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PPT – Chapter 37: Interference of Light Waves PowerPoint presentation | free to download - id: 34f96-NjcwN

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Chapter 37 Interference of Light Waves

Wave Optics

- Wave optics is a study concerned with phenomena

that cannot be adequately explained by geometric

(ray) optics - These phenomena include
- Interference Diffraction Polarization
- Interference
- In constructive interference the amplitude of the

resultant wave is greater than that of either

individual wave - In destructive interference the amplitude of the

resultant wave is less than that of either

individual wave - All interference associated with light waves

arises when the electromagnetic fields that

constitute the individual waves combine

37.1 Conditions for Interference

- To observe interference in light waves, the

following two conditions must be met - 1) The sources must be coherent
- They must maintain a constant phase with respect

to each other - 2) The sources should be monochromatic
- Monochromatic means they have a single wavelength

Producing Coherent Sources

- Light from a monochromatic source is used to

illuminate a barrier - The barrier contains two narrow slits
- The slits are small openings
- The light emerging from the two slits is coherent

since a single source produces the original light

beam

Diffraction

- From Huygenss principle we know the waves spread

out from the slits - This divergence of light from its initial line of

travel is called diffraction

Resulting Interference Pattern

- The light from the two slits forms a visible

pattern on a screen - The pattern consists of a series of bright and

dark parallel bands called fringes - Constructive interference occurs where a bright

fringe occurs - Destructive interference results in a dark fringe

Interference Patterns

- Constructive interference occurs at point P
- The two waves travel the same distance
- Therefore, they arrive in phase
- As a result, constructive interference occurs at

this point and a bright fringe is observed - The upper wave has to travel farther than the

lower wave to reach point Q - The upper wave travels one wavelength farther
- Therefore, the waves arrive in phase
- A second bright fringe occurs at this position

Interference Patterns, final

- The upper wave travels one-half of a wavelength

farther than the lower wave to reach point R - The trough of the bottom wave overlaps the crest

of the upper wave - This is destructive interference
- A dark fringe occurs

Youngs Double-Slit Experiment Geometry

- The path difference, ?, is found from the tan

triangle - ? r2 r1 d sin ? (37.1)
- This assumes the paths are parallel
- Not exactly true, but a very good approximation

if L gtgt d

Interference Equations

- For a bright fringe produced by constructive

interference, the path difference must be either

zero or some integral multiple of the wavelength - ? d sin ? bright m? (37.2)
- m 0, 1, 2, m is called the order number
- When m 0, it is the zeroth-order maximum
- When m 1, it is called the first-order maximum
- When destructive interference occurs, a dark

fringe is observed - This needs a path difference of an odd half

wavelength - ? d sin ?dark (m 1/2)? (37.3)

m 0, 1, 2,

Interference Equations, 2

- The positions of the fringes can be measured

vertically from the zeroth-order maximum - Assumptions
- L (m) gtgt d (mm), d (mm) gtgt ? (nm)
- Approximation
- ? is small and therefore the small angle

approximation tan ? sin ? can be used - y L tan ? L sin ? (37.4)

Interference Equations, final

- From Equation (37.2) sin ? m?/d and back

substitution into (37.4) gives - For bright fringes (37.7a)
- For dark fringes (37.7b)
- Youngs double-slit experiment provides a method

for measuring wavelength of the light - This experiment gave the wave model of light a

great deal of credibility - It was unthinkable that particles of light could

cancel each other in a way that would explain the

dark fringes

37.3 Intensity Distribution Double-Slit

Interference Pattern

- Note that the bright fringes in the interference

pattern do not have sharp edges - The equations developed give the location of only

the centers of the bright and dark fringes - We can calculate the distribution of light

intensity associated with the double-slit

interference pattern

Intensity Distribution, Assumptions

- Assumptions
- The two slits represent coherent sources of

sinusoidal waves - The waves from the slits have the same angular

frequency, ? - The waves have a constant phase difference, ?
- The total magnitude of the electric field at any

point on the screen is the superposition of the

two waves

Intensity Distribution, Electric Fields and

Phase Difference

- The magnitude of each wave at point P can be

found - E1 Eo sin ?t E2 Eo sin (?t

?) (37.8) - Both waves have the same amplitude, Eo
- The phase difference between the two waves at P

depends on their path difference ? r2 r1 d

sin ? (37.1) - A path difference of ? corresponds to a phase

difference of 2p radians - A path difference of ? is the same fraction
- of ? as the phase difference ? is of 2p
- Therefore (37.9)
- (37.9)

Intensity Distribution, Resultant Field

- The magnitude of the resultant electric field

comes from the superposition principle - EP E1 E2 Eosin ?t sin (?t ?)

(37.10) - Recall
- This allows us to write (37.10) as
- (37.11)
- EP has the same frequency as the light at the

slits - The magnitude of the field is multiplied by the

factor 2cos (? / 2)

Intensity Distribution, Equations

- The expression for the intensity comes from the

fact that the intensity of a wave is proportional

to the square of the resultant electric field

magnitude at that point - (37.12)
- Using equation (37.8), the intensity will be
- (37.13)
- (37.14)

Light Intensity, Graph

- The interference pattern consists of equally

spaced fringes of equal intensity - This result is valid only if L gtgt d and for small

values of ?

Multiple Slits, Intensity Graphs

- Figure shows I vs dsin?
- For three slits notice that The primary maxima

are nine times more intense than the secondary

maxima - The intensity varies as ER2
- For N slits, the primary maxima is N2 times

greater than that due to a single slit

Multiple Slits, Final Comments

- As the number of slits increases, the primary

maxima increase in intensity and become narrower - As the number of slits increases, the secondary

maxima decrease in intensity with respect to the

primary maxima - As the number of slits increases, the number of

secondary maxima also increases - The number of secondary maxima is always
- N 2 where N is the number of slits

37.5 Phase Changes Due To Reflection - Lloyds

Mirror

- An arrangement for producing an interference

pattern with a single light source - Waves reach point P either by a direct path or by

reflection - The reflected ray can be treated as a ray from

the source S behind the mirror - This arrangement can be thought of as a

double-slit source with the distance between

points S and S comparable to length d

Phase Changes Due To Reflection

- An interference pattern is formed
- The positions of the dark and bright fringes are

reversed relative to the pattern of two real

sources - This is because there is a 180 phase change

produced by the reflection - An electromagnetic wave undergoes a phase change

of 180 upon reflection from a medium of higher

index of refraction than the one in which it was

traveling - Analogous to a pulse on a string reflected from a

rigid support

Phase Changes Due To Reflection, final

- There is no phase change when the wave is

reflected from a boundary leading to a medium of

lower index of refraction - Analogous to a pulse on a string reflecting from

a free support

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