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Interference And Diffraction-Classical Physics-Handouts, Lecture notes of Classical Physics

University of Allahabad Classical Physics

This course includes alternating current, collisions, electric potential energy, electromagnetic induction and waves, momentum, electrostatics, gravity, kinematic, light, oscillation and wave motion. Physics of fluids, sun, materials, sound, thermal, atom are also included. This lecture includes: Interference, Diffraction, Constructive, Amplitude, Observing, Destructive, Phase, Uniform, Coherent, Illumination, Fringe

Typology: Lecture notes

2011/2012

Uploaded on 08/12/2012

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PHYSICS –PHY101 VU

102

destructive

constructive

Summary of Lecture 33 – INTERFERENCE AND DIFFRACTION

1. Two waves (of any kind) add up together, with the net result being the simple sum of the

two waves. Consider two waves, both of the same frequency, shown below. If they start

together (i.e. are with each other) then the net amplitude is increased. This is

called . But if they start at different times (i.e. are

with each other) th

in phase

constructive interference out of phase

en the net amplitude is decreased. This is called .destructive interference

In the example above, both waves have the same frequency and amplitude, and so the

resulting amplitude is doubled (constructive) or zero (destructive). But interference

occurs for any two waves even when their amplitudes and frequencies are different.

2. Although any waves from different sources interfere, if one

wants to observe the interference of light then it is necessary

to have a source of light. Coherent means that both

coherent

waves should have a fixed phase relative to each other. Even

with lasers, it is very difficult to produce coherent light from

two separate sources. Observing interference usually requires

taking two waves from a single source, with each going along a

different path. In the figure, an incoherent light source

illuminates the first slit. This creates a uniform and coherent

ill

12

umination of the second screen. Then waves from the slits

S and S meet on the third screen and create a pattern of

alternating light and dark fringes.

3. Wherever there is a bright fringe, constructive interference has occured, and wherever

there is a dark fringe, destructive interference has occured. We shall now calculate

where on the third

12

screen the interference is constructive. Take any point on the third

screen. Light reaches this point from both S and S , but it will take different amounts

of time to get there. Hence there will be a phase difference that we can calculate. Look

at the diagram below. You can see that light from one of the slits has to travel an extra

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constructive destructive

Summary of Lecture 33 – INTERFERENCE AND DIFFRACTION

Two waves (of any kind) add up together, with the net result being the simple sum of the two waves. Consider two waves, both of the same frequency, shown below. If they start together (i.e. are with each other) then the net amplitude is increased. This is called. But if they start at different times (i.e. are with each other) th

in phase constructive interference out of phase en the net amplitude is decreased. This is called destructive interference.

In the example above, both waves have the same frequency and amplitude, and so the resulting amplitude is doubled (constructive) or zero (destructive). But interference occurs for any two waves even when their amplitudes and frequencies are different.

Although any waves from different sources interfere, if one wants to observe the interference of light then it is necessary to have a coherent source of light. Coherent means that both waves should have a fixed phase relative to each other. Even with lasers, it is very difficult to produce coherent light from two separate sources. Observing interference usually requires taking two waves from a single source, with each going along a different path. In the figure, an incoherent light source illuminates the first slit. This creates a uniform and coherent ill 1 2

umination of the second screen. Then waves from the slits S and S meet on the third screen and create a pattern of alternating light and dark fringes.

Wherever there is a bright fringe, constructive interference has occured, and wherever there is a dark fringe, destructive interference has occured. We shall now calculate where on the third 1 2

screen the interference is constructive. Take any point on the third screen. Light reaches this point from both S and S , but it will take different amounts of time to get there. Hence there will be a phase difference that we can calculate. Look at the diagram below. You can see that light from one of the slits has to travel an extra

distance equal to sin , and so the extra amount of time it takes is ( sin ) /. There will be constructive interference if this is equal to , 2 , 3 , (remember that the time period is

d d c T T T

⋅ ⋅ ⋅ inversely related to the frequency, 1/ , and that ). We find that sin sin , where 1, 2,3, What about for destructive interference? Here the waves will

T c d (^) nT n d n c n

cancel each other if the extra amount of time is , 3 , 5 , The 2 2 2 condition then becomes sin ( 1 ). 2

T T T

d θ n λ

Example: two slits with a separation of 8.5× 10 m^ - create an interference pattern on a screen 2.3m away. If the 10 bright fringe above the central is a linear distance of 12 cm from it,

n =

1 1

what is the wavelength of light used in the experiment?

Answer: First calculate the angle to the tenth bright fringe using tan. Solving for gives, tan tan 0.

y L y m L m

θ −^ −

= ⎛^ ⎞^ = ⎛^ ⎞

⎜⎝ ⎟⎠ ⎜⎝ ⎟⎠ =^ 3.0. From this,D

sin sin(3.0 ) 4.4 10 440 (nanometres). 10

d (^) m nm n λ θ

⎛ × − ⎞ −

= = ⎜ ⎟ = × =

D

When a wave is reflected at the interface of two media, the phase will not change if it goes from larger refractive index to a smaller one. But for smaller to larger, there will be phase change of a half-wavelength. One can show this using Maxwell's equations and applying the boundary conditions, but this will require some more advanced studies. Instead let's just use this fact below.

We can get an interference pattern with a single slit provided its size is approximately equal to the wavelength of the light (neither too small nor too large).

Let's work out the condition necessary for diffraction of light from a single slit. With reference to the figure, imagine that a wave is incident from the left. It will cause secondary waves to be radiated from the edges of the slit. If one looks at angle , the extra distance that the wave emitted from the lower slit must travel is W sin. If this is a multiple of the wave

θ length , then

constructive interference will occur. So the condition becomes sin with 1, 2, 3... So, even from a single slit one will see a pattern of light and dark fringes

W m m

when observed from the other side.

Light with wavelength of 511 nm forms a diffraction pattern after passing through a single slit of width 2.2×10 m. Find the angle associated with (a) the first and (b) the second bright fr 1 1 9 6

1 1 9

inge above the central bright fringe.

SOLUTION: For 1, sin sin (1)(511^10 13. 2.20 10

For 2, sin sin (2)(511^10 2.20 10

m m^ m W m

m m^ m W

− − − −

− − −

⎛ ⎞ ⎛^ × ⎞

⎝ ⎠ ⎝ × ⎠

= = ⎛^ ⎞= ×

⎜⎝ ⎟⎠ ×

D

− (^6) m 27.

D

Diffraction puts fundamental limits on the capacity of telescopes and microscopes to separate the objects being observed because light from the sides of a circular aperture inter

min

feres. One can calculate that the first dark fringe is at 1.22 , where D is diameter of the aperture. Two objects can be barely resolved if the diffraction maximum of one obje

D

θ =^ λ

ct lies in the diffraction minimum of the second object. Clearly, the larger D is, the smaller the angular diameter separation. We say that larger apertures lead to better resolution.