new sample papers by cbse .......

http://cbseacademic.in/web_material/SQP/CLASS%20XII_2015_16/SQP-Physics%202016%20SQP.pdf

atom, nuclei & radioactivity notes

https://www.dropbox.com/s/7t9hvel390ia9su/atom%2C%20nuclei%20%26%20radioactivity.pdf?dl=0

interference in accordance to law of conservation of energy

interference of light

Interference of Light

 Interference of light is the phenomenon of redistribution of light energy in a medium on account of superimposition of light waves from two coherent sources. At the point where the resultant intensity of light is maximum interference is said to be constructive and at the points where the resultant intensity of light is minimum interference is said to be destructive.

Young�s Double Slit Experiment

The above experiment gives a rough experimental set up of Young's experiment.

Since light has a very small wavelength, we need two slits, which send out two continuous coherent waves. Since the two slits are placed very close, these waves overlap as shown below.

Explanation

 The condition for constructive interference and destructive interference

If y₁ = A Sin wt and y₂ = B Sin ( wt + f) represent the waves of two coherent sources with A and B as their respective amplitude and f the constant phase difference between the waves.

Then from superposition principle,

y = y₁ + y₂

= A sin wt + B sin( wt + f )

= sin wt( A+ B cos f) + cos wt B sin f

y = R sin (wt + f)

(On expanding the second term)

i.e. put a + cos f = R cos f

Since intensity I a (amplitude) ²

I f (a² + b² + 2ab cosf)

For constructive interference, intensity (I) is maximum when

cosf = +1 or f = 0, 2 p, 4p

f = 2n p where n=0,1,2

Now a path difference of l corresponds to the phase difference 2p. If x is the path difference, then

(Where f is the phase difference)

x = nl

For destructive inference, intensity should be minimum.

cos f = -1

f = p,3 p,5 pp�..

f = (2n - 1)p where n=1,2

Expression for Fringe Width in Interference

Let A and B be two fine slits, a small distance 'd' apart. Let them be illuminated by a monochromatic light of wavelength l.

MN in the screen is at a distance D from the slits AB. The waves from A and B superimpose upon each other and an interference pattern is obtained on the screen. The point C is equidistant from A and B and therefore the path difference between the waves will be zero and so the point C is of maximum intensity. It is called the central maximum.

For another point P at a distance 'x' from C, the path difference at P = BP - AP.

Now AB = EF = d, AE = BF = D

\D BPF

[Pythagoras theorem]

Similarly in D APE

(on expanding Binomially)

For bright fringes (constructive wavelength) the path difference is integral multiple of wavelength i.e., path difference is nl.

(x therefore represents distance of n^th bright fringe from C)

Now

and so on.

Therefore separation between the centers of two consecutive bright fringe is the width of a dark fringe.

Similarly for dark fringes,

The separation between the centers of two consecutive dark interference fringes is the width of a bright fringe.

The separation between the centers of two consecutive dark interference fringes is the width of a bright fringe.

All bright and dark fringes are of equal width as b₁ = b₂.

The intensity of all bright bands are the same. All dark bands also have same (zero) intensity. The intensity distribution Vs distance is shown as:

Conditions for sustained interference:

(1) The sources must be coherent (i.e., they must maintain a

constant phase relationship with one another).

(2) The sources must be monochromatic (i.e., of a single

wavelength).

(3) The linear superposition principle is applicable.

alpha, beta decay...... and universal gates

Expression for torque..... rotational motion

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https://www.google.co.in/url?sa=t&rct=j&q=&esrc=s&source=web&cd=2&cad=rja&uact=8&ved=0CCMQtwIwAWoVChMIxLqljZXuxwIVx5GOCh1IpQaa&url=http%3A%2F%2Fwww.youtube.com%2Fwatch%3Fv%3DxMqbZNJO_ws&usg=AFQjCNFYH_jzvBWDsFQlun_nSc1u4B7L1A&sig2=yMSEwCvHlCTaaMrOJORN2A&bvm=bv.102022582,d.c2E

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potential due to a dipole

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Draw a em wave propagating along x axis and give direction  of E and B.  Find the expression  for velocity  of light from maxwell equations.

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Solve these questions

QUESTION OF THE DAY.......... WITH ANSWER 16 AUG

SARVIR SINGH STARTING A NEW BLOG

Hello my dear students i have made this blog in which i will add different concepts of Physics.  you all can join it and ask questions also.....  hopefully  this will be fruitful  attempt......... SARVIR SINGH
                          DPS AMRITSAR