Wave Optics Diffraction (PDF)

Wave Optics Diffraction 𝐵𝑁 ∴ 𝑠𝑖𝑛 𝜃 = The phenomenon of bending of light round the sharp...

Wave Optics Diffraction 𝐵𝑁 ∴ 𝑠𝑖𝑛 𝜃 = The phenomenon of bending of light round the sharp 𝐴𝐵 𝐵𝑁 = 𝐴𝐵 𝑠𝑖𝑛 𝜃 --------(2) corners and spreading into the regions of the 𝐴𝑠 𝐴𝐵 = 𝑤𝑖𝑑𝑡ℎ 𝑜𝑓 𝑠𝑙𝑖𝑡 = 𝑎 and using equation geometrical shadow is called diffraction. (2) in equation (1) All types of waves exhibit diffraction. ∆𝒙 = 𝒂 𝒔𝒊𝒏 𝜽 ---------(3) Waves get diffracted only when the size of the To find the effect of all coherent waves at P, we have obstacle is comparable to the wavelength of the to sum up their contribution, each with a different light. phase. There are two types of diffraction 𝑺𝒍𝒊𝒕 𝑺𝒄𝒓𝒆𝒆𝒏 (i) Fresnel diffraction 𝑳𝟐 (ii) Fraunhofer diffraction 𝑳𝟏 𝑷 𝜽 Fresnel diffraction. In this type of diffraction, source 𝑨 𝒚 and screen are placed at finite distance from the slit. 𝑺 𝜽 𝒂 𝑪 𝑶 𝜽 𝜽 i.e wavefront that undergoes diffraction is either 𝑩 𝑵 spherical of cylindrical in nature. 𝑷𝒍𝒂𝒏𝒆 𝒘𝒂𝒗𝒇𝒓𝒐𝒏𝒕 𝑫 Fraunhofer diffraction. In this type of diffraction, source and screen are placed at infinite distance Position of central maximum from the slit. i.e wavefront that undergoes At the central point O of the screen, the angle 𝜃 is diffraction is plane in nature. zero. Hence the waves starting from all points of slit ------------------------------------------------------------------ arrive in the same phase. This gives maximum Single slit diffraction (Fraunhofer) intensity at the central point O. To produce fraunhofer diffraction, a point source Positions of secondary minima. 𝑆 is placed at the focus of a convex lens. Plane Consider that point P on the screen. wavefront 𝐴𝐵 emerging from the convex lens is Let path difference between rays starting from allowed to falls on the slit of width 𝑎. The diffracted edges A and B be 𝑛𝜆 , wavefront from the slit is brought onto a screen where 𝒏 = 𝟏, 𝟐, 𝟑, … …Then equation (3) becomes using another convex lens shown in the figure. 𝑎 sin 𝜃𝑛 = 𝑛𝜆 Let 𝜃 be the angle of diffraction for waves reaching 𝒏𝝀 𝐬𝐢𝐧 𝜽𝒏 = − − − (4) 𝒂 at point 𝑃 of screen and 𝐴𝑁 the perpendicular Such a point on the screen will be the position of dropped from 𝐴 on wave diffracted from 𝐵. secondary minimum. It is because, if the slit is The path difference between rays diffracted at assumed to be divided into 𝟐𝒏 equal parts, then the points 𝐴 and 𝐵 , is wavelets from the corresponding points of the two ∆𝑥 = 𝐵𝑃 − 𝐴𝑃 = 𝐵𝑁 ------(1) adjacent parts of the slit will posses path difference 𝐼𝑛 ∆𝐴𝑁𝐵 , ∠𝐴𝑁𝐵 = 90° 𝑎𝑛𝑑 ∠𝐵𝐴𝑁 = 𝜃 of 𝝀/𝟐 , i.e they will meet each other in opposite ISS- Physics Page 1 of 4  Wave Optics phase and minimum intensity is produced at the difference of 𝝀/𝟐 , i.e they will meet each other in point 𝑷 on the screen. opposite phase and wavelets from one part (say last If 𝜃𝑛 is very small, then part) remain not cancelled and produce the sin 𝜃𝑛 ≈ 𝜃𝑛 maximum intensity at 𝑷 on the screen. Therefore equ (4) becomes If 𝜃𝑛′ is very small, then 𝑛𝜆 sin 𝜃𝑛′ ≈ 𝜃𝑛′ 𝜃𝑛 = − − − (5) 𝑎 Therefore equ (9) becomes 𝑡ℎ If 𝑦𝑛 is the distance of the 𝑛 minimum from the (2𝑛 + 1)𝜆 centre of the screen, then 𝜃𝑛′ = − − − (10) 2𝑎 From ∆𝐶𝑂𝑃, we have If 𝑦𝑛 is the distance of the 𝑛𝑡ℎ minimum from the 𝑂𝑃 centre of the screen, then tan 𝜃𝑛 = 𝐶𝑂 From ∆𝐶𝑂𝑃, we have 𝑦𝑛 tan 𝜃𝑛 = − − − (6) 𝑂𝑃 𝐷 tan 𝜃𝑛′ = 𝐶𝑂 If 𝜃𝑛 is very small, then ′ 𝑦𝑛′ tan 𝜃𝑛 ≈ 𝜃𝑛 tan 𝜃𝑛 = − − − (11) 𝐷 Therefore equ (6) becomes If 𝜃𝑛 is very small, then 𝑦𝑛 𝜃𝑛 = − − − (7) tan 𝜃𝑛 ≈ 𝜃𝑛 𝐷 Therefore equ (11) becomes Comparing equ (5) and (7), we get 𝑦𝑛′ 𝑦𝑛 𝑛𝜆 𝜃𝑛′ = − − − (12) = 𝐷 𝐷 𝑎 𝒏𝝀𝑫 Comparing equ (10) and (12), we get 𝒚𝒏 = − − − (8) 𝒂 𝑦𝑛′ (2𝑛 + 1)𝜆 = Positions of secondary maxima. 𝐷 2𝑎 (𝟐𝒏 + 𝟏)𝝀𝑫 Consider that point P on the screen. 𝒚′𝒏 = − − − (13) 𝟐𝒂 Let path difference between rays starting from ----------------------------------------------------------------- edges A and B be (2𝑛 + 1)𝜆/2 , where 𝒏 = 𝟏, 𝟐, 𝟑, … …Then equation (3) becomes 𝜆 𝑎 sin 𝜃𝑛′ = (2𝑛 + 1) 2 (𝟐𝒏 + 𝟏)𝝀 𝐬𝐢𝐧 𝜽′𝒏 = − − − (9) 𝟐𝒂 Such a point on the screen will be the position of secondary maximum. It is because, if the slit is assumed to be divided into (𝟐𝒏 + 𝟏) equal parts, then the wavelets from the corresponding points of the two adjacent parts of the slit will posses path Width of secondary maximum and minimum ISS- Physics Page 2 of 4  Wave Optics Distance between two consecutive secondary minimum is equal to the width of a secondary 3𝜆Τ𝑎 maximum. 𝑆𝑒𝑐𝑜𝑛𝑑 𝑠𝑒𝑜𝑛𝑑𝑎𝑟𝑦𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝜆𝐷 2𝜆Τ𝑎 𝛽 = 𝑦𝑛 − 𝑦𝑛−1 = 𝐹𝑖𝑟𝑠𝑡 𝑠𝑒𝑐𝑜𝑛𝑑𝑎𝑟𝑦 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑎 (𝑛 − 1)𝜆𝐷 𝑛𝜆𝐷 𝜆 Τ𝑎 𝛽= − 𝐼 𝐶𝑒𝑛𝑡𝑟𝑎𝑙 𝑎 𝑎 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑂 𝝀𝑫 𝜷= − − − (𝟏) 𝒂 𝜆 Τ𝑎 Distance between two consecutive secondary 𝐹𝑖𝑟𝑠𝑡 𝑠𝑒𝑐𝑜𝑛𝑑𝑎𝑟𝑦 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 maximum is equal to the width of a secondary 2𝜆Τ𝑎 𝑆𝑒𝑐𝑜𝑛𝑑 𝑠𝑒𝑜𝑛𝑑𝑎𝑟𝑦𝑚𝑎𝑥𝑖𝑚𝑢𝑚 minimum. 3𝜆Τ𝑎 ′ 𝛽 = 𝑦𝑛′ − 𝑦𝑛−1 𝝀𝑫 𝝀𝑫 𝛽 = (2𝑛 + 1) − ( 2( 𝑛 − 1) + 1) -------------------------------------------------------------------- 𝟐𝒂 𝟐𝒂 𝝀𝑫 Angular fringe width 𝜷= − − − (2) 𝒂 𝐵2 From equations (1) and (2), it follows that in single 𝐷2 𝛽 slit diffraction, all secondary fringes are of same 𝐵1 width. 𝛽𝜃 𝐷1 ------------------------------------------------------------------ 𝑂 Width of the central maximum 𝐷 𝐷1 Width of the central maximum is equal to the 𝐵1 𝐷2 distance between the first secondary minimum formed on the two sides of the centre of the screen. Therefore, Angle subtended by the fringe (dark or bright) at the 𝜆𝐷 centre of the slit plane is known as angular width. 𝛽0 = 2𝑦1 = 2 𝑎 In the above figure 𝝀𝑫 𝜷𝟎 = 𝟐 − − − (3) 𝛽 𝜆𝐷 𝜆 𝒂 𝑡𝑎𝑛 𝛽𝜃 = = = 𝐷 𝑎𝐷 𝑎 Comparing (2) and (3), we get Since angle is very small, 𝜷𝟎 = 𝟐𝜷 𝜆 i.e; width of the central maximum is twice the width 𝑡𝑎𝑛 𝛽𝜃 = 𝛽𝜃 = 𝑎 of the secondary fringe. 𝝀 𝜷𝜽 = ------------------------------------------------------------------ 𝒂 ----------------------------------------------------------------- ISS- Physics Page 3 of 4  Wave Optics 𝟐𝝀 ************************************************ 𝑨𝒏𝒈𝒖𝒍𝒂𝒓 𝒘𝒊𝒅𝒕𝒉 𝒐𝒇 𝒕𝒉𝒆 𝒄𝒆𝒏𝒕𝒓𝒂𝒍 𝒎𝒂𝒙𝒊𝒎𝒖𝒎 = 𝒂 Angular position of first secondary minimum is equal to the half angular width of the central maximum. ----------------------------------------------------------------- Comparison of interference and diffraction pattern Interference Diffraction All fringes are of same The diffraction pattern width. has a central bright maximum which is twice as wide as the other maxima. All bright fringes are of Intensity of secondary same intensity (near to maxima decreases the central fringe) away from the centre, on either side. Contrast of interference Diffraction pattern has pattern is very good poor contrast Dark fringes are Regions of secondary perfectly dark. (If minima are not amplitudes of the waves perfectly dark. are same) Interference pattern Diffraction pattern is formed by the formed by the superposition of two superposition of wavefronts originating secondary wavelets from two coherent originating from same sources wavefront we get a maximum For a single slit of (not a null) for two width a, the first null of narrow slits separated the interference by a distance a. at angle pattern of 𝜆/𝑎. occurs at an angle of 𝜆/𝑎. ISS- Physics Page 4 of 4