The Electrodynamics of Rainbows

Transcription

1 Faculty of Science The Electrodynamics of University of Copenhagen Niels Bohr Institute Slide 1/17

2 Thinking Challenge: Explain the Physics of this Picture? Slide 2/17

3 Reflection and Transmission of Light in Waterdrops Law of Reflection : θ I = θ R Snells law : sinθ T sinθ I = n 1 n 2 n 1 sinθ I = n 2 sinθ T Slide 3/17 n 1 = n air 1 n 2 = n water 1.33 = n

4 Basic The blue light is reflected less because of dispersion Slide 4/17

5 Critical Angle of Basic With geometry one can calculate the critical angle θ C at which the light will leave the waterdrops k = h r = sin(θ I ) = n sin(θ T ) θ I = sin 1 (k) θ C = 4θ T 2θ I θ C k = 0 k maks = θ T = sin 1 ( k n 4 n 2 3 ) θ C maks 42 Slide 5/17

6 There is also a Secondary Rainbow Slide 6/17

7 Secondary The light is reflected two times because θ I is bigger The critical minimum angle is θ C = 51 The light that we see from secondary rainbows is refracted at the bottom of the waterdrop Slide 7/17

8 Double A double rainbow is when you see rainbows of different order at the same time Critical angles are beautifully shown by the dark area The secondary rainbow is less bright Is it all there is to say about the Thinking Challenge picture? Slide 8/17

9 Rainbow Around the Sun? Is there any suggestions to how this is possible? Slide 9/17

10 More Than Two Reflections? If the initial angle is larger, more reflections are possible Respectively 3 og 4 reflections of light in waterdrops The color of the outer ring of the rainbow changes Slide 10/17

11 Critical Angle of Tertiary & Quadrary The critical angles are respectively 318 and 404 or and Slide 11/17

12 Formation of Tertiary & Quadrary Light from waterdrops in front of the sun. The sun is in the middle Slide 12/17

13 Intensity of I have calculated the terminal intensity of the 4 different orders of rainbows: Order of Rainbow Terminal Intensity Relative to Initial Intensity Primary 8.78 % Secondary 3.52 % Tertiary 1.91 % Quadrary 1.20 % The low intensities makes highorder rainbows hard to see Larger surfaces and interference are aslo factors that makes it difficult to see them Slide 13/17

14 Research on Highorder The Thinking Challenge picture is from 2011 and is the first picture of a natural quadrary rainbow! It is possible to observe higher-order rainbows in laboratories by using extremely bright lasers In 1998 Ng et al. reported up to the 200th-order rainbow, by using an argon ion laser beam 1 1 Ng, P. H.; Tse, M. Y.; Lee, W. K. (1998). Observation of high-order rainbows formed by a pendant drop. Journal of the Optical Society of America B 15 (11): doi: /josab Slide 14/17

15 Thank you for your time! Slide 15/17

16 Calculation of Critical Angle of 3rd & 4th order Tertiary Quadrary θ C,3 = 8θ T 2θ I π θ C 16 n k = 0 k 2 maks,3 = θ C maks, θ C,4 = 2π + 2θ I 10θ T θ C 25 n k = 0 k 2 maks,4 = θ C min, General formular for k maks for a jth-order rainbow (j + 1) k maks,j = 2 n 2 (j + 1) 2 1 Slide 16/17

17 Calculation of Terminal Intensity α = cosθ T β = µ 1n 2 n 2 cosθ I µ 2 n 1 R = α β ( ) 2 ( 2 (α + β) 2 T 1 = αβ T 2 = (αβ) 1 2 α + β α 1 + β 1 The terminal intensity, of a jth-order rainbow equals: ) 2 T j = T 1 R j T 2 By inserting the above equations and k maks,j in T j, you get an expression which only depends on n and j: T j = 16(n 2 1) 2 (j + 1) 2 ( n ( n (j+1) 2 (n 2 1) j(j+2)n 2 (j+1) 2 (n 2 1) j(j+2)n 2 + j 2 (j + 2) (n 2 (j+1) 2 (n 2 1) j(j+2)n + 2 n 2 1 j(j+2) ) 2 ) 2 n 2 1 j(j+2) ) 4 n 2 1 j(j+2) Slide 17/17