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My Chapter 23 Lecture. Chapter 23: Reflection and Refraction of Light. Reflection Refraction Total Internal Reflection. § 23.2 Reflection of Light. - PowerPoint PPT Presentation
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MyChapter 23
Lecture
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Chapter 23: Reflection and Refraction of Light
•Reflection
•Refraction
•Total Internal Reflection
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§23.2 Reflection of Light
When light is reflected from a smooth surface the rays incident at a given angle are reflected at the same angle. This is specular reflection.
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Reflection from a rough surface is called diffuse reflection.
“Smooth” and “rough” are determined based on the wavelength of the incident rays.
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The angle of incidence equals the angle of reflection. The incident ray, reflected ray, and normal all lie in the same plane. The incident ray and reflected ray are on opposite sides of the normal.
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§23.3 Refraction of Light
When light rays pass from one medium to another they change direction. This is called refraction.
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Snell’s Law:
2211 sinsin θθ nn =
where the subscripts refer to the two different media. The angles are measured from the normal.
When going from high n to low n, the ray will bend away from the normal.
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The incident ray, transmitted ray, and normal all lie in the same plane.
The incident and transmitted rays are on opposite sides of the normal.
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Example (text problem 23.11): Sunlight strikes the surface of a lake. A diver sees the Sun at an angle of 42.0° with respect to the vertical. What angle do the Sun’s rays in air make with the vertical?
surfacen1 = 1.00; air
n2 = 1.33; water
Normal
42°
Transmitted wave
incident wave
θ1
( ) ( )
°==
°==
1.63
8920.0sin
42sin333.1sin00.1
sinsin
1
1
1
2211
θ
θ
θ
θθ nn
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§23.4 Total Internal Reflection
The angle of incidence for when the angle of refraction is 90° is called the critical angle θc.
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2c
22c1
2211
sin
90sinsin
sinsin
n
n
nnn
nn
=
=°==
θ
θθθ
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If the angle of incidence is greater than or equal to the critical angle, then no wave is transmitted into the other medium. The wave is completely reflected from the boundary.
Total internal reflection can only occur when the incident medium has a larger index of refraction than the second medium.
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Example (text problem 23.22): Calculate the critical angle for sapphire surrounded by air.
( ) ( )
°==
°==
4.34
565.0sin
90sin00.1sin77.1
sinsin
c
c
c
2211
θ
θ
θ
θθ nn
surface
n2 = 1.0; air
n1 = 1.77; sapphire
Normal
Transmitted wave
incident wave
θ2=90
θ1
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Summary
•The Laws of Reflection
•The Laws of Refraction
•Condition for Total Internal Reflection