occurs due to in accuracy in measurements and due to apparatus error.
Percentage error=
Percentage error =
Percentage error = 30.2%

The calculated results shows that the error in wavelength of red LASER is correct up to 68.8%.

Show that the relationship of voltage and amplitude of a current by Full wave rectification

Objectives:

Find the relationship of voltage and amplitude of a current by using bridge to convert the current into DC which is also called full wave rectification

Apparatus:

•  Step down transformer or digital power supply
•  1 Resistor
•  4 Diodes
•  Connecting wires
•  Digital multimeter
•  CRO

Full wave rectification:

A full-wave rectifier converts the whole of the input waveform of AC current to a continuous polarity DC current either positive or negative at its output. Full-wave rectification converts both polarizations of the input waveform to pulsating DC, and yields a higher average output voltage. This yields a uni-directional and the continuous pulses without any break.

Procedure:

First of all set the transformer or power supply to any desired voltage. The schematic diagram is shown in (Fig.2.1). Join the four diodes in such a way that it makes a square loop in which the two parallel diodes must act as forward bias and rest of two should act as reverse bias. Now join the wires to the circuit as given in a figure.
From the AC mains when a positive half cycle of AC current flows through the diode D1 and D3 it is half rectified, and when the other negative half cycle of AC current flows through the diodes D2 and D4the other negative cycle is rectified to positive wave pulse.
The total outcome of the one complete cycle of AC current gives us a two complete pulses of a DC current as an output. Check these signals by connecting CRO wire clippers at D2 and D4. The output signals are round saw tooth signal as a DC current.
Circuit Diagram:
(Fig.2.1)

Data tables and Calculations:
 No. of obs. Voltage (V) Amplitude (cm) Sweep time (ms) Resistor (Ω) 1. 6.05 4.0 1 404 2. 6.1 2.0 1 404 3. 6.2 1.0 1 404 4. 6.5 0.3 1 404 5. 9.1 3.0 1 404 6. 9.2 1.8 1 404 7. 9.5 0.3 1 404 8. 12.1 4.0 1 404 9. 12.2 2.0 1 404 10. 12.5 0.4 1 404

Results:

The experiments calculations shows that the amplitude and voltage for a full rectification is inversely proportional to each other. When we increased the voltage the amplitude dropped inversely.

Friday, 26 June 2015

Half wave rectification Lab Report

Show that the relationship between voltage and amplitude of a current by Half wave rectification

Objectives:
We will find the relationship between voltage and amplitude of current through resistor by using CRO.
Apparatus:
·         Step down transformeror digital power supply
·         A resistor
·         A diode
·         Some connecting wires
·         A digital multimeter
·         CRO
Rectification:
It is a process in which a sinusoidal wave form of AC current is converted into (one directional) DC current.
Theory:
A rectifier is an electrical device which converts an AC signal to DC signal. An AC signal consists of alternative positive and negative parts like a sine wave. However a DC signal has only the positive part. In the circumstance of half wave rectifier, when we apply the AC input to the circuit, the half of the wave is blocked. The other half reached in the output end, so the voltage is less. This is achieved using the single diode in a circuit having AC power supply, the output signal is uni-directional, this process is called rectification and its pulses are shown in the CRO as a single signal. Fig.1
Procedure:
First of all take a resistor and find its color coding resistance, (not necessary if already known). Take the diode and connect with resistor with the connecting wires in series. Turn on the step down transformer and connect two wires to lower terminal and higher terminal (high voltage requires large value of resistance). The p-type of the diode should be connected to the higher terminal of the battery, in our case, the higher terminal is up to 12V. When the circuit is established, turn on the CRO and connect its X or Y channel with wire clippers. Check the voltage by digital multimeter if it is not given on the transformer. Now connect the clippers across the resistor, this will generate the wave pulses separated by the distance equal to λ/2. Note down the corresponding voltage and calculate the amplitude shown on the screen of CRO, and draw a graph for the calculated data.
Circuit Diagram:

Data Tables and Calculations:
 No. of obs. Voltage (V) Amplitude (cm) Sweep time (ms) Resistor (Ω) 1. 3.0 1.0 50 120 2. 4.5 1.4 50 120 3. 6.0 1.8 50 120 4. 7.5 2.0 50 120 5. 9.0 2.4 50 120 6. 10.5 2.8 50 120 7. 12.0 3.2 50 120

Graph:

Conclusion:
Hence the voltage and amplitude in half wave rectification are directly proportional to each other, when we increased the voltage then the amplitude increased showing direct relation between them.