grade 8 light
TRANSCRIPT
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Colors of Light
module 6
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SoundLight
electromagnetic
transverse
travels very, very fast
mechanical
longitudinal
travels slower
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Electromagnetic Spectrum
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Transverse Wave
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speed of light (c)
3.00x108 m/s
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Behavior of Light
Reflection
Diffraction
Interference
Refraction
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Reflection
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Reflection
change of wave direction upon
hitting a reflecting surface
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0ri
R
N
I
R= reflecting rayr= angle of reflectionI= incident rayi= angle of incidenceN= normal
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Law of Reflection
States that waves are reflected from the boundary of the medium at the same
angle at which they strike it
θi = θr
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Refraction
the bending of light when it travels from
one medium to another of different optical densities
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Incident ray
Refracted Ray
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θi > θr
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Index of Refraction
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Known indices of refraction
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Water
Air
θi
θr
θi < θr
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Vacuum
Air
θi
θrθi > θr
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n = 1.360
n = 1.310
θi
θrθi < θr
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Air
Ruby
θi
θrθi > θr
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Problem Solving
• What is the speed of light in a diamond? (n = 2.147)
• n = c/v; v = c/n• v = c/2.147• v = 3.00x108 m/s / 2.147• v = 1.24 x 108 m/s
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Problem Solving
• The speed of light in ice is 2.29x108 m/s. What is the n of ice?
• n = c/v• n = c/2.29x108 m/s• n = 3.00x108 m/s / 2.29x108 m/s• n = 1.310
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Problem Solving
• The speed of light in a certain substance is 89% of its speed in water. What is the n of the substance?
• n of water = 1.360
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Problem Solving
• Speed in water• n = c/v; v = c/n• v = 3.00x108 m/s / 1.36• v =2.21 x108 m/s
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Problem Solving
• Speed in substance• 2.21 x108 m/s (89%) =
1.97 x108 m/s
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Problem Solving
• n of substance• n = c/v• n = 3.00x108 m/s / 1.97 x108 m/s• n = 1.528
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Vacuum
Unknown Substance
θi
θr
θi = 30° ni = 1.000
θr = 16° nr = ?
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Snell’s law/ Law of refraction
The ratio of the sines of the angles, where both angles are measured from the normal to the surface, is equal to
the inverse ratio of the two indexes of refraction.
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Air
Glass
θi
θr
sin θi = 0.707 ni = 1.310
sin θr = ? nr = 1.520
θr = ?
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Problem Solving
A ray of light strikes the seawater surface at an angle of incidence 35°. Compute the angle of refraction if the n of seawater is 1.33.
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Air
Seawater
θi
θr
θi = 35° ni = 1.0003
nr = 1.330θr = ?
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Solution
θr = 25.56°
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Air
Seawater
35°
25.56°
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Problem Solving
A ray of light traversing air enters another medium with an angle of incidence of 23° and a refracted angle of 14°. What is the n of the unknown medium?
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Air
Unknown
θi
θr
θi = 23° ni = 1.0003
nr = ?θr = 14°
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Solution
nr = 1.616
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Air
Seawater
35°
25.63°
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Electromagnetic Spectrum
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Dispersion
the separation of visible light or other electromagnetic waves into
different wavelengths
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Refractive indices of the Colors Light
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Refractive indices of the Colors Light
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Rainbows: Reflection
and Refraction
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Rainbows: Reflection
and Refraction
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