2. Diffraction at a single slit 11.3.1 Sketch the variation with angle of diffraction of the relative intensity of light diffracted at a single slit. 11.3.2 Derive the formula  =  /b for the position of the first minimum of the diffraction pattern produced at a single slit. 11.3.3 Solve problems involving single-slit diffraction. Be able to apply the formula  = /b. Topic 11: Wave phenomena 11.3 Diffraction

3. Diffraction at a single slit Sketch the variation with angle of diffraction of the relative intensity of light diffracted at a single slit.  If a wave meets a hole in a wall that is of comparable size to its wavelength, the wave will be bent through a process called diffraction.  If the aperture (hole, opening, etc.) is much larger than the wavelength, diffraction will be minimal to nonexistent. Topic 4: Oscillations and waves 4.5 Wave properties INCIDENT WAVE DIFFRACTED WAVE REFLECTED WAVE

4. Because of Huygen’s Principle, light passing through barriers do not continue as straight plane waves as in (a), but as spherical waves as in (b). This spreading out is called diffraction. Diffraction

5. Diffraction at a single slit Sketch the variation with angle of diffraction of the relative intensity of light diffracted at a single slit.  Huygen ’ s principle states “ Every point on a wavefront emits a spherical wavelet of the same velocity and wavelength as the original wave. ”  Note that because of Huygen ’ s principle waves can turn corners. Topic 4: Oscillations and waves 4.5 Wave properties

6. Diffraction at a single slit Sketch the variation with angle of diffraction of the relative intensity of light diffracted at a single slit.  The reason waves can turn corners is that the incoming wave transmits a disturbance by causing the medium to vibrate.  And wherever the medium vibrates it becomes the center of a new wave front as illustrated.  Note that the smaller the aperture b the more pronounced the effect. Topic 4: Oscillations and waves 4.5 Wave properties b = 12 b = 6 b = 2 b b b

7. Diffraction at a single slit Sketch the variation with angle of diffraction of the relative intensity of light diffracted at a single slit.  Huygen ’ s wavelets not only allow the wave to turn corners, they also interfere with each other. Topic 4: Oscillations and waves 4.5 Wave properties Constructive interference R E L A T I V E I N T E N S I T Y Destructive interference

8. EXAMPLE: If light is diffracted by a circular hole, a planar cross-section of the interference looks like the picture below. What will the light look like head-on? Diffraction at a single slit Sketch the variation with angle of diffraction of the relative intensity of light diffracted at a single slit. Topic 4: Oscillations and waves 4.5 Wave properties

9. Diffraction at a single slit Derive the formula  =  /b for the position of the first minimum of the diffraction pattern produced at a single slit.  Consider a single slit of width b.  From Huygen ’ s principle we know that every point within the slit acts as a wavelet.  At the central maximum we see that the distance traveled by all the wavelets is about equal, and thus has constructive interference.  Consider points x at one edge and y at the center of the slit:  At the 1 st minimum, the difference in distance (dashed) must be /2. Why? Topic 11: Wave phenomena 11.3 Diffraction b 1 s t m i n  x y Condition for destructive interference.

10. Diffraction at a single slit Derive the formula  =  /b for the position of the first minimum of the diffraction pattern produced at a single slit.  We will choose the midpoint of the slit (y) as our reference.  And we will call the angle between the reference and the first minimum .  We construct a right triangle as follows:  Why does the side opposite  equal /2? Topic 11: Wave phenomena 11.3 Diffraction  b 1 st min  x y b2b2 y  x  2 Condition for destructive interference.

11. Diffraction at a single slit Derive the formula  =  /b for the position of the first minimum of the diffraction pattern produced at a single slit.  From the right triangle we see that sin  = ( /2)/(b/2) sin  = /b.  Perhaps you recall that if  is very small (and in radians) then sin    (  in rad).  Finally… Topic 11: Wave phenomena 11.3 Diffraction b2b2 y  x  2 location of first minimum in single slit diffraction  = /b (  in radians)

12. Even a single slit can produce interference! This occurs because light waves from one part of the slit can interfere with light waves from another part of the slit.

13. Single Aperture Diffraction Pattern Effect of slit width Single Aperture Diffraction Pattern: Narrower Aperture

14. Effect of wavelength θ =λ/b θ =2λ/b

15. Derive the formula

16. If the path difference between rays from C and A to a point on a distant screen is λ, then the path difference between rays from A and B is λ/2. There will therefore be destructive interference between light from A and light from B. From the diagram, it is clear that Now consider the aperture to be made up of pairs of point sources, A and B, A’ and B’, etc, as shown in the next diagram. Light from all these pairs of points will also interfere destructively

17. As the angles are small, we can write

18. Red light from a laser is passed through a single narrow slit, as shown in Figure 1. A pattern of bright and dark regions can be observed on the screen which is placed several metres beyond the slit. (a) The pattern on the screen may be represented as a graph of intensity against distance along the screen. The graph has been started in outline in Figure 2. The central bright region is already shown. Complete this graph to represent the rest of the pattern by drawing on Figure 2. (b) State the effect on the pattern if each of the following changes is made separately. (i) The width of the narrow slit is reduced (ii) With the original slit width, the intense red source is replaced with an intense source of green light. IB Question

19. Topic 11: Wave phenomena 11.3 Diffraction INCIDENT WAVE DIFFRACTED WAVE REFLECTED WAVE Diffraction at a single slit Solve problems involving single-slit diffraction. Be able to apply the formula  = /b. PRACTICE: Sketch in the diffraction patterns in the double-slit breakwater with 5-m waves. Then map out MAX (10 m), MIN (-10 m) and 0 m points. 10 0 -10

20. Topic 11: Wave phenomena 11.3 Diffraction Diffraction at a single slit Solve problems involving single-slit diffraction. Be able to apply the formula  = /b.  Diffraction allows waves to turn corners.  All waves diffract- not just sound.

21. Topic 11: Wave phenomena 11.3 Diffraction Diffraction at a single slit Solve problems involving single-slit diffraction. Be able to apply the formula  = /b. Huygen says the wavelets will be spherical.

22. Topic 11: Wave phenomena 11.3 Diffraction Diffraction at a single slit Solve problems involving single-slit diffraction. Be able to apply the formula  = /b.

23. Topic 11: Wave phenomena 11.3 Diffraction Diffraction at a single slit Solve problems involving single-slit diffraction. Be able to apply the formula  = /b.  Caused by path length difference (PLD) along b.  1 st min  PLD = /2.  2 nd min  PLD = 3 /2. 1 st 2 nd  Central max  PLD = 0.  2 nd max  PLD =. Cent 2 nd

24. Topic 11: Wave phenomena 11.3 Diffraction Diffraction at a single slit Solve problems involving single-slit diffraction. Be able to apply the formula  = /b.  = /b d  tan  = d/D  tan    for small .  = d/D = /b.  d = D/b.