2nd midterm exam - university of rochester

2nd Midterm Exam Monday, March 26, 2012

12:30 – 1:45 pm, Hoyt Hall.

Final Exam:

Tuesday, May 8, 2012 Starting at 8:30 a.m.,

Hoyt Hall.

Next Week - Recitations will be dedicated to review solutions of

1st and 2nd Midterm Exams - No homework

- Wednesday, March 28: Relativity

Chapter 35 Diffraction and Polarization

35-5 Resolution of Telescopes and

35-6 Resolution of the Human Eye and Useful Magnification

35-7 Diffraction Grating

35-8 The Spectrometer and Spectroscopy

35-9 Peak Widths and Resolving Power for a Diffraction Grating

35-10 X-Rays and X-Ray Diffraction

35-11 Polarization

Units of Chapter 35

35-3 Circular Aperture Diffraction

Intensity

1.22 λD

−1.22 λD

sinθ

35-4 Limits of Resolution; Circular Apertures

The Rayleigh criterion states that two images are just resolvable when the center of one peak is over the first minimum of the other.

35-4 Limits of Resolution; Circular Apertures

θ = 1.22 λD

High Defini*on Technology Blu-ray Disc uses a "blue" (technically violet) laser, operating at a wavelength of 405 nm, to read and write data. The light is generated by a diode laser made of InGaN (Indium Gallium Nitride). Conventional DVDs and CDs use red and near-infrared lasers, at 650 nm and 780 nm, respectively.

The blue-violet laser's shorter wavelength makes it possible to store more information on a 12 cm CD/DVD-size disc. The minimum "spot size" on which a laser can be focused is limited by diffraction, and depends on the wavelength of the light and the diameter (or NA) of the lens used to focus it. By decreasing the wavelength, increasing the NA from 0.60 to 0.85, and making the cover layer thinner to avoid unwanted optical effects, the laser beam can be focused to a smaller spot. This allows more information to be stored in the same area. For Blu-ray Disc, the spot size is 580 nm.

Blu-ray offers more than five times the storage capacity of traditional DVDs and can hold up to 25GB on a single layer disc.

For telescopes, the resolution limit is:

35-5 Resolution of Telescopes and Microscopes; the λ Limit

For microscopes, assuming the object is at the focal point, the resolving power (RP) is given by

θ = 1.22 λD

RP = s = θ f = (1.22 λD) f

Diffraction sets an ultimate limit of resolution, which restricts the details that can be seen on an object

35-5 Resolution of Telescopes and Microscopes; the λ Limit

The λ Limit

It is not possible to resolve detail of objects smaller than the wavelength of the radiation being used.

f ≈ D 2 → RP ≈ 1.222

λ

The human eye can resolve objects that are about 1 cm apart at a distance of 20 m, or 0.1 mm apart at the near point.

This limits the useful magnification of a light microscope to about 500x–1000x.



Rayleigh criterion for diffraction-limited vision. Pupil diameter varies from 10 mm to 8 mm and for wavelength of 550 nm (green) the diffraction limit is:

This is another example of the remarkable performance of human senses: the most acute vision is close to the physical limits imposed by diffraction!

θ =1.22 λD

= [6 − 0.8]×10−4 rad

θeye = 5.0 ×10−4 rad best eye resolution

Resolving Power of our Eye

λ = 550 nm → RP ≈10 µm → ≈ 0.1mm

35-3 Double-Slit Diffraction

d = 0.50 mm; D = 0.04 mm

D D

d

d = 0.50 mm; D = 0.08 mm

35-3 Double-Slit Diffraction

A diffraction grating consists of a large n u m b e r o f e q u a l l y spaced narrow slits or lines.


d sinθ = m λ m = 0,1,2,... m is called the order of diffraction


d sinθ = m λ m = 0,1,2,... m is called the order of diffraction

Maxima of the diffraction pattern are


Example 35-8: Diffraction grating: lines.

Determine the angular positions of the first- and second-order maxima for light of wavelength 400 nm and 700 nm incident on a grating containing 10,000 lines/cm.


Example 35-9: Spectra overlap.

White light containing wavelengths from 400 nm to 750 nm strikes a grating containing 4000 lines/cm. Show that the blue at λ = 450 nm of the third-order spectrum overlaps the red at 700 nm of the second order.


Why the bright fringes are much sharper and narrower?

Consider waves leaving at an angle θ slightly different from its 1st order maximum

Δl = 1.001 λ à With two slits the interference will be almost entirely constructive: Iθ = I0

With 10,000 slits the wave through the first slit will interference destructively with the wave leaving 500 slits below:

λ + 500 (0.001 λ) = 1.5 λ


The diffraction factor appears as the envelope diffraction function.

Intensity in a Diffrac*on Gra*ng

Resolving Power of a Diffrac*on Gra*ng

Compact disk (CD) When you look at the surface of a music CD you see the colors of a

rainbow. Why?

land

pit

In a 80 min CD we have ≈ 7000 lines/cm à distance between lines is ≈1.4 µm.

The line spacing can be also estimated by the principal maxima of a diffraction grating:

sinθ = m λd

m = 0,1,2,... sinθ < 1 → d > mλ

λ ≅ 500 nm → for m = 3 we have d ≅ 1500 nm


These two sets of diagrams show the phasor relationships at the central maximum and at the first minimum for gratings of two and six slits.

As the number of slits becomes large, the width of the central maximum becomes very narrow. Half of the peak width is

The resolving power of a diffraction grating is the minimum difference between wavelengths that can be distinguished:


Δθ0 =λNd

R = λΔλ

= Nm

34-4 Intensity in the Double-Slit Interference Pattern

The electric field at the point P from the two slits is given by

where

.

E =E1 +

E2

E1 = E10 sin(ωt)E2 = E20 sin(ωt + δ )

δ = 2πλd sin(θ )

If l >> d, the two fields can be considered parallel

Δθ0

The angular half width zero order peak, that is, the separation between the m = 0 peak and the adjacent minimum, is found by setting

Nδ = 2π → Nδ = 2πλNd sinθ0 = 2π

sinθ0 = sinΔθ0 =λNd

→ sinΔθ0 ≅ Δθ0 =λNd

m = 0



Angular half width of the m-th peak.

Δθm = λNd cosθm

Δθm

m-th

The angular separation (Δθ) between two spectral lines differing in wavelength by Δλ is given by

mλ = d sinθ →Δ m

dλ⎛

⎝⎜⎞⎠⎟ =

md

⎛⎝⎜

⎞⎠⎟ Δλ

Δ(sinθ) = cosθ (Δθ)

⎧

⎨⎪

⎩⎪

Δθ = md cosθ

⎛⎝⎜

⎞⎠⎟ Δλ


Δθm = λNd cosθm

Δθ = Δλd cosθ

⎛⎝⎜

⎞⎠⎟Δλ

⎧

⎨

⎪⎪

⎩

⎪⎪

m-th

According to the Rayleigh criterion the two peaks are just resolvable when the maximum of one peak is directly over the first minimum of the other peak, that is,

Δθm

λNd cosθm

=Δλ

d cosθm

⎛

⎝⎜

⎞

⎠⎟ Δλ ⇒ R ≡ λ

Δλ= Nm


R ≡ λΔλ

= Nm

The ability to resolve two wavelength according to Rayleigh criterion is called resolving power (R) and is given by

Diffrac*on Gra*ng

35-‐9 Peak Widths and Resolving Power for a DiffracGon GraGng

Example 35-12: Resolving two close lines.

Yellow sodium light, which consists of two wavelengths, λ1 = 589.00 nm and λ2 = 589.59 nm, falls on a diffraction grating containing 7,500 lines/cm. Determine (a) the maximum order m that will be present for sodium light, and (b) the width of grating necessary to resolve the two sodium lines.

A spectrometer makes accurate measurements of wavelengths using a diffraction grating or prism.


Atoms and molecules can be identified when they are in a thin gas through their characteristic emission lines.


Spectroscopy

Spectroscopy

neon spectrum

X-Rays were discovered by Wilhelm Conrad Röntgen, a German physicist, in 1895. Around 1912, Max von Laue, another German physicist, discovered the diffraction of X-rays by crystals.

Thus X-rays are EM waves with wavelengths in the range

0.01 nm < λ < 10 nm

The wavelengths of X-rays are very short. Diffraction experiments are impossible to do with conventional diffraction gratings.

Crystals have spacing between their layers that is ideal for diffracting X-rays.


2d sinθ = mλ m = 1,2,...

Bragg equation


X-ray diffraction is now used to study the internal structure of crystals; this is how the helical structure of DNA was determined.


Light is polarized when its electric fields oscillate in a single plane, rather than in any direction perpendicular to the direction of propagation.

35-11 Polarization

Polarization

E =E0 cos

k ⋅ r − ωt⎡⎣ ⎤⎦

Ex and Ey are the amplitudes of the x and y directions, and ϕx and ϕy are the phases of the two components.

E =

E0,x cos k ⋅ z − ωt + φx[ ]E0,y cos k ⋅ z − ωt + φy⎡⎣ ⎤⎦

0

⎛

⎝

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟

Linear Polarization

If ϕx – ϕy = 0, π the two orthogonal components are in phase. Since the tip of the vector traces out a single line in the plane, this special case is called linear polarization. The direction of this line depends on the relative amplitudes of the two components.

Linear Polariza*on

E =

E0,x cos k ⋅ z − ωt + φx[ ]E0.y cos k ⋅ z − ωt + φy⎡⎣ ⎤⎦

0

⎛

⎝

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟

=E0,x±E0,y0

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟cos k ⋅ z − ωt[ ] φx − φy = 0, π

Circular Polariza*on

E0,x = E0.y = E0

φx − φy = ±π2

⎧

⎨⎪

⎩⎪

⇒E =


0

⎛

⎝

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟

= E0

cos k ⋅ z − ωt + φy ±π2

⎡⎣⎢

⎤⎦⎥

cos k ⋅ z − ωt + φy⎡⎣ ⎤⎦0

⎛

⎝

⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟

Circular Polariza*on

E =


0

⎛

⎝

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟

= E0

cos k ⋅ z − ωt + φy ±π2

⎡⎣⎢

⎤⎦⎥

cos k ⋅ z − ωt + φy⎡⎣ ⎤⎦0

⎛

⎝

⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟

E0,x = E0.y = E0

φx − φy = ±π2

⎧

⎨⎪

⎩⎪

Polarized light will not be transmitted through a polarized film whose axis is perpendicular to the polarization direction.

35-11 Polarization

When light passes through a polarizer, only the component parallel to the polarization axis is transmitted. If the incoming light is plane-polarized, the outgoing intensity is:

35-11 Polarization

This means that if initially unpolarized light passes through crossed polarizers, no light will get through the second one.

35-11 Polarization

Conceptual Example 35-14: Three Polaroids.

When unpolarized light falls on two crossed Polaroids (axes at 90°), no light passes through.

What happens if a third Polaroid, with axis at 45° to each of the other two, is placed between them?

35-11 Polarization

Light is also partially polarized after reflecting from a nonmetallic surface. At a special angle, called the polarizing angle or Brewster’s angle, the polarization is 100%:

.

35-11 Polarization

2nd midterm exam - university of rochester

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