9-7 Geometrical vs Physical Optics

Geometric optics versus physical optics Diffraction (Physical Optics) Single slit calculation Double slit calculation Snell’s law (Geometric optics)

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 without diverging When is this a valid assumption?

Field at any point is considered the sum of the contributions from fields at all other points in space (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?

Consider a slit of width d illuminated with a plane wave at normal incidence. What does its “shadow” look like at a distance L where L>>d?

? d L

Diffraction as considered by Huygen’s principle

Source: wikipedia

•B

d/2

E (x) =     

x d

A• x’

E (x)

+( ) L Since small changes to the value of the L

2

x

−

x



2

−ik

K (x , x) = e

denominator hardly affect the value of  E (x) the function √  2 2 L +(x−x ) 

=

E 0 L

= E 0

ikL

−

e

ikx2 /2L

−

e

 

d/2

I (x) L2

+ (x − x )2 

≈

L



(x − x ) 1+ 2L 



≈

L+

x

2L

+

xx



L

Since L>>x’ we can approximate this function by the first order expansion

u

This is such a common function in optics it has been given its own name!

K (x , x)dx 

ikxx /L  e dx

d/2

−

d −ikL −ikx2 /2L sinc e e L

    

  sinc(u) ≡

L2 + (x − x )2

d/2

”

is called a Kernel and is a common mathematical tool for expressing how a quantity at one point in space and time affect other points

s nu

 

L

≈

    

“

E (x )

−

    

 

 

d

= I 0 sinc L

2

  kxd

2L

  kxd

2L

Consider a slit of width d illuminated with a plane wave at normal incidence. What does its “shadow” look like at a distance L where L>>d?

? d L This is not predicted by geometric optics!

Based on the diffraction from a slit found using Huygen’s principle, how would you set up a calculation of the diffraction pattern for a double slit as shown?

•B

A

x’

•

d

A

d

•B

x

D L

L’ E A     

 E (x ) = √      

L2

D/2+d/2

−

E (x) =     

 

+ x2

−ik e

√ 

L2 +x2

E (x ) e L2 + (x + D/2)2     



ik

−

D/2−d/2

−

 

√ 

L2 +(x+D/2)2

D/2+d/2



dx +

 

D/2−d/2

E (x ) e L2 + (x − D/2)2     



ik

−

√ 

L2 +(x−D/2)2

 

Later we’ll learn some tricks that will allow us to evaluate E(x), For now consider the slits to be infinitely narrow…

dx



x’

•

d

A

d

•B

x

D L

L’ √  E (−D/2) −ik L E (x) =   e 2 2 L + (x + D/2)     

2

    

E 0 L2 + x2     

E (x)     

≈ √ 

E (x)     

≈

+(x+D/2)2



2

−ik(L+(x+D/2) /2L) e

2E 0 √  2 2 e−ik(L+(x L +x     

Considering infintesimal slits with uniform illumination E(x’)=E0

2

2

√  E (D/2) −ik L +   e 2 2 L + (x − D/2)     

2

2

+ e−ik(L+(x−D/2)

+D /4)/2L)

cos

  kxD

2L

/2L)



+(x−D/2)2

x’

•

d

A

•B

x

D

d

L

L’

2E 0 √  2 2 e−ik(L+(x L +x     

E (x)     

≈

I (x)

Considering infintesimal slits with uniform illumination E(x’)=E0

2

+D /4)/2L)

I 0

≈ √ 

2

L

2

2

+x

2

cos

cos

  kxD

2L

  kxD

2L

Why does the spoon appear bent in this image? Determine a relationship between the angle of the spoon and its apparent angle in a glass of water

nair

y 

d

d

sin =

t

=

nwater

d tan θi

tan θi

=

tan θt

≈

tan θw

=

i

d tan θt

sin θi sin θt

=

nair

h

nwater

=

=

y d’

h x h

tan θa

θw

d

x

tan θa tan θw

=

sin

θa

apparent distances get compressed in the direction normal to the interface

 

nair

nwater nair nwater

 

θi

θt

x nwater=1.33

nair=1

Using the law of reflection, determine the angle for a ray of light exiting a corner cube reflector as a function of the incident angle

Using the law of reflection, determine the angle for a ray of light exiting a corner cube reflector as a function of the incident angle

Some phenomena can only be understood by considering physical optics diffraction Many phenomena are well modeled by considering geometric optics refraction