9-7 geometrical vs physical optics

8/22/2019 9-7 Geometrical vs Physical Optics

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5.

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5.

Geometric optics versus physical optics

Diffraction (Physical Optics)

Single slit calculation

Double slit calculation

Snells law (Geometric optics)

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3/17

5.

Light is treated as traveling as rays

What are consequences of this?

Shadows are 1:1 mapping of the obstruction

Beams of light can propagate withoutdiverging

When is this a valid assumption?

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4/17

5.

Field at any point is considered the sum of thecontributions from fields at all other points inspace (i.e. light does not travel as rays)

What are the consequences of this?

Diffraction

Shadows do not have sharp edges

Beams of light diverge

When must we consider this formalism?

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5/17

5.

Consider a slit of width d illuminated with aplane wave at normal incidence. What does itsshadow look like at a distance L where L>>d?

5

d L

?


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5.

Diffraction as considered by Huygens principle

6

Source: wikipedia


7/17

E(x) =

d/2d/2

E(x)L2 + (x x)2

K(x, x)dx

E(x) =E0

LeikL

eikx2/2L

d/2d/2

eikxx/L

dx

I(x) = I0

d

L sinc2kxd

2L

E(x) = E0d

LeikL

eikx2/2Lsinc

kxd

2L

5. 7

d

L

Ax

B

x

Since small changes to the value of thedenominator hardly affect the value of

the function

L2 + (x x)2 L

K(x, x) = eik

L2+(xx)2

is called a Kernel and is a commonmathematical tool for expressing how

a quantity at one point in space andtime affect other points

L2 + (x x)2 L

1 +

(x x)

2L

L +

x

2L+xx

L

Since L>>x we can approximate thisfunction by the first order expansion

This is such a common function in

optics it has been given its own name!(pronounced sink)

sinc(u) s nu

u


8/17

5.

Consider a slit of width d illuminated with aplane wave at normal incidence. What does itsshadow look like at a distance L where L>>d?

8

d L

?

This is not predicted by geometric optics!


9/17

5.

Based on the diffraction from a slit found usingHuygens principle, how would you set up acalculation of the diffraction pattern for a

double slit as shown?

9

A

B


10/17

5.10

A

Bx

Ld

dD

x

L

E(x) =EA

L2 + x

2eik

L2+x2

Later well learn some tricks that will allow us to evaluate E(x),

For now consider the slits to be infinitely narrow

E(x) =

D/2+d/2

D/2d/2

E(x)L2 + (x + D/2)2

eikL2+(x+D/2)2dx +

D/2+d/2D/2d/2

E(x)L2 + (xD/2)2

eikL2+(xD/2)2dx


11/17

5.11

A

Bx

Ld

dD

x

L

E(x) =E(D/2)

L2 + (x + D/2)2

eikL2+(x+D/2)2 +

E(D/2)L2 + (x

D/2)2

eikL2+(xD/2)2

Consideringinfintesimal slits

with uniformillumination E(x)=E0

E(x)

2E0

L2

+x2eik(L+(x2+D2/4)/2L) cos

kxD

2L

E(x) E0

L2 + x2

eik(L+(x+D/2)2/2L) + eik(L+(xD/2)

2/2L)


12/17

5.12

A

Bx

Ld

dD

x

L

Consideringinfintesimal slits

with uniformillumination E(x)=E0

E(x)

2E0

L2 + x2eik(L+(x2+D2/4)/2L) cos

kxD

2L

I(x) I0L2 + x2

cos2kxD

2L


13/17

5.

Why does the spoon appearbent in this image?

Determine a relationship

between the angle of thespoon and its apparentangle in a glass of water

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14/17

5.14

a

it

nwater=1.33

nair=1

d

nair sin t = nwater sin i

dy

y = d tan i = dtan t

d

d=

tan i

tan t

sin i

sin t=

nair

nwaterh

x

w

tan a =x

h

tan w =x

h

nair

nwater

tan w = tan anair

nwater

apparentdistances get

compressed in thedirection normal to

the interface


15/17

5.

Using the law of reflection, determine the anglefor a ray of light exiting a corner cubereflector as a function of the incident angle

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5.

Using the law of reflection, determine the anglefor a ray of light exiting a corner cubereflector as a function of the incident angle

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17/17

5.

Some phenomena can only be understood byconsidering physical optics

diffraction

Many phenomena are well modeled byconsidering geometric optics

refraction

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9-7 geometrical vs physical optics

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