ADINA INSTITUTE OF SCIENCE AND TECHNOLOGY

Report on

“Polarisation”

SUBMITTED BY:

VAISHALI CHOURASIA

BRANCH : E.C.

POLARISATION

DEFINATION :

The phenomenon in which waves of light or other radiation are restricted in direction of vibration is known as polarisation.

CONCEPT OF POLARISATION LIGHT :

•The phenomenon of interference and diffraction shows that light is a form of wave motion.

•These phenomenon could not explain the types of wave motion i.e. whether the light waves are longitudinal or transverse or whether the vibrations are linear , circular , elliptical or torsional.

•These shortcomings were explained by phenomenon known as polarisation of light.

PLANE POLARISED LIGHT :The plane polarised light is the light in which the vibration take place only along one straight line perpendicular to the direction of light.

UNPOLARISED LIGHT : The unpolarised light is the light in which the vibration takes place along all possible straight lines perpendicular to the direction of propagation of light.

PLANE OF VIBRATION :The plane containing the direction of vibration and direction of propagation of light is called the plane of vibration.

PLANE OF POLARISATION : The plane containing the direction of propagation of light , but containing no vibrations is called the plane of polarisation. Plane of polarisation is always at right angles to the plane of vibration.

BREWSTER’S LAW : This law was discovered by Malus . Acc. To him , when an unpolarised light is incident on the surface of any transparent material, the reflected and refracted beams are partially plane polarised. For a particular angle of incident called as the “ polarising angle” the reflected light is completely plane polarised with the plane of vibration perpendicular to the plane of incident.

µ = tan ip

This is known as BREWSTER’S LAW.

ADINA INSTITUTE OF SCIENCE AND TECHNOLOGY

Report on

“IR Spectroscopy”

SUBMITTED BY:

VAISHALI CHOURASIA

BRANCH : E.C.

IR SPECTROSCOPY

Infrared spectroscopy deals with the interaction between a molecule and radiation from IR region of the EM spectrum ( IR region = 4000-400 cm-1 ). The cm-1 is the unit wave no. scale and is given by 1/( wavelength in cm ) .

IR radiation causes the excitation of the vibration of covalent bonds within that molecule . These vibrations include the stretching and bending modes.An IR spectrum shows the energy absorption one ‘scans’ the IR region of the EM spectrum .

In general terms it is convenient to split an ir spectrum into two appropriate regions.

  • 4000 – 1000 cm-1 known as the functional group region ,and
  • <1000 cm-1 known as the finger print region.

Most of the information that is used to interpret an IR spectrum is obtained from the functional group region. In practice , it is the polar covalent bonds that are IR ‘active’ and whose excitation can be observed in an IR spectrum.

In organic molecules these polar covalent bonds represent the functional group.Hence, the most useful information obtained from an IR spectrum is what functional group are present within the molecule. In the fingerprint region, the spectra tends to be more complex and much harder to assign.

ADINA INSTITUTE OF SCIENCE AND TECHNOLOGY

Report on

“Mechanics”

SUBMITTED BY:

VAISHALI CHOURASIA

BRANCH : E.C.

MECHANICS

Ques. From top of a tower , 60m high a bullet is fired at an angle of 20 degree up the horizontal with velocity 120m/sec. Determine

i)Time of flight

ii)Horizontal range of ground

iii)Maximum height of the bullet from the ground.

Soln. Formulae used

• Time of flight is t = 2VOSinϴ/g

• Horizontal range of ground R = Vicosϴt

i) Given Vo = 120 m/sec

g = 9.8 m/ sec

= 20 degree

Time of flight = 2VoSinϴ /g

= (2*120*sin20)/9.8

= 8.378

= 8.38 sec.

ii ) Horizontal range of ground

R = ViCosϴ*t

= 120*cos20*8.38

= 120*0.939*8.38

= 944.258 meters.

iii ) tanϴ= P/B

P = B*tanϴ

= 472*tan20

= 171.760

Therefore , height of the bullet from the ground = 171.160 + 60

= 231.760 m.

ADINA INSTITUTE OF SCIENCE AND TECHNOLOGY

Report on

“Mathematics”

SUBMITTED BY:

VAISHALI CHOURASIA

BRANCH : E.C.

MATHEMATICS

QUES. Find the maxima and minima of the function x3+y3-3x+12y+20. Also find the saddle point.

Soln. Given eqn f = x3+y3-3x+12y+20

∂f/∂x = 3x2 – 3 ∂f/∂y = 3y2 + 12

∂2f/∂x2 = 6x = r ∂2f/∂y2 = 6y = t

and

∂2f/∂x∂y = 6 = s

For maxima and minima

Put ∂f/∂x = 0 Put ∂f/∂y = 0

3x2 – 3 = 0 3y2 + 12 = 0

3x2= 3 3y2= -12

X = ± 1 y = ± 2i

Since the values of the variable is in the form of imaginary terms hence this case is not exist and question may be wrong.