Optics and Microscopy II

Graph of Fourier Optics resolution (Goodman) removed due to copyright restrictions.

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Huygens-Fresnel Principle

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Diffraction I

Single slit diffraction is a result of the interference of light due to its wave nature

Detector Screen

-3λd

-2λd

-λd

π 2π 3π

λd

2λd

3λd

Diff

ract

ed L

ight

Inte

nsity

Maximum Intensity

Figure by MIT OCW.

Diffraction I I

La's considm the simplest diffraction situation that of a fmite size slit:

The contributionofclement dy at position P can be calculated from Hsvgen's principle:

Diffraction Ill In the far field (Fm~mhoffea)limit, R3:3D

We cm appr~~ximater by R and the distance r can be approximated as a function of R y and E (usiny ofcosine):

The total field at P is:

Therefore, we have

E=-ED sin[(kD .!2 ) sin81sia(c~-kR)R (kD1)rin8

Diffraction IV

DefutmgP =(kD/2)sinO duIcu]atingimemtemitytPwe haw:

1 ED2 sinins 2 ( 0 )=-

2 (-1R (-1 p =I(0)sinc(P)

U a t e t h z t i n t h i s ; d e r i ~ l o e , w e h a v e i ~ m e a r ~ ~ @rmm!zldifbdm) as mtlas the vectar natweof the electric field-

Diffraction V

Microscopy imaging can be consider as the diffraction from a circular aperturewith a lens for focusing – diffraction results in “broadening” of the focal point.

Courtesy of Paul Padley.

Source: Padley, Paul. Figure 3 in "Diffraction from a Circular Aperture."Connexions, November 8, 2005, http://cnx.org/content/m13097/1.1/.

Recall from lecture 1, the interference of two plane waves:

Fourier Optics I

Recall the interference of two plane waves r r r r rE1(r , t ) = E cos(k1.r + wt) r r r r rE2 (r , t) = E cos(k2 .r + wt )

r r r r rI(r , t ) = 2I + 2I cos(k1 .r - k2 .r )

In the case where the waves incident symmetrically and looking at the intensity along the y axis

r k1 = k sinqx + k cosqy r k2 = k sinqx - k cosqy r r

= yy

The intensity has a simple distribution depend on angle q:

I(r r , t ) = 2I (1 + cos(2k cosqy))

Note that when angle is zero degree (light wave counter propagating), the highest 2frequency oscillation is observed at spatial frequency: 2k = 2p ( ) . When the waves arel

parallel, angle is 90 degree, the spatial frequency is zero (constant intensity light).

Fourier Optics II

Consider two point source at the focal plane of a lens, the light rays become collimated plane waves after the lens and interference is observed.

Fourier Optics III

What happen when the two sources coincide? Only parallel plane waves are generated.

Fourier Optics IV

What happen if the point sources are made further apart?

Resolution viewed from Fourier Optics

Light emission from any object in the specimen plane can be Decomposed into its Fourier components. Which Fourier component will pass the finite aperture of the objective lens? Low frequencies!

Resolution viewed from Fourier Optics II

NA = n sin(q )

What is the maximum frequency that can be pass? Consider the case of a very large lens (numerical aperture, NA approach one). The waves will approach counter propagating and the maximum frequency is:

2kmax = 2p ( )

l

Note that maximum spatial frequency is a function of wavelength.Shorter wavelength implies higher frequency (resolution) imaging.

At a given wavelength, we should expect a resolution of about l 2

Resolution viewed from Fourier Optics III

Point Spread function

2 l p

More quantitative analysis shows that:

2 2pl

pl

r aJ1(kasin ) J1( ) J1( ) J1( NAr)q = a = r =

f f

0.6lJ1 (x) = 0 at x = 3.83 � rmin =

NA

Resolution viewed from Fourier Optics VI

Graph of Fourier Optics resolution (Goodman) removed due to copyright restrictions.