my chapter 23 lecture

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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