ENE 451

1. ENE 451 Fundamental of Optical Engineering Lecture 5

2. Diffraction • Diffraction is any behavior of light which deviates from predictions of geometrical optics. • We are concerned about the transformations of wave from near field to far field and far field to near field. • A Guassian beam represents an example of diffraction.

3. Diffraction

4. Diffraction from a single slit • This may be called ‘Fraunhofer diffraction’.

5. Diffraction from a single slit

6. Diffraction from a single slit • If we consider PA for various situations such as • PA = 0 for sin(/2) = 0 unless /2 = 0. • PA = 0 for (ssin/λ)= ±, ±2, ±3… • PA = 0 for sin  = m1 λ/s where m = ±1, ±2, ±3,... • Also, /2  0: • Therefore, we have PA being maximum for  = 0 and PA = sP0/(λzL)

7. Diffraction from a single slit amplitude Intensity /2

8. Diffraction from a single slit

9. Example Determine PA vs. position on a screen 1.5 m away from a 1-dimensional aperture 5 m wide for an Ar laser beam (λ = 488 nm). Assume that the power density incident on the aperture is 106 W/m2.

10. Example • The far-field pattern from a single slit is imaged onto a screen. At a wavelength of 0.6 μm, the 3rd minimum is separated by a distance of 2 cm from the center of the diffraction pattern. • At what wavelength will the 4th minimum be displaced to that location? • If the distance from slit to screen is 50 cm, what is the slit width? • If the slit is replaced by a circular aperture of diameter equal to the slit width, what is the distance on the screen from the center of the pattern to the 1st minimum, for λ = 0.6 μm?

11. Circular aperture

12. Circular aperture • PA = 0 for  = 1.22λ/s ….. 1st order • PA = 0 for  = 2.23λ/s ….. 2nd order • PA = 0 for  = 3.24λ/s ….. 3rd order • For diffraction behind a circular aperature, m in single-slit has to be replaced by J (1st order Bessel-function).

13. Cylindrical lens

14. Spherical lens

15. Resolution • This refers to how we can distinguish or resolve 2 objects or features. • For example, case of telescope objective.

16. Resolution • Geometrical optics gives 2 infinitesimal dots seperated bhy a distance d f. • For diffraction theory, those dots change to Airy disc.

17. Rayleigh’s Criterion • This is use for resolvable images. • It may be arbitrary but useful. • This requires that the centers of the image patterns be no nearer than the angular radius of the Airy disc.

18. Rayleigh’s Criterion

19. Rayleigh criterion for microscope

20. Rayleigh criterion for microscope • Geometrical optics: • Diffraction theory

21. Rayleigh criterion for microscope • Numerical Aperture,

22. Summary of beam diffraction results

23. Example • The far-field pattern from a single slit is imaged onto a screen. At a wavelength of 0.6 m, the 3rd minimum is seperated by a distance of 2 cm from the center of diffraction pattern. • At what wavelength will the 4th minimum be displaced to that location? • If the distance from slit to screen is 50 cm, what is the slit width? • If the slit is replaced by a circular aperture of diameter equal to the slit width, what is the distance on the screen from the center of the pattern to the 1st minimum for λ = 0.6 m?

24. Example • Using the Rayleigh criterion, what is the spatial resolution of an f/2.5 lens at a wavelength of 0.63 μm?

25. Example • Consider a collimated beam incident on a screen containing a circular aperture of diameter d. Where will the image of the aperture be formed if the screen is located 6 cm to the left of a lens with f = 5 cm?

26. Example • Describe and sketch the power density profile of the image of a circular aperture with d = 20 μm, assuming parameter values given in previous examples. Repeat it with d = 0.1 μm.