Chapter 34

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Chapter 34

Visible spectrum: only a small window

Radiated (emitted) from sources

EK

and ^BK: 1. ^EK ⊥^BK:

EK

X^BK:The direction of travel of the wave 2.

EK

and ^BK always vary sinusoidally.

3.

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*No medium is required for all electromagnetic waves!

Consider partial derivatives

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364

Æ

(F=ma)

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Electromagnetic waves are polarized if their electric field vectors are all in a single plane, called the plane of oscillation.

n: index of refraction n=1 in air

Chromatic dispersion: The spreading of light

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θ>θC Æ total internal reflection

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We wish to use a glass plate with index of refraction n=1.57 to polarize light in air.

(a) At what angle of incidence is the light reflected by the glass fully polarized?

(b) What is the corresponding angle of refraction?

(a) θB=tan^-1(n/1)=tan^-11.57=57.5⁰ (b) θr =90⁰-θB =32.5⁰

Exercise:22,41,53,71

Chapter 35

Image: a reproduction derived form light

Two types: virtual image=exists only within the brain but never the less is said to exist at the perceived location.

Real image: can be formed on a surface, such as a card or a movie screen.

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Charles Barkley is 198 cm tall. How tall must a vertical mirror be if he is to be able o see his entire length in it?

h

a e b H

c

f ab=1/2he

bc=1/2ef

H=ab+bc

=1/2(he+ef) =1/2(198) =99cm

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Thin lens

The lens in air ^{1 ( 1)(1 1 )}

2

1 r

n r

f = − −

Images from lenses

Optical instruments

Simple magnifying lens

Exercisex:21,25,37,53

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Chapter 35

Interference: constructive and destructive

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N1 wavelengths in medium 1, the wavelength is

2

2 n

n

λ = λ _:

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(c) What is the phase difference in radians and in degrees ?

(a)

(b)

0.5<0.84<1 More constructive than destructive

(c) (0.84) (2π) = 5.3 (rad) (0.84) (360⁰) = 302 ⁰

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This proves that light is a wave

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Normally D >> d

Direct sunlight: partially coherent d is not small enough Æ incoherent

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or ⁾

2 ( 1

sinθ = m+ λ

d for m = 0,1,2,.... (minima)

Æ

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(Dark)

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Zebra stripe

Exercises:11,35,35,75

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Chapter 37

Newton: light is a stream of particles.

HuygenYoung: light wave

Diffraction can be easily explained in this way.

Fresnel’s theory: wave theory

S.D. Poisson, one of the Newtonians, pointed out the “strange result” that if Fresnel’s theories were correct, then light waves should flare into the

shadow region of a sphere as they pass the edge of the sphere, producing a bright spot at the center of the shadow.

The committee arranged a test and discovered that the predicted Fresnel bright spot,

its unexpected and counterintuitive predictions verified by experiment.

A clever and simplifying strategy: pairing up all the rays coming through the slit and then finding what conditions cause the waves of the rays in each pair to cancel each other.

First dark fringe

r1 and r2 in phase within the slit

r1 and r2 must be out of phase by λ/2 when they reach P1 to produce the first dark fringe.

To display this path length difference, we find a point b on r2 such that the path length from b to P1 matches the path length of r1. Then the path length difference

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between the two rays is the distance from the center of the slit to b.

When viewing screen C is near screen B, the diffraction pattern on C is difficult to describe mathematically.

If D>> a, then we can approximate rays r1 and r2 as being parallel, at angle θ to the central axis (fig. 37-4(b))

Æthe path difference=(a/2)sinθ

We can repeat this analysis for any other pair of rays originating at corresponding points in the two zones (at the middle points of the zones) and extending to point P1.

Common path difference

sin 2 2 2

θ λ

λ _⇒ a ₌

Discussion:

Begin with a>λ then narrow the slit while holding the wavelength constant. The extent of the diffraction is greater for a narrower silt.

=λ

a Æ sinθ=1Æθ= 90⁰

ÆThe bright fringe must cover the entire viewing screen!

Second dark fringes

Divide the slit into four zones of equal widths a/4.

a

D >> Æ a/4sinθ = λ/2 asinθ = 2λ (second minimum)

Æasinθ = mλ for m = 1,2,3,... (minima-dark fringes)

Intensity in single-slit diffraction

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(In radian)

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aÆ0 αÆ0 sinα/αÆ1⇒ ⇒ ⇒single-slit.

(a)

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m=1

~ 5 Ans: 9

(b)

~ 10 Ans: 4

A grating’s ability to resolve lines of different wavelengths depends on the width of the lines.

Path difference between the top and bottom rays is

hw

hw θ

θ =Δ

sinΔ

State without proof

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X-ray diffraction

2dsinθ=mλ for m=1,2,3,.... (Bragg’s law)

Exercises:19,56,67,80