Final Year Project Report

2021-2022

Full Unit – Final Report

BAND GAP OF LED USING NEWTON’S

RING METHOD

A report submitted in part fulfillment of the degree of

BSc Physics

1
 Table of Contents
➔ Introduction ………………………………………………….. 3

➔ Background Study ……………………………….………… 4-31

❖ Theory …………………………………………………. 4-25

● Semiconductors ………………………… 4
● P-N Junction Diode………………….. 8
● LED ………………………………………….. 14
● Visible Light Spectrum ……………… 17
● Band Gap Energy ……………………. 19
● Newton's Ring Method …………… 21

❖ Experiment ………………………………………… 26 - 31
● Aim …………………………………………… 26
● Apparatus ……………………………….. 26
● Diagram ………………………………….. 26
● Procedure …………………………………. 27
● Observations …………………………… 28
● Observation Table …………………… 29
● Calculation ……………………………… 31
● Graphs ……………………………………. 34
● Result ………………………………………. 36

➔ Conclusion …………………………………………………… 37

➔ References ………………………………………...…………. 38

2
● Introduction

Light emitting diodes (LEDs) are special semiconductor devices in which

recombination that takes place near the p-n junction results in light emission.
The emitted light, though not highly monochromatic, is treated
monochromatic on practical considerations, as line width is of the order of ±
10%.
Using a plano-convex lens of a known radius of curvature, Newton’s Rings are
observed by illuminating them with lights of different wavelengths obtained
by different LEDs.
Each kind of light gives a different set of Newton’s Rings.
Using the standard procedure the wavelengths of the illuminating lights are
determined. Then after using the relation between the wavelength and band
gap energy, the energy gaps of the materials of the respective LEDs are
determined and the values obtained are compared with the standard values.
Thus we can also find out the type of material which is used in making LEDs ;
as every semiconductor material has different values of energy gap.

3
● Theory

❖ Semiconductors :
A semiconductor material is one whose electrical properties lie in between
those of insulators and good conductors. Examples are : germanium and
silicon. In terms of energy bands, semiconductors can be defined as those
materials which have almost an empty conduction band and almost filled
valence band with a very narrow energy gap (of the order of 1 eV) separating
the two.

Types of Semiconductors:
Semiconductor may be classified as under:

A. Intrinsic Semiconductors
An intrinsic semiconductor is one which is made of the semiconductor
material in its extremely pure form. Examples of such semiconductors are:
pure germanium and silicon which have forbidden energy gaps of 0.72 eV and
1.1 eV respectively. The energy gap is so small that even at ordinary room
temperature; there are many electrons which possess sufficient energy to
jump across the small energy gap between the valence and the conduction
bands.

4
 Alternatively, an intrinsic semiconductor may be defined as one in which the
number of conduction electrons is equal to the number of holes.
Schematic energy band diagram of an intrinsic semiconductor at room
temperature is shown in fig. below -

B. Extrinsic Semiconductors
Those intrinsic semiconductors to which some suitable impurity or doping
agent or doping has been added in extremely small amounts
(about 1 part in 108) are called extrinsic or impure semiconductors.
Depending on the type of doping material used, extrinsic semiconductors can
be subdivided into two classes:
I ) N-type semiconductors &
II ) P-type semiconductors.

I. N-type Semiconductors
This type of semiconductor is obtained when a pentavalent material like
antimony (Sb) is added to pure Silicon crystal. As shown in Fig. below, each
antimony atom forms covalent bonds with the surrounding four Silicon atoms
with the help of four of its five electrons. The fifth electron is superfluous and
is loosely bound to the antimony atom.
Hence, it can be easily excited from the valence band to the conduction band
by the application of electric field or increase in thermal energy. It is seen

5
 from the above description that in N-type semiconductors, electrons are the
majority carriers while holes constitute the minority carriers.

II. P-type Semiconductors

This type of semiconductor is obtained when traces of a trivalent like boron
(B) are added to a pure silicon crystal. In this case, the three valence electrons
of a boron atom forms covalent bonds with four surrounding silicon atoms,
but one bond is left incomplete and gives rise to a hole as shown in Fig. below.
Thus, boron which is called an acceptor impurity causes as many positive
holes in a silicon crystal as there are boron atoms thereby producing a P-type
(P for positive) extrinsic semiconductor.
In this type of semiconductor, conduction is by the movement of holes in the
valence band.

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● Majority & Minority Charge Carriers

In a piece of pure germanium or silicon, no free charge carriers are available

at 0°K. However, as its temperature is raised to room temperature, some of
the covalent bonds are broken by heat energy and as a result, electron-hole
pairs are produced. These are called thermally-generated charge carriers.
They are also known as intrinsically-available charge carriers. Ordinarily,
their number is quite small. An intrinsic of pure germanium can be converted
into a P-type semiconductor by the addition of an acceptor impurity which
adds a large number of holes to it.
Hence, a P-type material contains a large number of positive holes—most of
them being the added impurity holes with only a very small number of
thermally generated ones & very small number of thermally generated
electrons.
Thus in a P-type material, the number of holes (both added and thermally
generated) is much more than that of electrons and hence, in such a material,
holes constitute majority carriers and electrons form minority carriers as
shown in Fig. below (a).
Similarly, in an N-type material, the number of electrons (both added
and thermally-generated) is much larger than the number of thermally
generated holes. Hence, in such a material, electrons are majority carriers
whereas holes are minority carriers as shown in Fig. below (b).

7
❖ P-N Junction Diode :
A p-n junction diode is a two-terminal or two-electrode semiconductor
device, which allows the electric current in only one direction while blocks
the electric current in opposite or reverse direction. If the diode is forward
biased, it allows the electric current flow. On the other hand, if the diode is
reverse biased, it blocks the electric current flow. P-N junction semiconductor
diode is also called as p-n junction semiconductor device.
In n-type semiconductors, free electrons are the majority charge carriers
whereas in p-type semiconductors, holes are the majority charge carriers.
When the n-type semiconductor is joined with the p-type semiconductor, a
p-n junction is formed. The p-n junction, which is formed when the p-type
and n-type semiconductors are joined, is called a p-n junction diode.

For designing the diodes, silicon is more preferred over germanium.The p-n
junction diodes made from silicon semiconductors work at higher
temperatures when compared with the p-n junction diodes made from
germanium semiconductors.

● Biasing of p-n junction semiconductor diode

The process of applying the external voltage to a p-n junction semiconductor
diode is called biasing. External voltage to the p-n junction diode is applied in
any of the two methods: forward biasing or reverse biasing.
If the p-n junction diode is forward biased, it allows the electric current to
flow. Under forward biased conditions, the p-type semiconductor is connected
to the positive terminal of the battery whereas; the n-type semiconductor is
connected to the negative terminal of the battery.

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 If the p-n junction diode is reverse biased, it blocks the electric current flow.
Under reverse biased conditions, the p-type semiconductor is connected to
the negative terminal of the battery whereas; the n-type semiconductor is
connected to the positive terminal of the battery.
The basic symbol of the p-n junction diode under forward bias and reverse
bias is shown in the below figure.

In the above figure, the arrowhead of a diode indicates the conventional

direction of electric current when the diode is forward biased
(from the positive terminal to the negative terminal). The holes which moves
from the positive terminal (anode) to the negative terminal (cathode) is the
conventional direction of current.

● Terminals of pn junction diode

Generally, terminal refers to a point or place at which any object begins or

ends. In a p-n junction diode, terminal refers to a point at which charge
carriers begin or end.The P-n junction diode consists of two terminals:
positive and negative. At the positive terminal, all the free electrons will end
and all the holes will begin whereas at the negative terminal all the free
electrons will begin and all the holes will end.

A. Terminals of diode under forward bias

In forward biased p-n junction diodes (p-type connected to positive terminal
and n-type connected to negative terminal), anode terminal is a positive
terminal whereas cathode terminal is negative terminal.
Anode terminal is a positively charged electrode or conductor, which supplies
holes to the p-n junction. In other words, anode or anode terminal or positive

9
 terminal is the source of positive charge carriers (holes), the positive charge
carriers (holes) begin their journey at anode terminal and travel through the
diode and end at cathode terminal.

Cathode is the negatively charged electrode or conductor, which supplies free

electrons to the p-n junction. In other words, cathode terminal or negative
terminal is the source of free electrons, the negative charge carriers (free
electrons) begin their journey at cathode terminal and travel through the
diode and end at anode terminal.
The free electrons are attracted towards the anode terminal or positive
terminal whereas the holes are attracted towards the cathode terminal or
negative terminal. And hence, maximum current flows through the diode.

B. Terminals of diode under reverse bias

If the diode is reverse biased (p-type connected to the negative terminal and
n-type connected to positive terminal), the anode terminal becomes a negative
terminal whereas the cathode terminal becomes a positive terminal.
Anode terminal or negative terminal supplies free electrons to the p-n
junction. In other words, the anode terminal is the source of free electrons,
the free electrons begin their journey at the negative or anode terminal and
fill the large number of holes in the p-type semiconductor. The holes in the

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p-type semiconductor get attracted towards the negative terminal. The free
electrons from the negative terminal cannot move towards the positive
terminal because the wide depletion region at the p-n junction resists or
opposes the flow of free electrons.

Cathode terminal or positive terminal supplies holes to the p-n junction. In

other words, the cathode terminal is the source of holes, the holes begin their
journey at the positive or cathode terminal and occupy the electron's position
in the n-type semiconductor. The free electrons in the n-type semiconductor
get attracted towards the positive terminal. The holes from the positive
terminal cannot move towards the negative terminal because the wide
depletion region at the p-n junction opposes the flow of holes. And hence,
ideally no current will flow through it.

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● V/I characteristics

● Forward Characteristics

When the diode is forward biased and applied voltage is increased from zero
hardly any current flows through the device in the beginning. It is so because
the external voltage is being opposed by the internal barrier voltage VB
whose value is 0.7 V for Si and 0.3 V for Ge. As soon as VB is neutralized,
current through the diode increases rapidly with increasing applied battery
voltage. It is found that as little a voltage as 1.0 V produces a forward
current of about 50 mA.

● Reverse characteristic

When the diode is reverse biased majority carriers are blocked and only
a small current (due to minority carriers) flows through the diode. As the
reverse voltage is increased from zero, the reverse current very quickly
reaches its maximum or saturation value I0 which is also known as leakage
current. It is of order of nano ampere (nA) for Si and micro ampere (µA) for
Ge. When reverse voltage VBR, the leakage current suddenly and
sharply increases, the curve indicating zero resistance at this point.

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● Advantages of p-n junction diode
➨It is much smaller and cheaper.
➨It needs only a small voltage to operate.
➨It requires no time to warm up to produce the current carriers since
it is a semiconductor, thus can be operated with a low voltage source.
➨It can be used as a switch.

● Disadvantages of p-n junction diode

➨P-n junction diode can not withstand very high reverse voltage.
➨It offers poor response.
➨It has reverse saturation current.
➨Noise level is high in semiconductor device at high frequencies.

13
❖ Light Emitting Diode (LED) :
A Light Emitting Diode (LED) is an optical semiconductor device that emits
light when voltage is applied. In other words, LED is an optical
semiconductor device that converts electrical energy into light energy.
The symbol of LED is similar to the normal p-n junction diode except that it
contains arrows pointing away from the diode indicating that light is being
emitted by the diode.

LEDs operate only in forward bias condition. To create an LED, the n-type
material should be connected to the negative terminal of the battery and
p-type material should be connected to the positive terminal of the battery. In
other words, the n-type material should be negatively charged and the p-type
material should be positively charged.
The construction of LED is similar to the normal p-n junction diode except
that gallium, phosphorus and arsenic materials are used for construction
instead of silicon or germanium materials.

When the Light Emitting Diode (LED) is forward biased, free electrons in the
conduction band recombines with the holes in the valence band and release
energy in the form of light.
The process of emitting light in response to the strong electric field or flow of
electric current is called electroluminescence.

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● Working of LED
When the Light Emitting Diode (LED) is forward biased, the free electrons
from the n-side and the holes from the p-side are pushed towards the
junction.
When free electrons reach the junction or depletion region, some of the free
electrons recombine with the holes in the positive ions. We know that positive
ions have less number of electrons than protons. Therefore, they are ready to
accept electrons. Thus, free electrons recombine with holes in the depletion
region. In the similar way, holes from p-side recombine with electrons in the
depletion region.

Because of the recombination of free electrons and holes in the depletion

region, the width of depletion region decreases. As a result, more charge
carriers will cross the p-n junction.
Some of the charge carriers from p-side and n-side will cross the p-n junction
before they recombine in the depletion region. For example, some free
electrons from n-type semiconductor cross the p-n junction and recombines
with holes in p-type semiconductor. In the similar way, holes from p-type
semiconductor cross the p-n junction and recombines with free electrons in
the n-type semiconductor. Thus, recombination takes place in depletion
region as well as in p-type and n-type semiconductor. The free electrons in the

15
 conduction band releases energy in the form of light before they recombine
with holes in the valence band. And thus the LED glows.

● Advantages of LED
➨The brightness of light emitted by LED is depends on the current flowing
through the LED. Hence, the brightness of LEDs can be easily controlled by
varying the current. This makes it possible to operate LED displays under
different ambient lighting conditions.
➨Light emitting diodes consume low energy.
➨LEDs are very cheap and readily available.
➨LEDs have longer lifetime.
➨LEDs operates very fast. They can be turned on and off in very less time.
➨LEDs do not contain toxic material like mercury which is used
in fluorescent lamps.
➨LEDs can emit different colors of light.

● Disadvantages of LED
➨LEDs need more power to operate than normal p-n junction diodes.
➨Luminous efficiency of LEDs is low.
➨Cannot withstand high reverse voltage.

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❖ Visible Light Spectrum :
Light is the electromagnetic radiation that occurs within a specific section of
the electromagnetic spectrum. The term essentially refers to the visible light,
it is the light that is distinguishable and visible to naked human eye and it is
also responsible for the sense of light.
The wavelengths of the visible light range between 400-700 nanometers, this
is between the infrared having longer wavelengths and the ultraviolet having
shorter wavelengths. The wavelength of the visible light indicates that its
frequency is approximately 430-750 terahertz (THz).
The light speed in the vacuum is 299,792,458 metres per second as per the
experiment. The visible light like other forms of the electromagnetic radiation
moves at this speed specifically in the vacuum.
In physics, the definition of light often refers to the electromagnetic radiation
having any wavelength, regardless of if it’s visible or not. The light exhibits
both particle nature and wave nature and the occurrence of this phenomenon
is described as light's dual nature.
Light doesn’t necessarily travel in the straight line but it travels in transverse
waves. The wave that is made of oscillation when moving and that occurs
perpendicular to direction of energy transfer is known as the transverse
waves.

● Wavelength

Wavelength is essentially the distance between two consecutive troughs or

two consecutive crests in the transverse wave. The wavelength is also used for
representing repeating patterns of travelling energies, like sound or light.
We know that light can be understood both as a particle and a wave. Photons
are the light particles which exist in the form of "packets" of electromagnetic

17
energy. On the other hand, waves are the form of energy where
electromagnetic radiation takes on when it is propagating.
Wavelength is the distance between two consecutive crests or two consecutive
troughs in a transverse wave.

Wavelength also represents a repeating pattern of any traveling energies, such

as light or sound. Wavelength is usually expressed by the units of nanometres
(nm). It is represented by the symbol λ (lambda).

Table of the Wavelengths of Various Colours

18
❖ Band Gap Energy :
A band gap is the distance between the valence band of electrons and the
conduction band. Essentially, the band gap represents the minimum energy
that is required to excite an electron up to a state in the conduction band
where it can participate in conduction. The lower energy level is the valence
band, and thus if a gap exists between this level and the higher energy
conduction band, energy must be input for electrons to become free. This
energy is called Band Gap Energy.It is denoted by Eg. The size and existence
of this band gap between conductors, semiconductors, and insulators is
shown in the figure below.

A semiconductor is a material with an intermediate-sized but non-zero band

gap that behaves as an insulator at absolute zero but allows thermal excitation
of electrons into its conduction band at temperatures that are below its
melting point.
The conductivity of intrinsic semiconductors is strongly dependent on the
band gap. The only available charge carriers for conduction are the electrons
that have enough thermal energy to be excited across the band gap and the
electron holes that are left off when such an excitation occurs.
Band-gap engineering is the process of controlling or altering the band gap of
a material by controlling the composition of certain semiconductor alloys,
such as GaAlAs, InGaAs, and InAlAs .The band-gap energy of semiconductors
tends to decrease with increasing temperature. When temperature increases,

19
the amplitude of atomic vibrations increases, leading to larger interatomic
spacing. LEDs usually emit photons with energy close to and slightly larger
than the band gap of the semiconductor material from which they are made.
Therefore, as the band gap energy increases, the LEDs colour changes from
infrared to red, red to violet, then to UV.
The table below lists the band gap energy of various semiconductors at room
temperature :-

As the band gap energy is understood to be the width of the energy gap
between conduction and valence band.
For single-photon processes, the optical wavelength corresponding to a given
band gap energy Eg can be calculated as

λ = hc/Eg
or numerically as 1.235 divided by the band gap energy in electron volts,
obtaining the band gap wavelength in units of micrometers.

20
❖ Newton's Ring Method :
Newton's rings is a phenomenon in which an interference pattern is created
by the reflection of light between two surfaces; a spherical surface and an
adjacent touching flat surface. It is named after Isaac Newton, who
investigated the effect in 1666. When viewed with monochromatic light,
Newton's rings appear as a series of concentric, alternating bright and dark
rings centered at the point of contact between the two surfaces. When viewed
with white light, it forms a concentric ring pattern of rainbow colors, because
the different wavelengths of light interfere at different thicknesses of the air
layer between the surfaces.

● Interference and it's type

When two or more waves having the same source and same frequency
(coherent waves) interact with each other, then the resulting phenomenon is
known as Interference.
According to the principle of superposition of waves – When two or more
waves of the same type are incident at the same point then the total
displacement at that point is equal to the vector sum of the displacements of
the individual waves. The superimposing waves can have constructive or
destructive interference.

21
 When crest of the first wave falls on the crest of second wave and trough of
the first wave falls on the trough of the second wave and then the resultant
wave is the vector sum of the amplitude of the two superimposing waves
which is equal to sum of the amplitude of two waves, this is known as
Constructive Interference.
When crest of the first wave falls on the trough of second wave and trough of
the first wave is falls on the crest of the second wave and the resultant wave is
the vector sum of the amplitude of the two superimposing waves which is
equal to the difference in amplitude of two waves, this is known as
Destructive Interference.
Constructive interference occurs when the phase difference between the
waves is a multiple of 2π, whereas destructive interference occurs when the
phase difference is an odd multiple of π.

● Formations of Bright and Dark fringes

When a parallel beam of monochromatic light is incident normally on a
combination of a plano-convex lens C and a glass plate P, as shown in fig.
below a part of each incident ray is reflected from the lower surface of the
lens, and a part, after refraction through the film between the lens and the
plate, is reflected back from the surface of the glass plate.

22
These two reflected rays are coherent, hence they will interfere and produce a
system of alternate dark and bright rings with the point of contact between
the lens and the plate at the center. And thus Newton’s rings are formed.

In general, the path difference between the reflected light beams which are
undergoing interference (for oblique incidence) is given by

∆ = 2µtcosθ − λ/2 ..………………. (1)

where the additional path difference of λ/2 is because one of the interfering
beams is reflected from film to glass surface.
Also, θ is the angle of incidence. For normal incidence θ=0° and hence, the
path difference will be

∆ = 2µt − λ/2 ……………………. (2)

23
 In the interference pattern bright fringe will be formed if the path difference
is equal to integral multiple of wavelength of light, i.e.,

∆ = 2µt − λ/2 = n ; n = 0, 1, 2, 3..

⇒ 2µt = ( n+½ ) λ ; n = 0, 1, 2, 3…. ………………….. (3)

For intensity minima (dark fringe), ∆ = ( n+½ ) λ ,and thus,

2µt = nλ n = 0, 1, 2, 3… …………………… (4)

● Relationship between ring diameter and wavelength

Let LOL' be the plano-convex lens placed on a glass plate. Plano-convex lens
appears as part of the circle of radius R.
Here, radius R is known as radius of curvature of plano-convex lens.
th
Suppose r is the radius of some n bright ring having thickness t.
Using the property of a circle. from above figure), we can write

EP × P F = P O × P Q,
2
⇒ r n = t × ( 2R − t )
2 2
⇒ r n = ( 2Rt − t )

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 Since R >> t, t2 can be neglected therefore
2
r n ≈ 2Rt ......…………. (5)

by using Eq.(4) and Eq.(5), we have

2
r n = nλR/µ .……………… (6)

th
Using rn = Dn /2, we can write following relation for diameter of n , ring as

2 2
D n = 2r n = nλR/µ ....….……….. (7)

th
Similarly,the diameter of some m dark fringe will be

2
D m = nλR/µ ....….……….. (8)

Subtracting Eq.(7) and Eq.(8), we can write following relation

2 2
λ = [D n − D m/4R(n − m)] × µ ..…………….. (9)

Above equation is used to find the wavelength of monochromatic light using

Newton ring’s method, in which material of refractive index µ is immersed
between plano-convex lens and glass plate.
If air is enclosed as thin film having µ=1, then Eq.(9) becomes

● An alternative and better method is to plot n ( no. of fringes) along the x-axis
2
and D on the y-axis. Then the slope of the straight line, tan (ϕ) will give the
wavelength as

λ = tan (ϕ) / 4R

25
● Experiment

❖ Aim :

To determine the wavelength of LEDs by Newton’s Ring method &

identifying the semiconductor material used in it using
Band gap energy Eg .

❖ Apparatus :

A plano-convex lens of large radius of curvature, optical arrangement

Newton's rings, plane glass plate, LEDs ( red,yellow,green,blue )
and a traveling microscope.

❖ Formula :

● The wavelength λ of light is given by the formula:

th
Where, Dn = diameter of n ring,
th
Dm = diameter of m ring,
m & n = an integer number (of the rings)
R = radius of curvature of the curved face of the plano-convex lens.

● The band gap energy Eg is given by

Eg = hc/λ
Where, λ = wavelength of light,
-34 2
h = planck's constant ( 6.626 × 10 m kg / s),
c = speed of light ( 3 × 10⁸ m / s).

26
❖ Diagram :

❖ Procedure :

● Level the traveling microscope table and set the microscope tube in a vertical
position. Find the vernier constant (least count) of the horizontal scale of the
traveling microscope.
● Clean the surface of the inclined glass plate G, the lens C and glass plate P.
● Place them in position as shown in above figure and as discussed in the
description of apparatus.
● Place the arrangement in front of the LEDs so that the height of the center of
the glass plate G is the same as that of the center of the LED.
● Adjust the position of the traveling microscope so that it lies vertically above
the center of lens C.
● Focus the microscope, so that alternate dark and bright rings are clearly
visible.
● Slide the microscope to the left till the cross wire lies tangentially at the
th
center of the n dark ring. Note the reading on the vernier scale of the
microscope.

27
● Slide the microscope backward with the help of the slow motion screw and
note the readings when the cross-wire lies tangentially at the center of the
dark rings, respectively.
● Keep on sliding the microscope to the right and note the reading when the
th
cross-wire again lies tangentially at the center of the 1st, 2nd and till the n
dark rings, respectively.
● Remove the plano-convex lens C and find the radius of curvature of the
surface of the lens in contact with the glass plate P accurately using a
spherometer.
● Find the diameter of each ring from the difference of the observations taken
on the left and right side of its center.
2 2
● Take any two diameters and perform the calculations for D n−D m (m<n) as
directed and calculate the value of wavelengths of the LEDs.
● Finally, calculate the Band Gap Energy Eg using the relation between the
wavelength and band gap energy and thereafter determine the semiconductor
material used in LED.

❖ Observations :

1. Least count of traveling Microscope =

smallest division on Main scale / Total no. of divisions on Vernier scale

= 0.05 cm / 50 div
= 0.001 cm

2. Radius of curvature of plano-convex lens R = 100 cm

3. Refractive index of air µ = 1

28
❖ Observation Table :

Red color LED :

Order Traveling Microscope Reading

Obs Of D D²
L.H.S R.H.S
No. Ring (cm) (cm)
MSR VSR TR MSR VSR TR

st
1) 1 5 24 5.024 4.7 42 4.742 0.282 0.0795

nd
2) 2 5 45 5.045 4.7 23 4.723 0.322 0.1036

rd
3) 3 5 64 5.064 4.7 06 4.706 0.358 0.1281

th
4) 4 5 83 5.083 4.6 89 4.689 0.394 0.1552

th
5) 5 5 97 5.097 4.6 72 4.672 0.425 0.18062

th
6) 6 5.1 09 5.109 4.6 62 4.662 0.447 0.1998

Green color LED:

Order Traveling Microscope Reading

Obs Of D D²
L.H.S R.H.S
No. Ring (cm) (cm)
MSR VSR TR MSR VSR TR

st
1) 1 5 21 5.021 4.7 64 4.764 0.257 0.0660

nd
2) 2 5 39 5.039 4.7 42 4.742 0.297 0.0882

rd
3) 3 5 55 5.055 4.7 22 4.722 0.333 0.1108

th
4) 4 5 72 5.072 4.7 13 4.713 0.359 0.1288

th
5) 5 5 93 5.093 4.7 03 4.703 0.390 0.1521

th
6) 6 5.1 00 5.100 4.6 93 4.693 0.407 0.1656

29
Yellow color LED :

Order Traveling Microscope Reading

Obs Of D D²
L.H.S R.H.S
No. Ring (cm) (cm)
MSR VSR TR MSR VSR TR

st
1) 1 5 07 5.007 4.7 81 4.781 0.401 0.0510

nd
2) 2 5 22 5.022 4.7 61 4.761 0.369 0.0681

rd
3) 3 5 44 5.044 4.7 39 4.739 0.335 0.0930

th
4) 4 5 61 5.061 4.7 26 4.726 0.305 0.1122

th
5) 5 5 80 5.080 4.7 11 4.711 0.261 0.1361

th
6) 6 5 97 5.097 4.6 96 4.696 0.226 0.1608

Blue color LED :

Order Traveling Microscope Reading

Obs Of D D²
L.H.S R.H.S
No. Ring (cm) (cm)
MSR VSR TR MSR VSR TR

st
1) 1 5 21 5.021 4.7 71 4.771 0.250 0.0625

nd
2) 2 5 32 5.032 4.7 49 4.749 0.283 0.0800

rd
3) 3 5 49 5.049 4.7 34 4.734 0.315 0.0922

th
4) 4 5 63 5.063 4.7 29 4.729 0.334 0.1115

th
5) 5 5 75 5.075 4.7 09 4.709 0.366 0.1339

th
6) 6 5 83 5.083 4.6 96 4.696 0.387 0.1497

30
❖ Calculations :

1. for, Red LED :

2 2
● λred = [D n − D m /4R(n − m)]
= [ 0.1806 - 0.1036 / 4×100×(5-2)]
= 0.077/1200
-5 -10
= 6.416 × 10 cm ～ 6416 × 10 m
= 6416 A°

● Eg = hc/λ
-34 8 -10
= [ 6.626 × 10 × ( 3 ×10 )] / 6416 × 10
-16
= 19.878 × 10 / 6416
-19
= 3.0981 × 10 J
-19
● Eg (eV) = Eg / 1.6 × 10
-19 -19
= 3.0918 × 10 / 1.6 × 10
-19
= 1.936 eV …...….. where, 1.6 × 10 eV(charge of electron )

∴ The Red LED is made up of Gallium Phosphide (GaP).

2. for, Green LED :

2 2
● λred = [D n − D m /4R(n − m)]
= [ 0.1521 - 0.0882 / 4×100×(5-2)]
= 0.0638 /1200
-5 -10
= 5.3175 × 10 cm ～ 5317 × 10 m
= 5317 A°

● Eg = hc/λ
-34 8 -10
= [ 6.626 × 10 × ( 3 ×10 )] / 5317 × 10
-16
= 19.878 × 10 / 5317
-19
= 3.7272 × 10 J
-19
● Eg (eV) = Eg / 1.6 × 10
-19 -19
= 3.7272 × 10 / 1.6 × 10
-19
= 2.329 eV …...….. where, 1.6 × 10 eV(charge of electrons )

∴ The Green LED is made up of Gallium Phosphide (GaP).

31
3. for, Yellow LED :
2 2
● λred = [D n − D m /4R(n − m)]
= [ 0.1361 - 0.0681 / 4×100×(5-2)]
= 0.068 /1200
-5 -10
= 5.666 × 10 cm ～ 5666 × 10 m
= 5666 A°

● Eg = hc/λ
-34 8 -10
= [ 6.626 × 10 × ( 3 ×10 )] / 5666 × 10
-16
= 19.878 × 10 / 5666
-19
= 3.5082 × 10 J
-19
● Eg (eV) = Eg / 1.6 × 10
-19 -19
= 3.5082 × 10 / 1.6 × 10
-19
= 2.190 eV …...….. where, 1.6 × 10 eV(charge of electron )

∴ The Yellow LED is made up of Aluminium Arsenide (AlAs).

4. for, Blue LED :

2 2
● λred = [D n − D m /4R(n − m)]
= [ 0.1339 - 0.0800 / 4×100×(5-2)]
= 0.0539 /1200
-5 -10
= 4.491 × 10 cm ～ 4491 × 10 m
= 4491 A°

● Eg = hc/λ
-34 8 -10
= [ 6.626 × 10 × ( 3 ×10 )] / 4491 × 10
-16
= 19.878 × 10 / 4491
-19
= 4.426 × 10 J
-19
● Eg (eV) = Eg / 1.6 × 10
-19 -19
= 4.426 × 10 / 1.6 × 10
-19
= 2.766 eV …...….. where, 1.6 × 10 eV(charge of electron )

∴ The Blue LED is made up of Gallium Nitride (GaN).

32
❖ Graphs :

● for, Red LED :

From the above graph , tan (ϕ) = m = 0.0246

● for, Green LED :

From the above graph , tan (ϕ) = m = 0.0202

33
● for, Yellow LED :

From the above graph , tan (ϕ) = m = 0.0221

● for, Blue LED :

From the above graph , tan (ϕ) = m = 0.0176

34
Calculations from Graph :-

● λ of Red LED :
λ = tan (ϕ) / 4R
= 0.0246 / 4 × 100
-5 -15
= 6.15 ×10 cm ～ 6150 × 10
= 4491 A°

● λ of Green LED :
λ = tan (ϕ) / 4R
= 0.0202 / 4 × 100
-5 -15
= 5.05 ×10 cm ～ 5050 × 10
= 5050 A°

● λ of Yellow LED :
λ = tan (ϕ) / 4R
= 0.0221 / 4 × 100
-5 -15
= 5.525 ×10 cm ～ 5525 × 10
= 5525 A°

● λ of Blue LED :
λ = tan (ϕ) / 4R
= 0.0176 / 4 × 100
-5 -15
= 4.4 ×10 cm ～ 4400 × 10
= 4400 A°

35
❖ Result :

● for Red LED

λ = 6416 A° ……….. by experiment

λ = 6150 A° ……….. by graph
Eg= 1.936 eV

● for Blue LED

λ = 4491 A° ……….. by experiment

λ = 4400 A° ……….. by graph
Eg= 2.766 eV

● for Green LED

λ = 5317 A° ……….. by experiment

λ = 5050 A° ……….. by graph
Eg= 2.329 eV

● for Yellow LED

λ = 5666 A° ……….. by experiment

λ = 5525 A° ……….. by graph
Eg= 2.190 eV

● The Red LED is made up of Gallium Phosphide (GaP).

● The Blue LED is made up of Gallium Nitride (GaN).

● The Green LED is made up of Gallium Phosphide (GaP).

● The Yellow LED is made up of Aluminium Arsenide (AlAs).

36
● Conclusion

The experiment shows the application of Newton’s rings to easily determine

the value of unknown wavelengths of light emitted by LEDs.
LED wavelengths are required for radiometry and photometry applications.
Also there is only a very small amount of error in the wavelength
measurement between the standard and experimental value.
Using an LED for illumination has an advantage that the power requirement
is only about 80mW (max). This stands very attractive against the huge power
requirement by sodium vapor lamps (Minimum is 35 watts).
This is helpful in the regular lab arrangement, especially during the failure of
mains electric power supply, as a 3V supply from dry cells will be enough to
drive LEDs.
It may be noted that it doesn’t require very high intensity light to secure
Newton’s rings. Also cost wise , a set of LEDs along with the power supply etc
works.
And also we can determine the Energy band gap of semiconductor material
using this method.

37
● References

1. Westfall, Richard S. (1980). Never at Rest, A Biography of Isaac Newton. Cambridge

University Press. p. 171. ISBN 0-521-23143-4.

2. Young, Hugh D.; Freedman, Roger A. (2012). University Physics, 13th Ed. Addison
Wesley. p. 1178. ISBN 978-0-321-69686-1.

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