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•Login to the Bison web Secured Login •Click Surveys •Click Arts and Sciences Course Evaluation •Read the information carefully and start the course evaluation process •Print and submit the evaluation confirmation page to course instructor If there is any problem during the course evaluation process, please the student should contact Mr. Girma at 202-806-9279 or [email protected] DO THE STUDENTS EVALUATIONS!

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DO THE STUDENTS EVALUATIONS!. Login to the Bison web Secured Login Click Surveys Click Arts and Sciences Course Evaluation Read the information carefully and start the course evaluation process Print and submit the evaluation confirmation page to course instructor - PowerPoint PPT Presentation

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Page 1: Login to the Bison web Secured Login Click Surveys Click Arts and Sciences Course Evaluation

•Login to the Bison web Secured Login•Click Surveys•Click Arts and Sciences Course Evaluation•Read the information carefully and start the course evaluation process•Print and submit the evaluation confirmation page to course instructor

If there is any problem during the course evaluation process, please the student should contact Mr. Girma at 202-806-9279 or [email protected]

DO THE STUDENTS EVALUATIONS!

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diffraction in films

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Diffraction

Diffraction is the bending of waves aroundobstacles or the edges of an opening.

Huygens’ principle

Every point on a wave front acts as a sourceof tiny wavelets that move forward with the samespeed as the wave; the wave front at a latterinstant is the surface that is tangent to thewavelets.

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Diffraction

The extent of the diffraction increases as the ratio of the wavelengthto the width of the opening increases.

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Diffraction

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Diffraction

This top view shows five sources of Huygens’ wavelets.

These wavelets interfere onthe right side

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Diffraction

Dark fringes forsingle slitdiffraction

K,3,2,1 sin == mW

mλθ

sinθ =λW

width of central bright fringe

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Diffraction

sinθ1 =λW

=0.1

θ1 = 0.1 rad

width of central bright fringe: 2y1= 0.08m

λ =400 nm

W=4x10-6

first dark fringe:

y1 =L tanθ1 ; Lθ1 =0.04m

L>>W>>λ: width = LλW

~

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Resolving Power

First minimum of a circular diffraction pattern

D

λθ 22.1sin =

diameter of hole

Width of bright circle =1.22Lλ

W

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Resolving Power

Three photographs of an automobile’s headlights, taken atprogressively greater distances.

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Resolving Power

Rayleigh criterion

Two point objects are just resolved when the first dark fringe inthe diffraction pattern of one falls directly on the central bright fringe in the diffraction patter of the other.

D

λθ 22.1min≈ If θ<θmin the diffraction patterns overlapIf θ<θmin the diffraction patterns overlap

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Diffraction

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X-Ray DiffractionX-rays are used to study materials (solid crystals)

DNA

Condition for diffraction λ<<d

Solids could not be studied with visible light, λ is too large.

X-rays have very small λ (10-10m < d~10-9m) and do not damage materials significantly.

In some cases “electrons” diffraction is used.

Condition for diffraction λ<<d

Solids could not be studied with visible light, λ is too large.

X-rays have very small λ (10-10m < d~10-9m) and do not damage materials significantly.

In some cases “electrons” diffraction is used.

From the diffraction patternit can be calculated thestructure of the material

From the diffraction patternit can be calculated thestructure of the material

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Polarization

In polarized light, the electric fieldfluctuates along a single direction.

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Polarization

Polarized light may be produced from unpolarized light withthe aid of polarizing material.

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Polarization

MALUS’ LAW

θ2cosoSS =

intensity beforeanalyzer

intensity afteranalyzer

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Polarization

Example 7 Using Polarizers and Analyzers

What value of θ should be used so the average intensity of the polarizedlight reaching the photocell is one-tenth the average intensity of theunpolarized light?

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Polarization

θ251 cos=

51cos =θ

o4.63=θ

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Polarization

When Polaroid sunglasses arecrossed, the intensity of the transmitted light is reduced to zero.

Two crossed polarizers (with perpendicular axes) transmitno light

θ2cosoSS =

zero

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Polarization

Conceptual Example 8 How Can a Crossed Polarizer andAnalyzer Transmit Light?

Suppose that a third piece of polarizing material is inserted between the polarizer and analyzer. Does light now reach thephotocell?

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Intensity of Polarized Light, Examples

•On the left, the transmission axes are aligned and maximum intensity occurs.•In the middle, the axes are at 45o to each other and less intensity occurs.•On the right, the transmission axes are perpendicular and the light intensity is a minimum.

Section 38.6

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Polarization by Reflection

•When an unpolarized light beam is reflected from a surface, the reflected light may be

– Completely polarized– Partially polarized– Unpolarized

•The polarization depends on the angle of incidence.– If the angle is 0°, the reflected beam is unpolarized.– For other angles, there is some degree of polarization.– For one particular angle, the beam is completely polarized.

Section 38.6

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Polarization by Reflection.

•The angle of incidence for which the reflected beam is completely polarized is called the polarizing angle, θp.

•Brewster’s law relates the polarizing angle to the index of refraction for the material.

•tan θp = n2/n1

•θp may also be called Brewster’s angle.

Section 38.6

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Polarization by Reflection, Partially Polarized Example

•Unpolarized light is incident on a reflecting surface.•The reflected beam is partially polarized.•The refracted beam is partially polarized

Section 38.6

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Polarization by Reflection, Completely Polarized Example

•Unpolarized light is incident on a reflecting surface.•The reflected beam is completely polarized.•The refracted beam is perpendicular to the reflected beam.•The angle of incidence is Brewster’s angle.

Section 38.6