1. 27.5 Diffraction

2. 27.5 Diffraction
Diffraction is the bending of waves around obstacles or the edges of an opening.

3. The Amount of Diffraction

4. Diffraction of Light

5. Diffraction Minima
Dark fringes for single-slit diffraction
                                   

6. Single-slit Diffraction Pattern

7. EXAMPLE 6: Single-slit Diffraction
Light passes through a slit and shines on a flat screen that is located L = 0.40 m away (see Figure 27.24). The width of the slit is 4.0×10-6 m. The distance between the middle of the central bright fringe and the first dark fringe is y. Determine the width 2y of the central bright fringe when the wavelength of the light in a vacuum is (a) l = 690 nm (red) and (b)l = 410 nm (violet).

8. Example 6, Page 834
Light passes through a slit and shines on a flat screen that is located L = 0.40 m away (see Figure 27.24). The width of the slit is 4.0×10-6 m. The distance between the middle of the central bright fringe and the first dark fringe is y. Determine the width 2y of the central bright fringe when the wavelength of the light in a vacuum is (a) l = 690 nm (red) and (b)l = 410 nm (violet).

9. 27.6. Resolving Power
The resolving power of an optical instrument is its ability to distinguish between two closely spaced objects.
Three photographs of an automobile’s headlights taken at progressively greater distances from the camera.

10. Diffraction from Circular Opening
The angle θ, in the picture locates the first circular dark fringe relative to the central bright region and is given by:

11. Rayleigh criterion
It is useful to have a criterion for judging whether two closely spaced objects will be resolved by an optical instrument. Figure 27.29a presents the Rayleigh criterion for resolution, first proposed by Lord Rayleigh (1842–1919):
Two point objects are just resolved when the first dark fringe in the diffraction pattern of one falls directly on the central bright fringe in the diffraction pattern of the other.

12. Example 7 The Human Eye Versus the Eagle’s Eye
A hang glider is flying at an altitude of H = 120 m. Green light (wavelength = 555 nm in vacuum) enters the pilot’s eye through a pupil that has a diameter of D = 2.5 mm. Determine how far apart two point objects must be on the ground if the pilot is to have any hope of distinguishing between them. An eagle’s eye has a pupil with a diameter of D = 6.2 mm. Repeat part (a) for an eagle flying at the same altitude as the glider.