CAUCHY’S CONSTANTS - Kar · 5. The given prism is placed on the prism table such that the parallel beam of light from collimator is made to fall on one of the reflecting surface

CAUCHY’S CONSTANTS.

Aim: To determine Cauchy’s constants ‘a’ & ‘b’ using prism and spectrometer.

Formula:

1. Direct Calculation.

𝑎 = 𝑛𝑏 − 𝑏

𝜆𝑏2 𝑏 =

𝑛𝑏− 𝑛𝑔1

𝜆𝑏2−

1

𝜆𝑔2

2. By Graph. 𝑎 = 𝑂𝐸 𝑖𝑠 𝑡ℎ𝑒 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡

𝑏 = 𝐴𝐵

𝐵𝐶 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑙𝑜𝑝𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑔𝑟𝑎𝑝ℎ.

nb is refractive index of blue light. λb is wave length of blue light.

ng is refractive index of green light. λg is wave length of green light

Principle: This experiment is based on the phenomenon of normal dispersion. i.e., when a ray of white light passes

through a prism it is separated into its constituent colors. According to Cauchy’s theory of normal dispersion,

𝑛 = 𝑎 + 𝑏

𝜆2 where, n is refractive index of the prism, λ is wave length of light

and á’ & ‘b’ are Cauchy’s constants.

Procedure:

1. Least count of the spectrometer is determined.

2. The prism table is levelled using spirit level.

3. The Telescope is adjusted to see the clear image of the distant object.

4. Keeping the telescope in line with the collimator, the slit is made narrow.

5. The given prism is placed on the prism table such that the parallel beam of light from collimator is made

to fall on one of the reflecting surface at almost grazing angle of incidence.

6. The telescope is turned to get the spectrum in the field of view.

7. Now the spectrum is adjusted for minimum deviation position.

8. Reading R1 for yellow-1 is recorded.

9. Similarly reading R1 for green, green-blue-2, blue and violet-1 is recorded.

10. By removing the prism, the spectrometer reading R2 is recorded for direct ray.

11. Refractive index for a particular colour is calculated using 𝑛 = 𝑆𝑖𝑛 (

𝐴+𝐷

2)

𝑆𝑖𝑛(𝐴

2)

12. Cauchy’s constants (a, b) for any two colours can be calculated using formula 1.

13. The Cauchy’s constants (a, b) can also be calculated by plotting a graph of n Vs 1/λ2

Result: Cauchy’s constants are found to be

1. a = ____________________ b = ______________________ m2 from direct calculation.

2. a = ____________________ b = ______________________ m2 from graph.

E

n

O

C B

A

a

OBSERVATIONS: -

Angle of Prism, A = 60°

Least Count of the Spectrometer = 𝑉𝑎𝑙𝑢𝑒 𝑜𝑓 𝑜𝑛𝑒 𝑀𝑆𝐷

𝑇𝑜𝑡𝑎𝑙 𝑛𝑜.𝑜𝑓 𝑉𝑆𝐷= ___________

Total reading = MSR + (CVD X LC)

To find ‘D’and ‘n’

Spectral line

Minimum

deviation

position

reading R1

Direct ray

reading

R2

Minimum

deviation

D = R1 - R2

Refractive Index

𝑛 = 𝑆𝑖𝑛 (

𝐴 + 𝐵2 )

𝑆𝑖𝑛𝐴2

Wavelength

λ x 10-10 m

1

𝜆2 𝑋 1012

Yellow – 1 5798 2.9829

Green 5461 3.3531

Blue 4358 5.2653

Violet 1 4047 6.1056

Substitution & Calculations:

S – Mercury Source

C – Collimator

P – Prism

B – Base of Prism

T – Telescope

D- Angle of minimum Deviation.

IMPEDANCE OF A SERIES RC CIRCUIT

Aim: To determine impedance of a series RC circuit and hence to calculate frequency of AC.

Formula:

1. 𝑍 = 𝑉𝑟𝑚𝑠

𝐼𝑟𝑚𝑠 = ________Ω

2. 𝑍2 = 𝑅2 + 1

𝐶2𝜔2

3. 𝑓 = 1

2𝜋𝑚= ________ Hz

Principle:

When an AC emf is applied to R-C circuit, the impedance is given by 𝑍 = 𝑉𝑟𝑚𝑠

𝐼𝑟𝑚𝑠 and also

𝑍2 = 𝑅2 + 1

𝐶2𝜔2 . When a graph of ‘Z2’ Vs 1

𝐶2 is drawn we get straight line having slope 𝑚 = 1

𝜔2 and

intercept R2. Instead of ‘Z2’ Vs 1

𝐶2 a graph of ‘Z’ Vs 1

𝐶 is drawn a straight line is obtained having slope

𝑚 = 1

𝜔 . Hence 𝜔 =

1

𝑚 and 𝑓 =

1

2𝜋𝑚.

Procedure:

1. Electrical connections are made as shown in figure.

2. Applied AC voltage (Vrms) is noted.

3. For capacitance 0.1µF, VR is noted. Irms is calculated by Irms = (VR/R) Impedance is calculated by

𝑍 = 𝑉𝑟𝑚𝑠

𝐼𝑟𝑚𝑠 This is experimental value of impedance.

4. Experiment is repeated for different values of capacitance.

5. The frequency of AC is calculated using formula 1.

6. A graph of ‘Z’ Vs 1

𝐶 is plotted. Slope of straight line ‘m’is found. Hence the frequency of AC is

calculated using the formula 2.

Result: Frequency of AC, f = ______________ Hz.

Where

Vrms – Applied AC Voltage,

Irms – RMS Value of Current

Z – Impedance

m -Slope of straight lien of Z Vs 1/C

f – frequency of AC = 50 Hz

C – Capacity of Capacitor.

Z

C B

A

1

𝐶

Vrms = 12 Volts R = 2000Ω

𝑺𝒍𝒐𝒑𝒆 = 𝒎 = 𝑨𝑩

𝑩𝑪

Trial C µF VR in Volts 𝑰𝒓𝒎𝒔 = 𝑽𝑹

𝑹 Amps 𝒁 =

𝑽𝒓𝒎𝒔

𝑰𝒓𝒎𝒔 Ω

𝟏

𝑪

1 0.1

2 0.2

3 0.3

4 0.4

5 0.5

6 0.6

7 0.7

Substitutions and Calculation:

Frequency of AC f = ____________ Hz

Where, C-Capacitance

AC =Vrms input voltage

T – Transformer

R- Resistance

VR – Voltage across Resistance

LOW PASS AND HIGH PASS FILTERS

Aim: To study the frequency response of RC low pass and high pass filters and hence to find the cut off

frequency.

Formula: Cut off frequency, 𝒇𝒄 = 𝟏

𝟐𝝅𝑹𝑪 = ________𝑯𝒛 Where R & C are Resistance and Capacitance

respectively.

Principle:

Filter: It is an electronic circuit which passes desired frequencies and rejects other frequencies.

Cut off Frequency: The frequency at which signal strength drops to 1

√2 times the maximum out put voltage.

Low pass Filter: It is an electronic circuit which allows (passes) the signals below the cut off frequency.

High pass Filter: It is an electronic circuit which allows the signals above the cut off frequency.

Procedure:

1. The oscillator output Vi is set for convenient value (say 8 or 10v)

2. The connections are made as shown in circuit 1 (Low pass Filter)

3. By varying the frequency, the output voltage Vo is recorded.

4. Gain 𝐺 = 𝑉𝑜

𝑉𝑖 is calculated.

5. A graph of gain Vs frequency is plotted.

6. A horizontal line corresponding to 𝐺𝑜

√2 is drawn. From this the cut off frequency fc is found.

7. The cut off frequency fc is verified by direct calculation.

8. The experiment is conducted for high pass filter as above (circuit II)

Result:

1. For low pass filter,

a. Cut off frequency, fc = ______________ Hz from graph

b. Cut off frequency, fc = ______________ Hz from direct calculation

2. For high pass filter,

a. Cut off frequency, fc = ______________ Hz from graph

b. Cut off frequency, fc = ______________ Hz from direct calculation

Observations:

Input voltage Vi = 10 or 6 Volt

Resistance R = 2200 Ω

Capacitance, C = 0.1 µf

Low pass filter: High Pass Filter

Frequency

Hz

Output voltage

Vo Volt

𝑮𝒂𝒊𝒏 𝑮

= 𝑽𝒐

𝑽𝒊

Frequency

Hz

Output voltage

Vo Volt

𝑮𝒂𝒊𝒏 𝑮

= 𝑽𝒐

𝑽𝒊

Vo C

R

Vo

C

R

Go

𝐺𝑜

√2

G

f fc fc

Go

𝐺𝑜

√2

G

f

NETWORK THEOREMS

1. MAXIMUM POWER TRANSFER

Resistance

RL Ω 100Ω 200Ω 300Ω 400Ω 500Ω 600Ω 700Ω 800Ω 900Ω 1000Ω

Voltage

VL Volts

Current

I mA

Power

P = VL I Watt

RL

R

R

1 0-50

Ba

(0-

30V)

m

A

V

RL

P

Maximum Power Transfer Theorem:

Aim: To verify maximum power transfer theorem.

Formula: Source Resistance 𝑅𝑠 = 𝑅1𝑅2

𝑅1+𝑅2 = ______________ Ω

Power is maximum when source resistance Rs = Load resistance RL

Principle: In any linear network power delivered by a source to a load resistance is maximum

when the load resistance is equal to internal resistance of the source.

Procedure:

1. Connections are made as shown in figure.

2. The source voltage is fixed to 15 V.

3. The resistances R1 & R2 are in the ratio 1:2

4. The load resistance is varied in steps of 100Ω to 1000Ω and corresponding current in

millimeter and voltage in voltmeter are recorded.

5. Power is calculated for each load resistance.

6. A graph of power against the load resistance is drawn and RL is noted.

Result: It is verified that power is maximum when RL = ______ Ω = Rs = _________Ω

THEVENIN’S THEOREM

Circuit ! Circuit 2

VE = 15 V R1 = 300Ω

R2 = 600Ω

Vth = 10 V

Rth = 200 V

Load

Resistance

RLΩ

Current I mA Current I` mA

RL R2

R1 0-25 mA

Ba

(0-30V)

mA

RL

Rth

0-25 mA

I`

Ba

(Vth)

mA

THEVENIN’S THEOREM

Principle: Any linear network containing linear impedances can be replaced by an equivalent

circuit, Consisting of an impedance with an equivalent voltage source in series.

Formula; The venin’s voltage 𝑉𝑡ℎ = 𝐸𝑅2

𝑅1+ 𝑅2= _______𝑉𝑜𝑙𝑡𝑠

The venin’s Resistance 𝑅𝑡ℎ = 𝑅1𝑅2

𝑅1+𝑅2= ____Ω

Procedure:

1. Connections are made as shown in circuit 1

2. Resistances R1 and R2 are taken in the ratio 1:2

3. The source voltage E is adjusted to 15 V

4. The load resistance is varied in steps of 100Ω and corresponding currents I are

recorded.

5. For the given circuit The venin’s voltage and The venin’s resistance is calculated using

above mentioned formula.

6. Connectors are made as shown in circuit 2. Source voltage is adjusted to Vth keeping

the value of Rth the currents I ‘for different load resistances are noted.

Result: It is observed that I = I`

FIELD ALONG THE AXIS OF THE CIRCULAR COIL

Observations:

Current through the coil I = _____________ amp

No. of turns in the coil used, N = __________

Radius of the coil, r = circumference of the coil/ 2π = ____________ m

From the graph, the value of tanθ at x = r/2 = ______

Permeability of free space µo = 4π x 10-7 H/m

Dist in

Cm

Right side Deflections Left side Deflections

θ1 θ2 Mean θ Tan θ θ1 θ2 Mean θ Tan θ

Calculations: The Horizontal component of Earth’s magnetic field 𝐵𝐻 = 4µ𝑜𝑁𝐼

√125

1

𝑟𝑡𝑎𝑛𝜃 Tesla

Result: The horizontal component of Earth’s magnetic field at the given place is

BH = _____________ Tesla

A +

B

a

Circular Coil Compas

s Scale

K R

h

tan

θ

𝑋

= 𝑟

O

FIELD ALONG THE AXIS OF THE CIRCULAR COIL

Aim: To determine the Horizontal component of Earth’s magnetic field BH by studying the

variation of magnetic field along the axis of a circular coil carrying current.

Apparatus: Circular coil with arms, magnetic compass, ammeter, rheostat, battery etc.

Formula: The Horizontal component of Earth’s magnetic field 𝐵𝐻 = 4µ𝑜𝑁𝐼

√125

1

𝑟𝑡𝑎𝑛𝜃 Tesla

Where r is the radius of the coil, N is number of turns used, I is current through the

circuit, θ is mean deflection, µo is the permeability of free space.

The Field along the axis of the coil at a distance x from the center of the coil is 𝑩 =µ𝒐𝑵𝑰𝒓𝟐

𝟐(𝒓𝟐+𝒙𝟐)𝟐𝟑

Procedure: Coil is connected to an ammeter, battery, key and rheostat in series as shown in the

circuit diagram. The compass box is placed at the center of the coil and the coil with the

perpendicular arms is rotated so that the magnetic needle at the center of the coil is parallel to

the plane of the coil. The compass box is now rotated until the aluminum pointer reads 0° - 0°.

With the compass box at the center of the coil (x=0), the current is adjusted to give a

deflection of about 60°. The readings of the ends of the pointer are noted as θ1 and θ2. The mean

deflection θ is found.

Next the compass box is moved towards right along the axis of the coil in steps of 2cm

at a time. In each position deflections are noted and mean deflection is calculated. Readings are

taken at different positions until θ is reduced to about 30°.

The experiment is repeated by shifting the compass box towards left from the center of

the coil. Readings are recorded as shown in the tabular column.

If the coil is having N turns, the field along the axis of the coil is given by 𝑩 =µ𝒐𝑵𝑰𝒓𝟐

𝟐(𝒓𝟐+𝒙𝟐)𝟐𝟑

Put x = r/2 in the relation B = BH tan θ, we get 𝐵𝐻 = 4µ𝑜𝑁𝐼

√125

1

𝑟𝑡𝑎𝑛𝜃

A graph of tan θ verses distance x is plotted. From the graph the value of tan θ

at x=r/2 is found out. The value of BH is calculated in the above relation.

Result: It is found that the value of BH is _____________

CHARGING AND DISCHARGING OF A CAPACITOR.

Aim: To determine the capacity of a capacitor by studying variation of voltage during charging and discharging.

Apparatus: DC Power Supply, Capacitor, Resistor, Voltmeter, plug Keys, stop watch, Connecting wires, etc.

Formula: 𝐶 = 𝑇1

2

0.693 𝑅𝐹𝑎𝑟𝑎𝑑 Or 𝐶 =

𝑡

𝑅= _________ Farad

By graph, 𝐶 = −1

𝑅 𝑋 𝑆𝑙𝑜𝑝𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑔𝑟𝑎𝑝ℎ = ___________ Farad Where slope = -a/b

Observation: Resistance R = ___________Ω

Charging. (K1 Closed & K2 Open) Discharging (K1 Open & K2 Closed)

Time in t

Second

Voltmeter

(V) reading

in volt

0

5

10

15

20

25

30

35

40

45

50

55

60

Time in t

Second

Voltmeter

(V) reading

in volt

0

5

10

15

20

25

30

35

40

45

50

55

60

K1

K2

Ba

C

R

- +

Where,

Ba - Battery

K1 & K2 – Plug Keys

R – Resistance

C – Capacitor (electrolytic)

V - Voltmeter

CHARGING AND DISCHARGING OF A CAPACITOR.

Aim: To study the voltage variation on capacitor during charging and discharging and hence to find the time

constant.

Apparatus: Capacitor (2200µF, 4700µF), resistor (10kΩ), Battery (12V), Plug keys, stop clock.

Procedure: 1. Charging of a capacitor:

Circuit connections are made as shown in figure. To study the voltage across the capacitor during

charging, the key k1 is closed and simultaneously and simultaneously a stop clock is started. The voltmeter

readings are noted at intervals of 5 seconds till the capacitor attains maximum voltage (equal to applied voltage)

readings are tabulated.

A graph of voltage across the capacitor against time t is plotted. From the graph, the time T1/2

required to attain Vmax/2 is noted.

The time constant T of the RC circuit can be calculated using the formula 𝑇 = 𝑇1

2

0.673

Or T = 0.632 Vmax

Which is also equal to the theoretical value of T = RC or C = T/R

1. During Charging

The time constant T of the RC circuit can be calculated

using the formula

𝑇 = 𝑇1

2

0.673 or T = 0.632 Vmax

Which is also equal to the theoretical value of T = RC

or C = T/R

2. During Discharging

The time constant T of the RC

Circuit can be calculated using the formula

𝑇 = 𝑇1

2

0.693 or T = 0.368 Vmax

Which is also equal to the theoretical

value of T = RC or C = T/R.

t in sec

Vmax/2

Vmax

𝑇12

O

t in sec

Vmax/2

Vmax

𝑇12

O

AIR WEDGE

1. To find fringe width β

Fringe

No.

Traveling microscope

reading R1

Fringe

No.

Travelling microscope

Reading R2

Fringe width

𝜷 = 𝑹𝟏~𝑹𝟐

𝟐𝟎

0 20

5 25

10 30

15 35

Mean β = ____________ cm = ______________m

II to find length of air wedge (L)

Reading corresponding to line of contact L1 = ____________ cm,

Reading corresponding to wire L2 = ____________ cm,

Length of air wedge, L1 ~ L2 = _________ cm = __________ m

Calculation:

𝒕 = 𝝀𝑳

𝟐𝜷

L

Plane Glass

Plates

Microscop

e

Reflector

45°

wire

Observation:

Least count of travelling microscope =

𝑽𝒂𝒍𝒖𝒆 𝒐𝒇 𝟏 𝑴𝑺𝑫

𝑵𝒐.𝒐𝒇 𝒅𝒊𝒗𝒊𝒔𝒊𝒐𝒏𝒔 𝒐𝒏 𝒗𝒆𝒓𝒏𝒊𝒆𝒓 𝒔𝒄𝒂𝒍𝒆 = ________ cm

T R = MSR + (CVD X LC)

Air Wedge

Aim: Determination of thickness of thin wire using interference at a wedge shaped air

Film.

Formula: 𝒕 = 𝝀𝑳

𝟐𝜷 Where λ is wavelength of sodium light = 5893 x 10-10 m

β is fringe width & L is length of air wedge.

Principle:

When a monochromatic beam of light incident normally on the air wedge of

thickness ‘t’, it suffers reflection at the top and bottom surfaces of the air film. The

reflected light rays are coherent and undergo interference producing equidistant

alternative bright and dark fringes parallel to the object in an air wedge. The thickness of

the object is directly proportional to wavelength, the length of air wedge and inversely

proportional to bandwidth. Hence by measuring the bandwidth and length of air wedge

the thickness of an object can be determined.

Procedure:

1. Air wedge is formed between two optically flat glass plates due to the wire whose

thickness is to be determined.

2. The reflector is adjusted for the proper incidence of light on the air wedge. The

eyepiece of traveling microscope is adjusted to see the clear view of interference

fringes.

3. Least count of traveling microscope is determined.

4. The vertical cross wire of the traveling microscope is made to coincide with one of

the dark fringes at the left side and is considered as zero fringe, the reading

corresponding to it is noted.

5. By moving the cross wire, the readings corresponding to every five fringes upto 35

fringes are noted & fringe width β is calculated.

6. The microscope readings corresponding to line of contact between two glass plates

L1 and to the wire L2 are noted. The length of air wedge is given by L = L1~L2.

7. The thickness of wire ‘t’ is determined using the above mentioned formula.

Result:The thickness of given wire ‘t’ = _____m.

Biprism

I To find the band width

Distance of

Eyepiece

form slit ‘m’

Fringe

No.

Micrometer

reading

á’cm

Fringe

No.

Micrometer

reading

‘b’cm

Width of

‘10’fringes

x = (a~b)

cm

Mean

‘x’cm

Band

width β

=(x/10)cm

0 10

2 12

4 14

6 16

8 18

I;mages of two virtual

sources

Micrometer reading Distance between two sources

d = (R1~R2) cm Image I1 R1 cm Image I2 R2 cm

Maginified image

Diminished image

𝑑 = √𝑑1𝑑2 = ______ cm

B

A

N

M

S d

c

b

a

D

Fig-1

S – Monochromatic Source

A, B – Virtual Sources

Abc – Biprism

E – Eye piece

MN- Screen

d1 d2

Fig-2 A Fig-2 B

Observations:

Least count of micrometer= 𝑉𝑎𝑙𝑢𝑒 𝑜𝑓 1 𝑃𝑆𝐷

𝑁𝑜.𝑜𝑓 𝑑𝑖𝑣𝑖𝑠𝑖𝑜𝑛𝑠 𝑜𝑛 𝐻𝑆 = _______ cm

TR = PSR +( HSR x LC)

Biprism

Aim: Determination of the wave length ‘λ’ of a monochromatic light using biprism.

Formula: 𝝀 = 𝛽

𝐷 √𝑑1𝑑2 mm

Where β is the band width measured at a distances D meter. d1 & d2 are the distance

between two coherent sources in magnified and diminished forms.

Principle: When a monochromatic light falls on refracting edge of a biprism, two virtual images of the

line sources are formed by division of wave front. These two forms the coherent sources and interfere

forming interference fringes.

Procedure: I To find Fringe width: Fig 1

1. Least count of micrometer fixed to eyepiece is found.

2. The slit, biprism & eyepiece are made collinear. This gives interference fringes.

3. The eyepiece stand is fixed at a distance of about 40cm (D) from slit.

4. The cross wire of eye piece is coincided to one of the fringes on the left side of the fringe

system and it is considered as zero fringe. The reading corresponding to it is noted.

5. By moving the micrometer drum the reading corresponding to every five fringes up to 18

fringes is noted and band width β is calculated as explained in the tabular column

II To find the distance between two virtual coherent sources: Fig 2

1. The lens upright is introduced between eyepiece and Biprism.

2. The lens and eyepiece are adjusted for clear view of interference fringes.

3. By moving the lens upright towards the biprism, the readings corresponding to magnified

images are noted (R1 & R2) and d1 = (R1~R2) is calculated.

4. By moving the lens upright towards the eyepiece, the readings corresponding to diminished

images are noted (R1 & R2) and d2 = (R1~R2) is calculated.

5. Wavelength of monochromatic light is calculated using the above formula.