Physics 1202: Lecture 26 Today’s Agenda Announcements: –Midterm 2: Friday Nov. 6… –Chap. 18, 19, 20, and 21 No HW for this week (midterm)No HW for this.

Physics 1202: Lecture 26Today’s Agenda

• Announcements:

– Midterm 2: Friday Nov. 6…

– Chap. 18, 19, 20, and 21

• No HW for this week (midterm)No HW for this week (midterm)

• Optics – Interference

– Diffraction

Interference

A wave through two slits

Screen

P=d sin

d

In Phase, i.e. Maxima when P = d sin = nOut of Phase, i.e. Minima when P = d sin = (n+1/2)

The Intensity

We can rewrite intensity at point Pin terms of distance y

Using this relation, we can rewrite expression for the intensity at point P as function of y

Constructive interference occurs at

where m=+/-1, +/-2 … (destructive at m= ±1/2, ±3/2 …

Change of Phase Due to Reflection

Lloyd’s mirror P2

P1

S

I

LMirror

The reflected ray (red) can be considered asan original from the image source at point I.Thus we can think of an arrangement S and I as a double-slit source separated by the distance between points S and I.

An interference pattern for this experimentalsetting is really observed …..but dark and bright fringes are reversed in order

This mean that the sources S and I are different in phase by 1800

An electromagnetic wave undergoes a phase change by 1800 uponreflecting from the medium that has a higher index of refraction thanthat one in which the wave is traveling.

Change of Phase Due to Reflection

1800 phase change

n1 n2

n1<n2

n1 n2

no phase change

n1>n2

Interference in Thin Films

Air

Air

Film t

12

1800 phase change no phase

changeA wave traveling from air toward filmundergoes 1800 phase change upon reflection.The wavelength of light n in the medium with refraction index n is

The ray 1 is 1800 out of phase with ray 2 which is equivalent to a path difference n/2.The ray 2 also travels extra distance 2t.

Constructive interference

Destructive interference

Chapter 26 – Act 1

Estimate minimum thickness of a soap-bubble film (n=1.33) thatresults in constructive interference in the reflected light if the film isIlluminated by light with =600nm.

A) 113nm B) 250nm C) 339nm

ProblemConsider the double-slit arrangement shown in Figure below, where the slit separation is d and the slit to screen distance is L. A sheet of transparent plastic having an index of refraction n and thickness t is placed over the upper slit. As a result, the central maximum of the interference pattern moves upward a distance y ’. Find y’

where will thecentral

maximum be now ?

Solution

Corresponding path length difference:

Phase difference for going though plastic sheet:

Angle of central max is approx: Thus the distance y’ is:

gives

Phase Change upon Reflection from a Surface/Interface

Reflection fromOptically Denser Medium (larger n)

Reflection fromOptically Lighter Medium (smaller n)

by analogy to reflection of traveling wave in mechanics

180o Phase Change No Phase Change

Examples :constructive: 2t = (m +1/2) n

destructive: 2t = m n

constructive: 2t = m n

destructive: 2t = (m +1/2) n

ApplicationReducing Reflection in Optical Instruments

Diffractio

n

Experimental Observations:(pattern produced by a single slit ?)

First Destructive Interference:

(a/2) sin = ± /2

sin = ± /a

mth Destructive Interference:

(a/4) sin = ± /2

sin = ± 2/a

Second Destructive Interference:

sin = ± m /a m=±1, ±2, …

How do we understand this pattern ?

See Huygen’s Principle

So we can calculate where the minima will be !

sin = ± m /a m=±1, ±2, …

Why is the central maximum so much stronger than the others ?

So, when the slit becomes smaller the central maximum becomes ?

Phasor Description of Diffraction

Can we calculate the intensity anywhere on diffraction pattern

?

= = N

/ 2 = y sin () /

= N = N 2 y sin () / = 2 a sin () /

Let’s define phase difference () between first and last ray (phasor)

1st min.

2nd max.

central max.

(a/ sin = 1: 1st min.

Yes, using Phasors !Let take some arbitrary point on the diffraction pattern

This point can be defined by angle or by phase difference between first and last ray (phasor)

The arc length Eo is given by : Eo = R

sin (/2) = ER / 2R

The resultant electric field magnitude ER is given (from the figure) by :

ER = 2R sin (/2) = 2 (Eo/ ) sin (/2) = Eo [ sin (/2) / (/2) ]

I = Imax [ sin (/2) / (/2) ]2

So, the intensity anywhere on the pattern :

= 2 a sin () /

Other Examples

What type of an object would create a diffraction pattern shown on the left, when positioned midway between screen and light source ?• A penny, …• Note the bright spot at the center.

Light from a small source passes by the edge of an opaque object and continues on to a screen. A diffraction pattern consisting of bright and dark fringes appears on the screen in the region above the edge of the object.

Fraunhofer Diffraction(or far-field)

Incoming wave

Lens

Screen

Fresnel Diffraction(or near-field)

Incoming wave

Lens

Screen

P

(more complicated: not covered in this course)

Resolution(single-slit aperture)

Rayleigh’s criterion:• two images are just resolved WHEN:

When central maximum of one image falls on the first minimum of another image

sin = / a

min ~ / a

Diffraction patterns of two point sources for various angular separation of the sources

Resolution(circular aperture)

min = 1.22 ( / a)

Rayleigh’s criterionfor

circular aperture:

EXAMPLE

A ruby laser beam ( = 694.3 nm) is sent outwards from a 2.7-m diameter telescope to the moon, 384 000 km away. What is the radius of the big red spot on the moon?

a. 500 mb. 250 mc. 120 md. 1.0 kme. 2.7 km min = 1.22 ( / a)

R / 3.84 108 = 1.22 [ 6.943 10-7 / 2.7 ]

R = 120 m !

EarthMoon

Two-Slit Interference Pattern with a Finite Slit Size

Idiff = Imax [ sin (/2) / (/2) ]2

Diffraction (“envelope” function):

= 2 a sin () /

Itot = Iinter . Idiff

Interference (interference fringes):

Iinter = Imax [cos (d sin / ]2

smaller separation between slits => ?

smaller slit size => ?The combined effects of two-slit and single-slit interference. This is the pattern produced when 650-nm light waves pass through two 3.0- mm slits that are 18 mm apart. Animation

Example

The centers of two slits of width a are a distance d apart. Is it possible that the first minimum of the interference pattern occurs at the location of the first minimum of the diffraction pattern for light of wavelength ?

da

a

1st minimum interference:d sin = /2

1st minimum diffraction:a sin =

The same place (same ) : /2d = /a

a /d =

No!

ApplicationX-ray Diffraction by crystals

Can we determine the atomic structure of the crystals, like proteins, by analyzing X-ray

diffraction patters like one shown ? A Laue pattern of the enzyme

Rubisco, produced with a wide-band x-ray spectrum. This

enzyme is present in plants and takes part in the process of

photosynthesis.

Yes in principle: this is like the problem of determining the slit separation (d)

and slit size (a) from the observed pattern, but much much more

complicated !

Determining the atomic structure of crystalsWith X-ray Diffraction (basic principle)

2 d sin = m m = 1, 2, ..Crystalline structure of sodium chloride (NaCl). length of the cube edge is a = 0.562 nm.

Crystals are made of regular arrays of atoms

that effectively scatter X-ray

Bragg’s Law

Scattering (or interference) of two X-rays from the

crystal planes made-up of atoms