Fisica Generale - Alan Giambattista, Betty McCarty Richardson Copyright © 2008 – The McGraw-Hill Companies s.r.l. 1 Chapter 23: Reflection and Refraction

$Page 1: Fisica Generale - Alan Giambattista, Betty McCarty Richardson Copyright © 2008 – The McGraw-Hill Companies s.r.l. 1 Chapter 23: Reflection and Refraction$
1

Fisica Generale - Alan Giambattista, Betty McCarty Richardson

Copyright © 2008 – The McGraw-Hill Companies s.r.l.

Chapter 23: Reflection and Refraction of Light

•Huygens’s Principle

•Reflection

•Refraction

•Total Internal Reflection

•Polarization by Reflection

•Formation of Images

•Plane Mirrors

•Spherical Mirrors

•Thin Lenses

2



§23.1 Huygens’s Principle

A set of points with equal phase is called a wavefront.

3



A ray points in the direction of wave propagation and is perpendicular to the wavefronts. Or a ray is a line in the direction along which light energy is flowing.

4



Huygens’s principle: At some time t, consider every point on a wavefront as a source of a new spherical wave. These wavelets move outward at the same speed as the original wave. At a later time t+t, each wavelet has a radius vt, where v is the speed of propagation of the wave. The wavefront at t+t is a surface tangent to the wavelets.

5



Geometric optics is an approximation to the behavior of light that applies when interference and diffraction are negligible. In order for diffraction to be negligible, the sizes of objects must be large compared to the wavelength of light.

6



§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.

7



Reflection from a rough surface is called diffuse reflection.

“Smooth” and “rough” are determined based on the wavelength of the incident rays.

8



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.

9



§23.3 Refraction of Light

When light rays pass from one medium to another they change direction. This is called refraction.

10



Snell’s Law

2211 sinsin 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.

11



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.

12



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

1

1.63

8920.0sin

42sin333.1sin00.1

sinsin

1

1

1

2211

nn

13



§23.4 Total Internal Reflection

The angle of incidence for when the angle of refraction is 90° is called the critical angle.

1

2

221

2211

sin

90sinsin

sinsin

n

n

nnn

nn

c

c

14



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.

15



Example (text problem 23.22): Calculate the critical angle for sapphire surrounded by air.

4.34

565.0sin

90sin00.1sin77.1

sinsin

1

2211

c

c

nn

surface

n2 = 1.0; air

n1 = 1.77; sapphire

Normal

Transmitted wave

incident wave

2=90

1

16



§23.5 Polarization by Reflection

Brewster’s angle is the angle of incidence for which the reflected light is completely polarized.

Light is totally polarized when the reflected ray and the transmitted ray are perpendicular.

i

tB

BtBtBi

ttii

n

n

nnn

nn

tan

cos90sinsin

sinsin

17



Example (text problem 23.32): (a) Sunlight reflected from the still surface of a lake is totally polarized when the incident light is at what angle with respect to the horizontal?

1.53

33.100.1

33.1tan

air

water

B

B n

n

The angle is measured from the normal, so 90 - 53.1 = 36.9 is the angle from the horizontal.

18



(b) In what direction is the reflected light polarized?

Example continued:

It is polarized perpendicular to the plane of incidence.

19



Example continued:

(c) Is any light incident at this angle transmitted into the water? If so, at what angle below the horizontal does the transmitted light travel?

9.36

6000.0sin

sin333.11.53sin00.1

sinsin

2

2

2

2211

nnFrom Snell’s Law:

The angle is measured from the normal, so 90 - 36.9 = 53.1 is the angle from the horizontal.

20



§23.6 Formation of Images

An image is real if light rays from a point on the object converge to a corresponding point on the image.

A camera lens forms a real image.

21



Your eye focuses the diverging rays reflected by the mirror.

The light rays appear to come from behind the mirror.

An image is virtual if the light rays from a point on the object are directed as if they diverged from a point on the image, even though the rays do not actually pass through the image point.

22



Example (text problem 23.35): A defect in a diamond appears to be 2.00 mm below the surface when viewed from directly above that surface. How far beneath the surface is the defect.

Surface

Actual location of defect

Air

n2 =1.00

Diamond

n1 = 2.4191

2

1

2

y

y’

23



The angles 1 and 2 are related by Snell’s Law:

2211 sinsin nn

The actual depth of the defect is y and it appears to be at a depth of y’. These quantities are related by:

12 tantan yy

Example continued:

24



Dividing the previous two expressions gives:

2211 coscos ynyn

As long as you are directly above the defect and its image, the angles 1 and 2 are nearly 0°. Rays from only a narrow range of angles will enter your eye. The above expression simplifies to:

1

2

21

n

n

y

y

ynyn

(general result)

Example continued:

25



The actual depth of the defect in the diamond is then

mm. 84.4mm 00.200.1

419.2

2

1

y

n

ny

Example continued:

26



§23.7 Plane Mirrors

A point source and its image are at the same distance from the mirror, but on opposite sides of the mirror.

Treat an extended object as a set of point sources.

27



Example (text problem 23.41): Entering a darkened room, Gustav strikes a match in an attempt to see his surroundings. At once he sees what looks like another match about 4 m away from him. As it turns out, a mirror hangs on one of the walls. How far is Gustav from the wall with the mirror?

The image seems 4 m away, but the mirror is only 2 m away since the rays will appear to come from a point 2 m behind the mirror.

28



§23.8 Spherical Mirrors

A convex (or diverging) mirror curves away from the observer.

Principal axis

vertexCenter of curvature

The focal point

29



A ray parallel to the principle axis is reflected, and it appears to have come from point F, the focal point of the mirror.

For a convex mirror, the focal point is on the axis and is located a distance 0.5R behind the mirror, where R is the radius of curvature.

30



Drawn in green, red, and blue are the principal rays.

1. A ray parallel to the principal axis is reflected as if it came from the focal point. (green)

2. A ray along a radius is reflected back upon itself. (red)

3. A ray directed toward the focal point is reflected parallel to the principal axis. (blue)

31



For the pencil in the previous figure, the image is upright, virtual, smaller than the object, and closer to the mirror than the object.

32



Principal axis

vertexCenter of curvature

The focal point

A concave (or converging) mirror curves toward the observer.

33



1. A ray parallel to the principal axis is reflected through the focal point. (green)

2. A ray along a radius is reflected back upon itself. (red)

3. A ray along the direction from the focal point to the mirror is reflected parallel to the principal axis. (blue)

Drawn in green, red, and blue are the principal rays.

34



The magnification is defined as .sizeobject

size image

h

hm

An inverted image has m<0 and an upright image has m>0.

The expression for magnification can also be written as

p

qm where p is the object distance and

q is the image distance.

35



The mirror equation:

fqp

111

where f is the focal length of the mirror. f<0 when the focal point is behind the mirror.

36



Example (text problem 23.46): An object 2.00 cm high is placed 12.0 cm in front of a convex mirror with a radius of curvature of 8.00 cm. Where is the image formed?

fqp

111

where p = 12.0 cm, f = -0.5R = -4.00 cm, and q is the unknown image distance. Solving gives q = -3.00 cm. The image is behind the mirror.

37



§23.9 Thin Lenses

A diverging lens will bend light away from the principle axis.

A converging lens will bend light toward the principal axis.

38



39



Magnification:

The thin lens equation:

p

q

h

hm

fqp

111

40



Example (text problem 23.64): A diverging lens has a focal length -8.00 cm.

(a) What are the image distances for objects placed at various distances from the lens? Is the image real or virtual? Upright or inverted? Enlarged or diminished?

Object distance

Image distance

Real / virtual?

Upright / inverted?

Enlarged/ diminished

5 cm -3.08 cm Virtual upright Diminished





41



(b) If the object is 4.00 cm high, what is the height of the image?

Object distance

Image distance

Magnification Image height

5 cm -3.08 cm 0.616 2.46 cm

8 cm -4.00 cm 0.500 2.00 cm

14 cm -5.09 cm 0.364 1.45 cm

16 cm -5.33 cm 0.333 1.33 cm

20 cm -5.71 cm 0.285 1.14 cm

Example continued:

42



Summary

•The Laws of Reflection

•The Laws of Refraction

•Condition for Total Internal Reflection

•Condition for Total Polarization of Reflected Light

•Real/virtual Images

•Mirrors (plane & spherical)

•Thin Lenses

Documents

Fisica Generale - Alan Giambattista, Betty McCarty Richardson Copyright © 2008 – The McGraw-Hill Companies s.r.l. 1 Chapter 23: Reflection and Refraction