Diffraction Theory 1 2 , 2 4 3 5 1 + 2 - PowerPoint PPT Presentation

Diffraction Theory 1

2

𝐹 Ԧ 𝑠, 𝑢 𝑠 2 𝑠 4 𝑠 3 𝑠 5 𝑠 1 + 𝐹 2 Ԧ 𝑠, 𝑢 + 𝐹 5 Ԧ 𝑠, 𝑢 𝐹 Ԧ 𝑠, 𝑢 = 𝐹 1 Ԧ 𝑠, 𝑢 + 𝐹 3 Ԧ 𝑠, 𝑢 + 𝐹 4 Ԧ 𝑠, 𝑢 + … = ෍ 𝐹 𝑗 Ԧ 𝑠, 𝑢 𝑗 𝑠 1 𝑓 𝑗 𝑙 𝑠 1 − 𝜕 𝑢 + 𝜁 1 + 𝐹 0,2 + 𝐹 0,3 + 𝐹 0,4 + 𝐹 0,5 𝐹 0,1 𝑓 𝑗 𝑙 𝑠 2 − 𝜕 𝑢 + 𝜁 2 𝑓 𝑗 𝑙 𝑠 3 − 𝜕 𝑢 + 𝜁 3 𝑓 𝑗 𝑙 𝑠 4 − 𝜕 𝑢 + 𝜁 4 𝐹 Ԧ 𝑠, 𝑢 = 𝑓 𝑗 𝑙 𝑠 5 − 𝜕 𝑢 + 𝜁 5 + ⋯ 𝑠 2 𝑠 3 𝑠 𝑠 5 4 𝐹 0,𝑗 𝑓 𝑗 𝑙 𝑠 𝑗 −𝜕 𝑢 + 𝜁 𝑗 = ෍ 𝑠 𝑗 3 𝑗

Huygens-Fresnel Principle 4

𝐹 0,𝑗 𝑓 𝑗 𝑙 𝑠 𝑗 − 𝜕 𝑢 + 𝜁 𝑗 𝐹 𝑍, 𝑎, 𝑢 = ෍ 𝑠 𝑗 𝑗 𝑗 𝑧, 𝑨 ෍ ඵ 𝑒𝑧 𝑒𝑨 𝑗 𝐹 0 𝑧, 𝑨 𝑏𝑞𝑓𝑠𝑢𝑣𝑠𝑓 𝑓 𝑗 𝑙 𝑠 𝑧, 𝑨 − 𝜕 𝑢 + 𝜁 𝑧, 𝑨 = ඵ 𝑒𝑧 𝑒𝑨 𝑠 𝑧, 𝑨 𝑏𝑞𝑓𝑠𝑢𝑣𝑠𝑓 𝑧 𝑍 𝑠 𝑡 𝑨 𝑎 𝑡 2 + 𝑍 − 𝑧 2 + 𝑎 − 𝑨 2 𝑠 𝑧, 𝑨 = 5

Fresnel Diffraction 1 + 𝑍 − 𝑧 2 + 𝑎 − 𝑨 2 𝑡 2 + 𝑍 − 𝑧 2 + 𝑎 − 𝑨 2 = 𝑡 1 + 𝜂 2 𝑠 𝑧, 𝑨 = = 𝑡 𝑡 2 𝑡 2 𝜂 2 ≡ 𝑍 − 𝑧 2 + 𝑎 − 𝑨 2 𝑡 2 𝑡 2 = 𝑡 1 + 𝜂 2 2 − 𝜂 4 8 + 𝜂 6 16 − 5 𝜂 8 128 + ⋯ 𝑙 𝑡 𝑛𝑏𝑦 𝜂 4 Fresnel Approximation ≪ 𝜌 8 𝑠 𝑧, 𝑨 ≅ 𝑡 1 + 𝜂 2 = 𝑡 1 + 𝑍 − 𝑧 2 + 𝑎 − 𝑨 2 2 𝑡 2 2 𝑡 2 2 = 𝑡 + 𝑍 2 + 𝑎 2 + 𝑧 2 + 𝑨 2 − 𝑍 𝑧 + 𝑎 𝑨 2 𝑡 𝑡 2 𝑡 6

𝑙 𝑡 𝑛𝑏𝑦 𝜂 4 Fresnel Approximation ≪ 𝜌 8 𝑠 𝑧, 𝑨 ≅ 𝑡 + 𝑍 2 + 𝑎 2 + 𝑧 2 + 𝑨 2 − 𝑍 𝑧 + 𝑎 𝑨 2 𝑡 𝑡 2 𝑡 𝐹 0 𝑧, 𝑨 𝑓 𝑗 𝑙 𝑠 𝑧, 𝑨 − 𝜕 𝑢 + 𝜁 𝑧, 𝑨 𝐹 𝑍, 𝑎, 𝑢 = ඵ 𝑒𝑧 𝑒𝑨 𝑠 𝑧, 𝑨 𝑏𝑞𝑓𝑠𝑢𝑣𝑠𝑓 𝑡 + 𝑍 2 +𝑎 2 + 𝑧 2 +𝑨 2 𝐹 0 𝑧, 𝑨 𝑍 𝑧+𝑎 𝑨 𝑓 𝑗 𝑙 − − 𝜕 𝑢 + 𝜁 𝑧, 𝑨 ≅ ඵ 𝑒𝑧 𝑒𝑨 2 𝑡 𝑡 2 𝑡 𝑡 𝑏𝑞𝑓𝑠𝑢𝑣𝑠𝑓 7

𝑙 𝑡 𝑛𝑏𝑦 𝜂 4 Fresnel Approximation ≪ 𝜌 8 𝑠 𝑧, 𝑨 ≅ 𝑡 + 𝑍 2 + 𝑎 2 + 𝑧 2 + 𝑨 2 − 𝑍 𝑧 + 𝑎 𝑨 2 𝑡 𝑡 2 𝑡 In addition to Fresnel Approximation: Fraunhofer Diffraction also known as Far-Field Diffraction 𝑙 𝑛𝑏𝑦 𝑧 2 + 𝑨 2 ≪ 𝜌 2 𝑡 Fraunhofer Approximation 𝑛𝑏𝑦 𝑧 2 + 𝑨 2 ≪ 1 𝜇 𝑡 8

𝑆 2 ≡ 𝑡 2 + 𝑍 2 + 𝑎 2 𝑡 2 + 𝑍 − 𝑧 2 + 𝑎 − 𝑨 2 𝑆 2 − 2 𝑍 𝑧 − 2 𝑎 𝑨 + 𝑧 2 + 𝑨 2 𝑠 𝑧, 𝑨 = = 1 + −2 𝑍 𝑧 − 2 𝑎 𝑨 + 𝑧 2 + 𝑨 2 ≅ 𝑆 − 𝑍 𝑧 + 𝑎 𝑨 = 𝑆 𝑆 2 𝑆 𝐹 0 𝑧, 𝑨 𝑓 𝑗 𝑙 𝑠 𝑧, 𝑨 − 𝜕 𝑢 + 𝜁 𝑧, 𝑨 𝐹 𝑍, 𝑎, 𝑢 = ඵ 𝑒𝑧 𝑒𝑨 𝑠 𝑧, 𝑨 𝑏𝑞𝑓𝑠𝑢𝑣𝑠𝑓 𝑗 𝑙 𝑆 − 𝑍 𝑧 + 𝑎 𝑨 𝐹 0 𝑧, 𝑨 − 𝜕 𝑢 + 𝜁 𝑧, 𝑨 𝑆 ≅ ඵ 𝑓 𝑒𝑧 𝑒𝑨 𝑆 𝑏𝑞𝑓𝑠𝑢𝑣𝑠𝑓 = 𝑓 𝑗 𝑙 𝑆 −𝜕 𝑢 𝐹 0 𝑧, 𝑨 𝑓 𝑗 𝜁 𝑧,𝑨 𝑓 − 𝑗 𝑙 𝑍 𝑧 + 𝑎 𝑨 ඵ 𝑒𝑧 𝑒𝑨 𝑆 𝑆 𝑏𝑞𝑓𝑠𝑢𝑣𝑠𝑓 9

Fraunhofer Diffraction 𝑧 𝑆 2 ≡ 𝑡 2 + 𝑍 2 + 𝑎 2 𝑍 𝑠 𝑆 𝑡 𝑨 𝑎 𝐹 𝑍, 𝑎, 𝑢 = 𝑓 𝑗 𝑙 𝑆 −𝜕 𝑢 𝐹 0 𝑧, 𝑨 𝑓 𝑗 𝜁 𝑧, 𝑨 𝑓 − 𝑗 𝑙 𝑍 𝑧 + 𝑎 𝑨 ඵ 𝑒𝑧 𝑒𝑨 𝑆 𝑆 𝑏𝑞𝑓𝑠𝑢𝑣𝑠𝑓 10

Illumination at the Aperture: In the examples to follow, we will consider a flat wavefront at normal incidence on the aperture Inside the aperture 𝐹 0 { 𝐹 0 𝑧, 𝑨 𝑓 𝑗 𝜁 𝑧, 𝑨 = Outside the aperture 0 𝐹 𝑍, 𝑎, 𝑢 = 𝐹 0 𝑓 𝑗 𝑙 𝑆 −𝜕 𝑢 𝑓 − 𝑗 𝑙 𝑍 𝑧 + 𝑎 𝑨 ඵ 𝑒𝑧 𝑒𝑨 𝑆 𝑆 𝑏𝑞𝑓𝑠𝑢𝑣𝑠𝑓 11

Apertures considered here: 1. Single Slit 2. Double Slit 3. Rectangular Aperture 4. Circular Aperture 12

𝑧 1. Single Slit 𝑒 𝑨 𝑧 𝑡𝑗𝑜 𝜄 = 𝑍 𝑆 𝑍 𝑠 𝑆 𝑡 𝜄 𝑍 2 + 𝑡 2 𝑆 ≡ 𝑎 𝑨 𝑒 2 + ൗ 𝐹 𝑍, 𝑎, 𝑢 = 𝐹 0 𝑓 𝑗 𝑙 𝑆 −𝜕 𝑢 𝑓 − 𝑗 𝑙 𝑍 𝑆 𝑧 𝑒𝑧 න 𝑆 𝑒 2 − ൗ 13

1. Single Slit, cont. 𝐹 𝑍, 𝑎, 𝑢 = 𝐹 0 𝑓 𝑗 𝑙 𝑆 − 𝜕 𝑢 𝑒 𝑡𝑗𝑜𝑑 𝑙 𝑍 𝑒 2 𝐽 ≡ 𝐹 𝑆 2 𝑆 𝐽 0 ≡ 𝐹 0 2 𝐽 𝑍, 𝑎 = 𝐽 0 𝑡𝑗𝑜𝑑 2 𝑙 𝑍 𝑒 2 𝑆 2 𝑒 2 2 𝑆 𝑍(𝑛𝑛) 𝑙 𝑍 𝑛 𝑒 geometrical zeros at 𝑒 = 50 µ𝑛 = 𝑛 𝜌 shadow 2 𝑆 𝜇 = 0.6 µ𝑛 𝑍 1 with 𝑛 = ±1, ±2, ±3 𝑡 = 1 𝑛 𝑆 ≅ 1 𝑛 𝑛 = 𝑛 𝜇 𝑆 𝑍 𝑍 𝑍 −1 𝑒 𝐽 𝐽 0 𝑎 ൗ 𝑡𝑗𝑜 𝜄 𝑛 = 𝑍 𝑆 = 𝑛 𝜇 𝑛 𝑒 14

Mathematica 15

2. Double Slit 𝑧 𝑏 2 + ൗ 𝑒 2 ൗ 𝑒 𝑏 2 − ൗ 𝑒 2 ൗ 𝑏 𝑨 −𝑏 2 + ൗ 𝑒 2 ൗ 𝑒 −𝑏 2 − ൗ 𝑒 2 ൗ 16

𝑧 𝑡𝑗𝑜 𝜄 = 𝑍 𝑆 𝑍 𝑠 𝑆 𝑡 𝜄 𝑍 2 + 𝑡 2 𝑆 ≡ 𝑎 𝑨 −𝑏 2+ ൗ 𝑒 2 𝑏 2+ ൗ 𝑒 2 ൗ ൗ 𝐹 𝑍, 𝑎, 𝑢 = 𝐹 0 𝑓 𝑗 𝑙 𝑆 −𝜕 𝑢 𝑓 − 𝑗 𝑙 𝑍 𝑓 − 𝑗 𝑙 𝑍 𝑆 𝑧 𝑒𝑧 + 𝑆 𝑧 𝑒𝑧 න න 𝑆 −𝑏 2− ൗ 𝑒 2 𝑏 2− ൗ 𝑒 2 ൗ ൗ = 𝐹 0 𝑓 𝑗 𝑙 𝑆 −𝜕 𝑢 𝑒 𝑡𝑗𝑜𝑑 𝑙 𝑍 𝑒 2 𝑑𝑝𝑡 𝑙 𝑎 𝑏 𝑆 2 𝑆 2 𝑆 17

𝐹 𝑍, 𝑎, 𝑢 = 𝐹 0 𝑓 𝑗 𝑙 𝑆 −𝜕 𝑢 𝑒 𝑡𝑗𝑜𝑑 𝑙 𝑍 𝑒 2 𝑑𝑝𝑡 𝑙 𝑍 𝑏 𝑆 2 𝑆 2 𝑆 𝐽 𝑍, 𝑎 = 4 𝐽 0 𝑡𝑗𝑜𝑑 2 𝑙 𝑍 𝑒 𝑑𝑝𝑡 2 𝑙 𝑍 𝑏 𝐽 0 ≡ 𝐹 0 2 2 𝑆 2 𝑒 2 2 𝑆 2 𝑆 𝑏 𝑒 Mathematica 18

3. Rectangular Aperture 𝑧 𝑏 𝑐 𝑨 19

𝐹 𝑍, 𝑎, 𝑢 = 𝐹 0 𝑓 𝑗 𝑙 𝑆 −𝜕 𝑢 𝑓 − 𝑗 𝑙 𝑍 𝑧 + 𝑎 𝑨 ඵ 𝑒𝑧 𝑒𝑨 𝑆 𝑆 𝑏𝑞𝑓𝑠𝑢𝑣𝑠𝑓 𝑐 2 𝑏 2 ൗ ൗ = 𝐹 0 𝑓 𝑗 𝑙 𝑆 −𝜕 𝑢 𝑓 − 𝑗 𝑙 𝑍 𝑓 − 𝑗 𝑙 𝑎 𝑆 𝑧 𝑒𝑧 𝑆 𝑨 𝑒𝑨 න න 𝑆 −𝑏 2 −𝑐 2 ൗ ൗ = 𝐹 0 𝑓 𝑗 𝑙 𝑆 −𝜕 𝑢 𝑐 𝑡𝑗𝑜𝑑 𝑙 𝑍 𝑐 𝑏 𝑡𝑗𝑜𝑑 𝑙 𝑎 𝑏 𝑆 2 𝑆 2 𝑆 20

𝐽 𝑍, 𝑎 = 𝐽 0 𝑡𝑗𝑜𝑑 2 𝑙 𝑍 𝑐 𝑡𝑗𝑜𝑑 2 𝑙 𝑎 𝑏 2 𝑆 2 𝑆 𝐽 0 ≡ 𝐹 0 2 2 𝑆 2 𝑏 2 𝑐 2 𝑍 𝑎 21

Emission of Semiconductor Laser 22

4. Circular Aperture 𝑧 = 𝜍 𝑡𝑗𝑜 𝜒 𝑧 𝜍 𝑏 𝜒 𝑨 𝑨 = 𝜍 𝑑𝑝𝑡 𝜒 23

Observation Plane 𝑍 𝑟 𝑍 = 𝑟 𝑡𝑗𝑜 Φ Φ 𝑎 𝑎 = 𝑟 𝑑𝑝𝑡 Φ 24

𝐹 𝑍, 𝑎, 𝑢 = 𝐹 0 𝑓 𝑗 𝑙 𝑆 −𝜕 𝑢 𝑓 − 𝑗 𝑙 𝑍 𝑧 + 𝑎 𝑨 ඵ 𝑒𝑧 𝑒𝑨 𝑆 𝑆 𝑏𝑞𝑓𝑠𝑢𝑣𝑠𝑓 𝑍 𝑧 + 𝑎 𝑨 = 𝑟 𝑡𝑗𝑜 Φ 𝜍 𝑡𝑗𝑜 𝜒 + 𝑟 𝑑𝑝𝑡 Φ 𝜍 𝑑𝑝𝑡 𝜒 = 𝑟 𝜍 𝑑𝑝𝑡 Φ 𝑑𝑝𝑡 𝜒 + 𝑡𝑗𝑜 Φ 𝑡𝑗𝑜 𝜒 = 𝜍 𝑟 𝑑𝑝𝑡 𝜒 − Φ 𝑒𝑧 𝑒𝑨 = 𝜍 𝑒𝜒 𝑒𝜍 𝑏 2𝜌 𝐹 𝑟, Φ, 𝑢 = 𝐹 0 𝑓 𝑗 𝑙 𝑆 −𝜕 𝑢 𝑒𝜒 𝑓 − 𝑗 𝑙 𝜍 𝑟 𝑑𝑝𝑡 𝜒 − Φ න 𝜍 𝑒𝜍 න 𝑆 𝑆 0 0 Due to axial symmetry, we can choose: Φ = 0 25

A couple of integrals to solve: 𝑏 2𝜌 𝐹 𝑟, Φ, 𝑢 = 𝐹 0 𝑓 𝑗 𝑙 𝑆 −𝜕 𝑢 𝑒𝜒 𝑓 − 𝑗 𝑙 𝜍 𝑟 𝑑𝑝𝑡 𝜒 න 𝜍 𝑒𝜍 න 𝑆 𝑆 0 0 26

2𝜌 1 𝑒𝜒 𝑓 𝑗 𝑣 𝑑𝑝𝑡 𝜒 ≡ 𝐾 0 𝑣 Bessel function 2 𝜌 න of order zero 0 𝑏 2𝜌 𝐹 𝑟, Φ, 𝑢 = 𝐹 0 𝑓 𝑗 𝑙 𝑆 −𝜕 𝑢 𝑒𝜒 𝑓 − 𝑗 𝑙 𝜍 𝑟 𝑑𝑝𝑡 𝜒 න 𝜍 𝑒𝜍 න 𝑆 𝑆 0 0 27

𝑣 ≡ − 𝑙 𝑟 𝑆 𝜍 𝑏 𝐹 𝑟, Φ, 𝑢 = 𝐹 0 𝑓 𝑗 𝑙 𝑆 − 𝜕 𝑢 𝜍 𝑒𝜍 𝐾 0 − 𝑙 𝑟 2 𝜌 න 𝑆 𝜍 𝑆 0 2 𝑆 𝛽 ≡ −𝑙 𝑟 𝜍 𝑒𝜍 = α 𝑒α 𝜍 𝑙 𝑟 𝑆 −𝑙 𝑟 𝑏 𝑆 2 = 𝐹 0 𝑓 𝑗 𝑙 𝑆 − 𝜕 𝑢 𝑆 2 𝜌 න α 𝑒α 𝐾 0 α 𝑆 𝑙 𝑟 0 28

𝛽 න 𝛽 𝐾 0 𝛽 𝑒𝛽 ≡ 𝛽 𝐾 1 𝛽 0 29

−𝑙 𝑟 𝑏 𝑆 2 𝐹 𝑟, Φ, 𝑢 = 𝐹 0 𝑓 𝑗 𝑙 𝑆 − 𝜕 𝑢 𝑆 2 𝜌 න α 𝑒α 𝐾 0 α 𝑆 𝑙 𝑟 0 2 −𝑙 𝑏 𝑟 = 𝐹 0 𝑓 𝑗 𝑙 𝑆 − 𝜕 𝑢 𝑆 −𝑙 𝑏 𝑟 2 𝜌 𝐾 1 𝑆 𝑙 𝑟 𝑆 𝑆 𝜌 𝑏 2 2 𝐾 1 𝑙 𝑏 𝑟 = 𝐹 0 𝑓 𝑗 𝑙 𝑆 − 𝜕 𝑢 𝑆 𝑙 𝑏 𝑟 𝑆 𝑆 30

2 2 𝐾 1 𝑙 𝑏 𝑟 𝐽 0 ≡ 𝐹 0 2 𝑆 2 𝑆 2 𝜌 𝑏 2 2 𝐽 𝑟, Φ = 𝐽 0 𝑙 𝑏 𝑟 𝑆 𝐽 𝐽 0 ൗ 𝑙 𝑏 𝑟 𝑆 31