grade 8 light

Colors of Light

module 6

SoundLight

electromagnetic

transverse

travels very, very fast

mechanical

longitudinal

travels slower

Electromagnetic Spectrum

Transverse Wave

speed of light (c)

3.00x108 m/s

Behavior of Light

Reflection

Diffraction

Interference

Refraction

Reflection

Reflection

change of wave direction upon

hitting a reflecting surface

0ri

R

N

I

R= reflecting rayr= angle of reflectionI= incident rayi= angle of incidenceN= normal

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

Refraction

the bending of light when it travels from

one medium to another of different optical densities

Incident ray

Refracted Ray

θi > θr

Index of Refraction

Known indices of refraction

Water

Air

θi

θr

θi < θr

Vacuum

Air

θi

θrθi > θr

n = 1.360

n = 1.310

θi

θrθi < θr

Air

Ruby

θi

θrθi > θr

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

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

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

Problem Solving

• Speed in water• n = c/v; v = c/n• v = 3.00x108 m/s / 1.36• v =2.21 x108 m/s

Problem Solving

• Speed in substance• 2.21 x108 m/s (89%) =

1.97 x108 m/s

Problem Solving

• n of substance• n = c/v• n = 3.00x108 m/s / 1.97 x108 m/s• n = 1.528

Vacuum

Unknown Substance

θi

θr

θi = 30° ni = 1.000

θr = 16° nr = ?

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.

Air

Glass

θi

θr

sin θi = 0.707 ni = 1.310

sin θr = ? nr = 1.520

θr = ?

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.

Air

Seawater

θi

θr

θi = 35° ni = 1.0003

nr = 1.330θr = ?

Solution

θr = 25.56°

Air

Seawater

35°

25.56°

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?

Air

Unknown

θi

θr

θi = 23° ni = 1.0003

nr = ?θr = 14°

Solution

nr = 1.616

Air

Seawater

35°

25.63°

Electromagnetic Spectrum

Dispersion

the separation of visible light or other electromagnetic waves into

different wavelengths

Refractive indices of the Colors Light

Refractive indices of the Colors Light

Rainbows: Reflection

and Refraction

Rainbows: Reflection

and Refraction