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Lecture physics a2 diffraction and spectroscopy huynh quang linh

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Tiêu đề Diffraction And Spectroscopy
Trường học University of Physics
Chuyên ngành Physics
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P Incident Wave (wavelength l) y L a Lecture 3 Diffraction & Spectroscopy We will discuss N split interference (diffraction) and single split diffraction Multi slit Interference, I = 4I1cos 2(f/2) (Fo[.]

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Incident Wave (wavelength l ) y

L a

Lecture 3:

Diffraction

&

Spectroscopy

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We will discuss N-split interference (diffraction) and single-split diffraction

Multi-slit Interference, I = 4I 1 cos 2 (f/2)

(For point sources, I 1 = constant.)

and

Single-slit Diffraction, I 1 (q)

(For finite sources, I 1 = I 1 (q).)

to obtain

Total Interference Pattern, I = 4I 1 (q)cos 2 (f/2)

(2 slits)

(Remember how f is related to q: f/2p = d/l = (dsinq)/l  d q/l)

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Multiple-slit Interference formula

Diffraction Gratings

Optical Spectroscopy

Spectral Resolution

Single-Slit Diffraction

Interference + Diffraction

Applications: X-ray Crystallography

Content

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General properties of N-Slit Interference

The positions of the principal maxima of the intensity patterns always occur at f = 0, 2p, 4p, [f is the phase between adjacent slits]

(i.e., dsinq = ml, m = 0, 1, 2,…).

• The principal maxima become taller and narrower as N increases

• The intensity of a principal maximum is equal to N 2 times the maximum intensity from one slit The width of a principal maximum goes as 1/N.

• The # of zeroes between adjacent principal maxima is equal to N-1

The # of secondary maxima between adjacent principal maxima is N-2.

-2p

I 0

16I1

N=4

-2p

I 0

25I1

N=5

-2p

I

0

9I1

N=3

0 10

20 25

0 h5( ) x

10

10 -l/ d 0 l/ x d

f q

0

5

9

0

g( ) x

10

f q -l/ d 0 l/ d 10 0 10

0

10 16

0 h( ) x

10

10 -l/ d 0 l/ x d

f q

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exercise 1 Light interfering from 10 equally spaced slits initially illuminates

a screen Now we double the number of slits, keeping the

spacing constant

What happens to the net power on the screen?

a stays the same b doubles c increases by 4

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exercise 1 Light interfering from 10 equally spaced slits initially illuminates

a screen Now we double the number of slits, keeping the

spacing constant

What happens to the net power on the screen?

a stays the same b doubles c increases by 4

If we double the number of slits, we expect the net

power on the screen to double How does it do this…

The location and number of the principle maxima (which

have most of the power) does not change.

The principle maxima become 4x brighter.

But they also become only half as wide.

Therefore, the net power (integrating over all the peaks) increases two-fold, as we would expect.

We will soon see that we often use such an array of slits (also called a “diffraction grating”) to perform very precise

metrology, e.g, spectroscopy, crystallography, etc

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N-Slit Interference – Summary

 The Intensity for N equally spaced slits is given by:

L

y

and d

sin

=

l

q l

q l

d p

f 2

2 1

) 2 / sin(

) 2 / sin(

=

f

f

N I

*

* Note: we can not be able to use the small angle approximations if d ~ l.

y

L

d

q

 As usual, to determine the pattern at the screen

(detector plane), we need to relate f to q or y = Ltan q :

**

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Example 1

In an N-slit interference pattern, at what angle qmin does the

intensity first go to zero? (In terms of l , d and N.)

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Example 1

In an N-slit interference pattern, at what angle qmin does the

intensity first go to zero? (In terms of l , d and N.)

But f = 2p(d sinq)/l  2pd q/l = 2p/N Therefore, qmin  l /Nd

As the illuminated number of slits increases, the peak widths decrease!

This is a general feature: Wider slit features  narrower patterns

in the “far field”

2 1

sin( / 2) sin( / 2)

N

N

f

  has a zero when Nfmin/2 = p, or fmin = 2p/N.

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Optical spectroscopy – how we know

about the world

• Quantum mechanics  definite energy levels, e.g.,

of electrons in atoms or molecules

• When an atom transitions between energy levels 

emits light of a very particular frequency.

• Every substance has it’s own “signature” of what

colors it can emit.

• By measuring the colors, we can determine the

substance, as well as things about it’s surroundings (e.g., temperature, magnetic fields), whether it’s

moving (via the Doppler effect), etc.

Optical spectroscopy is invaluable in materials

research, engineering, chemistry, biology, medicine…

But how do we precisely measure wavelengths???

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