Light: Reflection, Refraction, and Interference in Optics

Light: Reflection, Refraction, and Interference in Optics
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Explore the fascinating world of optics with topics ranging from the principles of reflection and refraction to the intricate phenomenon of interference. Delve into the concepts of index of refraction, chromatic dispersion, rainbows, and more. Discover the secrets behind the blue-green color of a Morpho's wing and the shifting colors of inks. Uncover the principles of Huygens and the laws of refraction, all illustrated with captivating images.

  • Optics
  • Reflection
  • Refraction
  • Interference
  • Huygens principle

Uploaded on Aug 14, 2024 | 1 Views


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  1. 11

  2. Sections 1. (reflection) (refraction) 2. (interference) 3. (diffraction)

  3. 11-1 Reflection and Refraction ( ) 1 1 n n = = (reflectio n) sin sin (refractio n) 2 2 1 1

  4. The Index of Refraction The Stealth Aircraft F-117A

  5. Chromatic Dispersion prisms and gratings

  6. Rainbows

  7. 11-2 Interference () What produces the blue-green of a Morpho s wing? How do colorshifting inks shift colors?

  8. Huygens principle All points on a wavefront serve as point sources of spherical secondary wavelets. After a time t, the new position of the wavefront will be that of a surface tangent to these secondary wavelets. Fig. 35- 2

  9. Law of Refraction from Huygens principle

  10. Fig. 35-3 10 35-

  11. v v = = = = t t 1 2 1 1 ec hg v v 1 2 2 2 = sin (for triangle hce) 1 1 hc = sin (for triangle hcg) 2 2 hc sin sin v v c v = = = 1 1 1 n Index of Refraction: 2 c v 2 2 c v = = and n n 1 2 1 2 sin sin c n c n n n = = 1 1 2 = sin sin n n Law of Refraction: 1 1 2 2 2 2 1 11 35-

  12. Phase Difference, Wavelength and Index of Refraction

  13. Wavelength and Index of Refraction v c v c = = = n n n n v c n c = = = = f f n n n The frequency of light in a medium is the same as it is in vacuum 35- 13

  14. Phase Difference Since wavelengths in n1 and n2 are different, the two beams may no longer be in phase Fig. 35-4 L L Ln Ln = = = Number of wavelengths in : 1 n N 1 1 n L 1 1 n L = = = Number of wavelengths in : 2 n N 2 2 n 2 2 n Ln Ln L ( ) = = Assuming : 2 2 n n N N n n 2 1 2 1 2 1 = 1/2 wavelength destructive interference N N 2 1 35- 14

  15. Ex.11-1 35-1 wavelength 550.0 nm n2=1.600 and L = 2.600 m

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