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### L.-Interference.

posted Jul 1, 2016, 3:07 AM by Upali Salpadoru   [ updated Jul 13, 2016, 4:16 PM ]
Reflectitive diffraction and Interference.
Sunlight getting reflected by a CD.

In the case of a water wave, which is a transverse progressive wave,  the particles only bob up and down. Fig.1 The blue wave moves to the right and take the position shown in red.

In order for the blue wave to take the position shown by the red, some particles should move up while others go down. Try to determine which way the particles shown in black will move.

1………  2……..   3……… 4………… Fig.2 Traveling of the bllue wave to the red.

The second position of the particles re shown by the blue dots. While numbers 1,and 3 have gone up numbers 2 and 4 have come down.

This is not wave interference. Interference occurs when two waves trvelling in a particular medium blend to form a single wave.

In this figure the red wave merges with the blue wave . What happens to the particles now? A particle cannot move up and down at the same time.

We have to find the sum of the two amplitudes considering the sign. Fig.2.  Merging of a wave due to interference.

At  ….A…...    1 + -2 = -1

At …..B……    0 + 2  =  2

At …..C …..    -0.5 + - 2 = - 2.5

We have to take these readings and draw the combined wave. Fig.3 The resulting wave after interference.

Constructive and destructive interference.

When two waves merge, the amplitude has to change. Considering the red wave in Fig.3 at A +1 has become -2; this is a case of a crest meeting a trough. This is destructive interference as the energy of the particle has been reduced due to the opposing nature of the two waves. At C particle has been pushed further down. This is constructive interference.

For a better understanding, let us study Fig.4. S

Fig.4. Constructive and destructive interference.

As two waves travel towards each other at a particular time the red numbered points will meet the corresponding blue numbers. Red no. 1 will have to merge with blue no.1. The red no. 2 merges with blue no.2 and so on. Let us see what the results would be.

 No. Red amplitude Blueamplitude. Resultingamplitude. Type of interference. 1 +1 -2 -1 Destructive 2 -1 +1 0 Destructive 3 +1 +0.8 +1.8 Constructive. 4. 5. 6. 7. 8.

In this table 3 rows have been completed. It will be interesting to do the rest on your own.

When a crest meets a crest or a trough meets a trough there is constructive interference.
When the opposing phases meet there is destructive interference.

Phase in a wave.
Phase describes the position of a point in the wave form. This can be given in degrees. A complete cycle is really a circle. This is illustrated in the Fig.5.

Fig.5 Phase in a wave.

A redcrest and the blue trough forms a complete cycle and the total horizontal distance covered is the wave length.

In Fig 5B. the crest and the trough are brought in a vertical line. In C the cycle is divided into 4 right angles. This is how a cycle gets 360 degree.
There is a phase difference between the points a, b, c and d. Can you find them?
a to b = 90
a to c= 180
a to d = 270 and  'a' to 'a' back again 360.

In 1801 Thomas Young 1773-1829 separated light to spread from 2 points and observed the bright and dark lines formed on a distant screen. He was able to show that light behaves as a wave contrary to the corpuscular theory of Sir Isaac Newton.

He was also able to measure the wave length of light.

The experiment is even used today in a modified form with a laser beam.  Light rays from S1 and S2, shown in red, meet at N1. As N1 is the first antinodal point , the distance from S2 to N could be only 1 wavelength longer. This difference is shown by ‘p’.

It is not easy to measure ‘p’ directly but if we project the interference pattern to a distance ‘x’ will become larger.

Now let us consider these two triangles and try to calculate ‘P’ by measuring ‘y’. In the two coloured triangles, S1BS2,  and SNC,

Angle S1BS2  = angle NCS..  (Right angles.)

Angle S1S2B is approximately equal to SNC (Corresponding angles)

Then, angle θ of yellow triangle becomes equal to angle θ of the blue triangle.

Therefore the two triangles are similar.

Then  P / d  =  y / SN

But SN is approximately equal to L.

Then  P/d = y/L   That is path difference = dy / L

If we consider the first anti node then this value is the wavelength.

Another way to derive this would be:-
Sine θ of yellow = sine θ blue
ie. nP /d = y / SN (n is no. of the nodal point) These are really anti nodal points.
nP = dy/L ( But SN is approximately = L)
If the place value P is given in nodes we have to use (n-1/2) P = dy / L.

Diffraction and Interference. Diffraction through many slits is similar to two slit diffraction. Parallel beam of a coloured light passing through many slits more than 20 per mm. produces very sharp of bright and dark areas, This method can separate the colours in white light better than a prism.

The equation used in this respect is almost the same as for double slit diffraction.

If there are ‘n’ slits in the length ‘d’

The size of the gap ‘e’ will be      e =  d/n

If we take m1 the path difference = λ

Sine Θ =  λ / e

Then considering the larger triangle

λ / e   =   sy/L

Λ    = exy / L

Reflective diffraction

The colours we see in a CD or a DVD is due to reflective refraction. In a CD there are microscopic pits arranged in grooves which are 1.6 micrometers (10-8) apart, This amounts to 625 lines per mil meter.

"Binary data (0's and 1's) are encoded on a compact disc in the form of pits in the plastic substrate which are then coated with an aluminum film to make them reflective.The data is detected by a laser beam which tracks the concentric circular lines of pits. The pits are 0.8 to 3 micrometers long and the rows are separated by 1.6 micrometers."

Courtsey: Georgia State University. Velocity of sound in air  330ms-1
Coloured spectrum

Q.1.0 A sound wave is produced by the 2 speakers which are 1.5 m apart.  The loudness of the sound was tested10m away. Dark dots show loud notes. Find the following:-

1. Path difference.

2. Wave length.

3. Frequency.

Q.2.0
Light from two sources 0.5mm. apart  forms bright parallel fringes on a screen due to interference 5.mm apart.The distance to screen was 4 m. Find the wavelength of this mono coloured light.
What would be the colour of light?
Q.3.0

CD has 625 tracks per mm.  A laser beam gives m1 diffraction at 23°   Find the wave length and the colour of the light.

Q.4.0

A ray of light was projected from 2 sources 1.0 mm apart for a distance of 10 m.The distance to the 3rd dark area in the fringes was 1 cm. Find the wave length and colour of the light.

Q.5.0

Monochromatic Light comes from 2 sources 0.2mm apart. A screen 1.5 m away produced fringes of 5 bright parallel lines on 1 cm .

1. Wavelength.

2. Frequency.

3. Colour.

1.0         PD / 1.5  =  1/ 4  (1m. is the distance between 2 antinodes)

.                       PD =  0.25 x1.5

=0.375 m.    (corrected to).38m)

1.2       0.375. M.

1.3     Frequency = velocity / wavelength

F  = 330 / 0.375…………………=    880Hz.
2.0 .λ / d  = y/L

λ  = dy /L

= 0.5x10-3 x 5x10-3 / 4

= 2.5x10-6 / 4

= 6.25x10-7m     =625 nm.

Colour red.

3.0    λ  = sine 23° x1/625 x10 -3

= 6.25 x10-7m. =625 nm.

4.0 (m-.1/2).Lam =  dy /L

(3-1/2)= 1x10-3  x1x10-2/ 2.5x10

=1x10-5 /25 =  4x10-7  m.  =  400 nm.  colour Blue.