3 Diffraction.pptxvjzhvkjzh hdfdfhvbxxmcnb

diffraction

1. Diffraction

2. Contents: Diffraction definition Types of diffraction Fraunhofer diffraction at single slit Width of central maxima Diffraction grating Resolving power

3. Diffraction: It is defined as the bending of waves around the corners of an obstacle or aperture into the region of   the geometrical shadow of the obstacle . For diffraction to occur the size of the obstacle or aperture must be comparable to the wavelength of the incident waves. Diffraction is due to interference between the secondary waves starting from the same wave front. https://www.youtube.com/watch?v=4bCUTLWyicM

4. Types of diffraction: Fresnel diffraction Fraunhofer diffraction

5. Fresnel diffraction: When the source of light and screen are at finite distance from the aperture or obstacle, the wave fronts are spherical. This type of diffraction is called the Fresnel diffraction.

6. Fraunhofer diffraction: A form of diffraction, in which light source and the screen are considered as at infinite distances or at great distance from the diffracting object. The diffraction is called Fraunhofer diffraction. T he resultant wave fronts are considered as planar.

7. Fraunhofer diffraction at single slit:

8. Let us consider a source ‘S’ acts as a focus of convex lens L1. light from S falls on L1 and passes parallelly then, incident on slit AB of width ‘d’. After passing the slit diffraction pattern is seen on the screen. Due to the superposition of the lights from different point sources on AB we can observe that, i ) Maximum intensity at center O called central maxima. ii) Alternate dark and bright bands on both sides of the central maxima. These are called secondary maxima and secondary minima .

9. Central maxima : When light from point source A and point source B reach at O on the screen, Distance travel by wave are equal. So, Path difference is zero. Thus, there is maxima at point O called central maxima.

10. Secondary minima: Letus draw a perpendicular AN from point A on the diffracted light BY. Here, BN gives the path difference. Suppose θ be the angle of deviation of diffracted ray from the original path. Then, BAN = θ In ∆ BAN , Sinθ = BN = d sinθ Suppose, BN = λ where, λ is the wavelength of light  

11. From geometry, CM = λ/2 This shows that light emitting from every source of upper half AC is at path difference λ/2 or phase difference 180 out of phase with corresponding point source of lower half BC. During the superposition waves are destructively interfere, we get minimum intensity at point P. Similarly, we get secondary minima at point P when, BN = 2λ, 3λ, 4λ, 5λ ….. In general, we get secondary minima when , BN = n λ where, n = 1,2,3,4….

12. Therefore, For secondary minima, dsin = n λ …………………(1)  

13. Secondary maxima: Let BN =λ/2 , During superposition, light from all point source of AB are constructively interfere. Therefore, we get maxima at point P on the screen. Similarly, let BN = 3λ/2 AB can be divided into three equal parts, light from every point source of first part path difference λ/2 with light from every point source of second part. During superposition they are destructively interfere. Due to the light source from the remaining part we get maxima at P on the screen.

14. Similarly, we get maxima for, BN = 3λ/2, 5λ/2 , 7λ/2 , 9λ/2 ……. I n general, BN = (2n+1) where, n = 1,2,3, …. For maxima, dsin = (2n+1) Intensity variation is shown by the figure below, https://www.youtube.com/watch?v=h9xx3pXFW-g  

15. Width of the central maxima:

16. The width of central maximum is the distance between the first secondary minimum on the either sides of the central maxima. For the secondary minima we have, dsin = n For n = 1 Sin For small angle, sin ………. ( i ) If y be the distance of first m inimum from central maxima and D is the distance between slit and screen then, from figure, tan =  

17. For small angle, tan = ……… (ii) From ( i ) and (ii) ,we have, y = The width of central maximum is the distance between first minima on either side. So, Width of central maxima ( = 2y = Also, angular width for central maxima = 2  

18. Width of secondary maxima and secondary minima: Width of secondary maxima or minima(  

19. Diffraction grating: An arrangement of a large number of very narrow equidistant slits is called diffracting grating. If the slits are parallel to each then the grating is called plane diffraction grating. It can be made by drawing a large number of straight and equidistant parallel lines or scratches over a thin glass plate. The lines acting as obstacles are known as opacities of definite width ‘b’ and spacing between them is acts as slits of width ‘a’ called transparencies. The sum a + b is called grating element . If N be the number of lines per centimeter in the grating then, a + b = cm https://www.youtube.com/watch?v=F6dZjuw1KUo  

20. The light diffracted through the grating in the direction of incident light superpose constructively at the center of the screen O producing a maxima called center maxima of the grating. Suppose light is diffracted a an angle , to the direction of incident light as in fig. Consider wavelets passing through the corresponding points A and C on two successive slits of the grating. The path difference between two wavelets from A and C is AN = which is same for all the pair of wavelets from other corresponding points on these two slits and is also same for all successive pairs of slits in the grating. If the path difference is equals one wavelength all the wavelets from all slits are in phase at point P on the screen. Then, = 1 This is the condition for first order maximum.  

21. Similarly, If the path difference is equal to 2 wavelets superpose constructively in a direction producing second order maximum. Then, ……… (ii) In general, for order maximum, ……..(iii) where, n = 0,1,2,3,….. The variation intensity in the diffraction pattern with is as shown in figure below.  

22. Resolving Power: The resolving power of an optical instrument is the ability of an optical instrument to produce distantly separate images of two close objects.

23. Resolving power of a microscope: The resolving power of the microscope is the reciprocal of the minimum separation between two point objects seen distinctly. If is the semi-vertical angle , the least distance between two objects which can be distinguished is given by,  

24. Resolving power of a telescope: It is the reciprocal of the smallest angular separation between two distinct objects whose images are separated in the telescope. The angular separation is given by, Where, is the angle subtended at the objective , is the wavelength of the light used, and D is the diameter of the objective of telescope. The number 1.22 appears due to diffraction through the circular aperture of the telescope.  

25. Resolving power of telescope = = So, the resolving power of a telescope can be increased by increasing the diameter of the objective of telescope and decreasing the wavelength of light  

26. Short questions: What is the difference between Fresnel’s and Fraunhofer’s diffraction? Why is the diffraction of sound waves more evident in daily experience than that of light waves? Radio waves diffract around building but light waves do not why? What is the difference between interference and diffraction? What is the diffraction of light? Diffraction grating is better than two slit setup for measuring the wavelength of monochromatic light. Explain.

27. What happens to Fraunhofer single slit diffraction pattern if the whole apparatus is immersed into water? Describe what happens to single slit diffraction when the width of the slit is less than the wavelength of waves?

28. Long questions: Discuss the diffraction pattern produced in a single slit. What is diffraction grating? Discuss the diffraction pattern produced in a diffraction grating. What is Fraunhofer diffraction? Show that the width of central maxima is inversely proportional to the distance between two slits.

29. Numerical Questions: How wide is the central diffraction peak on a screen 3.5 m behind a 0.010 mm slit illuminated by 500 nm light? Monochromatic light from a distant source is incident on slit 0.75 mm wide. On a screen 2m away, the distance from the central maxima of diffraction pattern to the first minima is measured to be 1.35 mm. Calculate the wavelength of light.

30. A Diffraction grating has 400 lines per mm and is illuminated normally by mono chromatic light of wavelength m. Calculate The grating spacing The angle at which the first principal order or maximum is seen the number of diffraction maxima obtained.  

31. A parallel beam of sodium light is incident normally on a diffraction grating. The angle between the two first order spectra on either side of normal is . Assuming that the wavelength of light is , find the number of rulings per mm on the grating?  

32. Multiple choice questions: 1.To observe diffraction, the size of an aperture Should be much larger than the wavelength Should be exactly half of wavelength should be the same order as wavelength Has no relation to wavelength

33. 2.A single slit diffraction pattern is obtained using a beam of red light. What happens if the red light is replaced by blue light? There is no change in the diffraction pattern Diffraction fringes become narrower and crowded together Diffraction fringes become broader and farther apart The diffraction pattern disappears

34. Ans for mcq : C b