Diffraction of light

Engineering Physics, Chapter 3

1. CHAPTER 3. DIFFRACTION OF LIGHT By Dr. Vishal Jain Assistant Professor

2. Diffraction of Light Diffraction refers to various phenomena that occur when a wave encounters an obstacle or a slit. It is defined as the bending of light around the corners of an obstacle or aperture into the region of geometrical shadow of the obstacle. Diffraction pattern of red laser beam made on a plate after passing through a small circular aperture in another plate Thomas Young's sketch of two-slit diffraction, which he presented to the Royal Society in 1803.

3. Fresnel's Diffraction Fraunhofer diffraction Cylindrical wave fronts Planar wave fronts Source of screen at finite distance from the obstacle Observation distance is infinite. In practice, often at focal point of lens. Move in a way that directly corresponds with any shift in the object. Fixed in position Fresnel diffraction patterns on flat surfaces Fraunhofer diffraction patterns on spherical surfaces. Change as we propagate them further ‘downstream’ of the source of scattering Shape and intensity of a Fraunhofer diffraction pattern stay constant. Classification of Diffraction Diffraction phenomena of light can be divided into two different classes

4. Fraunhofer Diffraction at a Single Slit Let us consider a slit be rectangular aperture whose length is large as compared to its breath. Let a parallel beam of wavelength be incident normally upon a narrow slit of AB. And each point of AB send out secondary wavelets is all direction. The rays proceeding in the same direction as the incident rays are focused at point O and which are diffracted at angle θ are focused at point B. The width of slit AB is small a. The path difference between AP and BP is calculated by draw a perpendicular BK. According to figure the path difference the phase difference …………….. eq 1 …………….. eq 2

5. According to Huygens wave theory each point of slit AB spread out secondary wavelets which interfere and gives diffraction phenomena. Let n be the secondary wavelets of the wave front incident on slit AB . The resultant amplitude due to all equal parts of slit AB at the point P can be determine by the method of vector addition of amplitude. This method is known as polygon method For this construct a polygon of vector that magnitude A o represent the amplitude of each wavelets and direction of vector represented the phase of each wavelets nɸ = δ δ /2 δ /2 N A B C δ 2 δ n δ r nɸ = δ A B R R/2 R/2

6. Now a perpendicular CN is draw from the center C of an arc on the line A, which will divide the amplitude R in two parts from triangle ACN and BCN By assuming polygon as a arc of a circle of radius r we can calculate the angle AC=BC = r so By putting the values of r By assuming δ /2 = α = π / λ a sin θ and the intensity I is given by

7. Intensity distribution by single slit diffraction Central Maxima For the central point P on the screen θ = 0 and hence α = 0 Hence intensity at point P will be Hence intensity at point P will be maximum Principal Minima For the principal minima intensity should be zero Where n = 1, 2, 3,4…… n = 0

8. Secondary Maxima To find out direction of secondary maxima we differentiate intensity equation with respect to α and equivalent to zero This is the condition for secondary maximas and can be solved by plotting a graph between y= tan α and y= α as shown The point of intersection of two curves gives the position of secondary maxima. The positions are α 1 = 0, α 2 = 1.43 π , α 3 = 2.46 π , α 4 = 3.47 π ,..

9. Intensity distribution by single slit diffraction Central maxima Secondary maxima’s Principal Minima’s

10. Width of the central maximum The width of he central maximum can be derived as the separation between the first minimum on either side of the central maximum. If he first maximum is at distance x from the central maximum then x x D f We know that From the diagram θ 1 If θ is very small sin θ 1 = tan θ 1 = θ 1 …………….. eq 1 ….. eq 2 ….. eq 3 ….. eq 4

11. Diffraction Grating A diffraction grating is an arrangement equivalent to a large number of parallel slits of equal widths and separated from one another by equal opaque spaces . Construction Diffraction grating can be made by drawing a large number of equidistant and parallel lines on an optically plane glass plate with the help of a sharp diamond point. The rulings scatter the light and are effectively opaque, while the unruled parts transmit light and act as slits. The experimental arrangement of diffraction grating is shown They are two type refection and transmission gratings

12.  A very large reflecting diffraction grating An incandescent light bulb viewed through a transmissive diffraction grating.

13. Theory for transmission grating (resultant intensity and amplitud e ) Let AB be the section of a grating having width of each slit as a, and b the width of each opaque space between the slits. The quantity (a + b) is called grating element, and two consecutive slits separated by the distance (a + b) are called corresponding points. The schematic ray diagram of grating has been shown in figure Let a parallel beam of light of wavelength λ incident normally on the grating using the theory of single slit & Huygens principle, the amplitude of the wave diffracted at angle θ by each slit is given by …………………… eq 1

14. Diffraction by n parallel slit at an angle θ is equivalent to N parallel waves of amplitude R That emitted from each slit s 1 , s 2 , s 3 ….. s N Where α = π / λ (a sin θ ), These N parallel waves interfere and gives diffraction pattern consisting of maxima and minima on the screen. The path difference between the waves emitted from two consecutive slits given by. The corresponding Phase Difference Thus there are N equal waves of equal amplitude and with a increasing phase difference of δ …………………… eq 2 …………………… eq 3 …………………… eq 4

15. To find the resultant amplitude of these N parallel waves we use the vector polygon method. Waves from each slit is represented by vectors where its magnitude represented by amplitude and direction represents the phase. Thus joining the N vectors tail to tip we get a polygon of N equal sides and the angle between two consecutive sides is δ The phase difference between waves from first to last slit is N δ obtained by drawing the tangents at A and B N δ N δ /2 N δ /2 A B C δ 2 δ N δ r N δ A B R N R N N

16. Diffraction by n parallel slit at an angle θ is equivalent to N parallel waves of amplitude RN. Consider a triangle ACN and DCN C A N D δ /2 δ /2 R/2 R/2 Since AC=CD we can rewrite …………………… eq 5 In triangle ABC …………………… eq 6 Here

17. So the resultant intensity …………………… eq 7 The above equation gives the resultant intensity of N parallel waves diffracted at an angle θ . The resultant intensity is the product of two terms Due to diffraction from each slit Due to Interference of N slits Intensity distribution by diffraction Grating Central Maxima Hence intensity at point P will be maximum Principal Minima Where m = 1, 2, 3,4…… 1. Due to Diffraction from Each Slit

18. Secondary Maxima 2. Due to Interference of N slits Principal Maxima’s Position of Principal maxima’s obtained when Where n= 0, 1, 2, 3……….. Then sinN β is also equal to zero and becomes indeterminate so by using L’ Hosptal Rule Hence the intensity of principal maxima is given by

19. Manima’s The intensity will be minimum when I = 0 i.e. sinN β = 0 but sin β ≠ 0 N β = ±p π here p = 1, 2, 3………..(N-1)(N+1)…….(2N-1)(2N+1)…….. i.e. p ≠ N, 2N…… hence Secondary Maxima’s To find out direction of secondary maxima we differentiate intensity equation with respect to α and equivalent to zero The solution of the above equation except p=±n π gives the position of the secondary maxima’s

20. Intensity of Secondary Maxima’s The position of secondary maxima is given by using this equation a right triangle can be formed as shown N β (1+N 2 tan 2 β ) 1/2 A B C Ntan β From figure As increase in number of slit the number of secondary maxima also increases

21. Intensity distribution by D iffraction Gratings

22. Formation of Spectra with Diffraction Grating With White Light With Monochromatic Light

23. Characteristics of Grating Spectra If the angle of diffraction is such that, the minima due to diffraction component in the intensity distribussion falls at the same position of principal maxima due to interference component, then the order of principal maxima then absent. If m th order minima fall on n th order principal maxima then Now we consider some cases If b=a, i.e. width of opaque space in equal to width of slit then from equation 3. n = 2m since m=1, 2, 3 …. Then n = 2 nd , 4 th , 6 th ….spectra will be absent B. If b=2a, i.e. width of opaque space in equal to width of slit then from equation 3. n = 3m since m=1, 2, 3 …. Then n = 3 rd , 6 th , 9 th ….spectra will be absent 1. ABSENT SPECTRA …………… eq 1 …………… eq 2 …………… eq 3

24. 2. Maximum Number of Order Observed by Grating Principal maximum in grating spectrum is given by Maximum possible angle of diffraction is 90 degree therefore So Q.1. A plane transmission grating has 6000 lines/cm. Calculate the higher order of spectrum which can be seen with white light of wavelength 4000 angstrom Sol. Given a+b =1/6000, Wavelength 4000X 10 -8 cm As we know that gratings equation written as For maximum order Maximum order will be 4 th …………….. eq 1 …………….. eq 2

25. 3. Width of principal maxima The angular width of principal maxima of nth order is defined as the angular separation between the first two minima lying adjacent to principal maxima on either side θ n 2 δθ n θ n - δθ n θ n + δθ n A O If θ n is the position of nth order principal maxima θ n + δθ n, θ n - δθ n are positions of first minima adjacent to principal minima then the width of nth principal maxima = 2 δθ n From the Grating Equation n th order maxima And the position of minima is given by Hence equation rewritten as …………… eq 1 … eq 2 On dividing eq 2 by eq 1 If d θ n is very small than cosd θ n = 1, sind θ n = d θ n

26. 4. Dispersive Power of Diffraction Gratings For a definite order of spectrum, the rate of change of angle of diffraction θ with respect to the wavelength of light ray is called dispersive power of Grating. Dispersive Power = d θ /d λ We know that gratings equation Also written as …………….. eq 1 …………….. eq 2 …………….. eq 3

27. 5. Experimental demonstration of diffraction grating to determine wavelength

28. Resolving Power

29. Rayleigh Criterion for Resolution Lord Rayleigh (1842-1919) a British Physicist proposed a criterion which can manifest when two object are seen just separate this criterion is called Rayleigh’s Criterion for Resolution Well Resolved Just resolved Not resolved

30. Resolving power of a telescope Resolving power of telescope is defined as the reciprocal of the smallest angle sustained at the object by two distinct closely spaced object points which can be just seen as separate ones through telescope. Let a is the diameter of objective telescope as shown in fig and P 1 and P 2 are the positions of the central maximum of two images. According to Rayleigh criterion these two images are said to be separated if the position of central maximum of the second images coincides with the first minimum or vice versa. P 1 P 2 A B d θ d θ d θ a The path difference between AP 2 and BP2 is zero and the path difference between AP 1 and BP 1 is given by If d θ is very small sin d θ = d θ C For rectangular aperture For circular aperture …………….. eq 1 ……… eq 2

31. Resolving power of a Diffraction Grating The resolving power of a grating is the ability to separate the spectral lines which are very close to each other. When two spectral lines in spectrum produced by diffraction grating are just resolved, then in this position the ratio of the wavelength difference and the mean wave length of spectral lines are called resolution limit of diffraction Grating Q d θ Let parallel beams of light of wavelength λ and λ +d λ be incident normally on the diffraction grating. If nth principal maxima of λ and λ +d λ are formed in the direction of θ n , θ n +d θ n respectively For the principal maximum by wavelength λ the gratings equation written as for wavelength d λ We know that for minima By eq 2 and 3 θ n λ +d λ δθ n A λ P …………….. eq 1 .. eq 2 .. eq 3 .. eq 4

32. Thanking You