Resolution

1. Resolution

2. Resolution • Sketch the variation with angle ofdiffraction of the relative intensity oflight emitted by two point sourcesthat has been diffracted at a singleslit. • State the Rayleigh criterion for imagesof two sources to be just resolved including a circular aperture) • Describe the significance of resolutionin the development of devicessuch as CDs and DVDs, the electronmicroscope and radio telescopes.

3. Diffraction from a single slit n = 2 n = 1 θ b bsinθ = nλ

4. Diffraction from a single slit bsinθ = nλ (θb = λ) (small angle approximation)

5. Diffraction from a circular aperture θ = 1.22λ (angle of first minimum radians) b

6. Rayleigh criterion • Two sources are just resolved if the central maximum of the diffraction pattern of one source falls on the first minimum of the other. i.e. θ = 1.22λ/b (radians)

7. Distant star through a circular aperture

8. Two distant stars • Images merge

9. Observing two stars • Two stars separated by distance s and lie a distance d from earth s θ d

10. Observing two stars • Angular separation = s/d (radians) s θ d

11. Observing two stars • The stars can be resolved by a telescope provided Angular separation = s/d (radians) = 1.22λ/b (small angle approximation in radians) s θ d

12. Example • The camera of a spy satellite orbiting 200km above the earth’s surface has a diameter of 35cm. What is the smallest distance the camera can resolve on the surface of the earth (assume λ = 500 nm)

13. Example • The camera of a spy satellite orbiting 200km above the earth’s surface has a diameter of 35cm. What is the smallest distance the camera can resolve on the surface of the earth (assume λ = 500 nm) Angular separation = s/d = 1.22λ/b

14. Example • The camera of a spy satellite orbiting 200km above the earth’s surface has a diameter of 35cm. What is the smallest distance the camera can resolve on the surface of the earth (assume λ = 500 nm) Angular separation = s/d = 1.22λ/b

15. Example • The camera of a spy satellite orbiting 200km (2 x 105 m) above the earth’s surface has a diameter of 35cm (0.35m) . What is the smallest distance the camera can resolve on the surface of the earth (assume λ = 500 nm (5 x 10-7 m)) Angular separation = s/d = 1.22λ/b

16. Example • The camera of a spy satellite orbiting 200km (2 x 105 m) above the earth’s surface has a diameter of 35cm (0.35m) . What is the smallest distance the camera can resolve on the surface of the earth (assume λ = 500 nm (5 x 10-7 m)) s = 1.22dλ/b =1.22(2 x 105 x 5 x 10-7)/0.35 = 0.34m

17. Another example The headlights of a car are 2m apart. The pupil of the human eye has a diameter of 2mm. Supposing wavelength of light of 500nm what is the maximum distance at which two headlights are seen as distinct?

18. Another example The headlights of a car are 2m apart. The pupil of the human eye has a diameter of 2mm. Supposing wavelength of light of 500nm what is the maximum distance at which two headlights are seen as distinct? Angular separation = s/d = 1.22λ/b

19. Another example The headlights of a car are 2m apart. The pupil of the human eye has a diameter of 2mm. Supposing wavelength of light of 500nm what is the maximum distance at which two headlights are seen as distinct? Angular separation = s/d = 1.22λ/b

20. Another example The headlights of a car are 2m apart. The pupil of the human eye has a diameter of 2mm (2 x 10-3 m). Supposing wavelength of light of 500nm (5 x 10-7 m) what is the maximum distance at which two headlights are seen as distinct? Angular separation = s/d = 1.22λ/b

21. Another example The headlights of a car are 2m apart. The pupil of the human eye has a diameter of 2mm (2 x 10-3 m). Supposing wavelength of light of 500nm (5 x 10-7 m) what is the maximum distance at which two headlights are seen as distinct? d = sb/1.22λ = (2x2 x 10-3)/1.22x5 x 10-7 d= 7000m

22. Let’s do an experiment!

23. Let’s try some questions! Page 270 Qs 1,2,3,4 and 7