Section 16.5 Temporal Interference of Sound Waves: Beats
17.2 Reflection and Refraction of Light
Figure17.3shows a beam of light of wavelengthλ1and speedv1represented by a light ray traveling in a straight line in medium 1. The beam encounters the smooth boundary surface (or interface) of the transparent medium 2, which is more dense than medium 1. Part of the incident light is reflected by the surface and another part penetrates medium 2 with wavelengthλ2and speedv2. Unless the incident beam is perpendicular to the surface, the ray that enters medium 2 is bent at the boundary and is said to be refracted.
Incident ray Reflected ray
Refracted ray Medium 1
Medium 2
Normal
θ1
θ2 1
2
1 Interface
1 1
θ ′=θ
Fig. 17.3 An incident ray in medium 1 is reflected from the interface and maintains the same speedv1, while the refracted ray is bent toward the normal and propagates in medium 2 with a speedv2< v1
In Fig.17.3, the incident, reflected, and refracted rays are all in a plane perpendic- ular to the boundary surface. In addition, the incident, reflected, and refracted rays make anglesθ1,θ1, andθ2,respectively, with the normal to the boundary surface.
Moreover,v1andv2are the speeds of the light rays in media 1 and 2, respectively.
Experiments and theory prove the following two laws:
Spotlight
• θ1=θ1 (Law of reflection) (17.1)
• v2sinθ1=v1sinθ2 (Law of refraction) (17.2) The speed of lightvin any material is less than its speed in vacuum c. It is found that the value ofv slightly depends on the wavelengthλ.Also, it is convenient to define a dimensionless quantity known as the index of refraction n of a material as follows:
n= c
v (17.3)
Since v is always less than c, then n>1 for any material and n=1 for vacuum.
Table17.1lists the indices of refraction for various materials.
As light crosses an interface between two media, its speedv and wavelengthλ change, but its frequency f remains the same. This can be understood by considering a normal incidence of light and treating light as photons, each with energy E=h f . If f changes, then energy will pile up at the interface, which is a mechanism that cannot take place under the laws of Physics. Since the relationv=λf must be satisfied in both media of Fig.17.3, and since the frequency f of the incident and refracted rays must be the same, then:
v1=λ1 f and v2=λ2 f (17.4)
If the media 1 and 2 have indices of refraction n1and n2,respectively, then Eq.17.3 leads to:
n1=c/v1 and n2=c/v2 (17.5)
17.2 Reflection and Refraction of Light 565
Table 17.1 Some indices of refractiona
Medium Index of refraction n
Vacuum Exactly 1
Airb 1.000 29
Carbon dioxideb 1.000 45
Water 1.333
Acetone 1.360
Ethyl alcohol 1.361
Sugar solution (30%) 1.38
Glycerin 1.473
Sugar solution (80%) 1.49
Benzene 1.501
Ice 1.309
Fused quartz 1.46
Polystyrene 1.49
Crown glass 1.52
Sodium chloride 1.544
Flint glass 1.66
Heaviest flint glass 1.89
Cubic zirconium 2.20
Diamond 2.419
Gallium 3.50
aFor light with a wavelength of 589 nm traveling in a vacuum.bAt 0◦C and 1 atm
Using Eq.17.4with Eq.17.5will give:
λ1
λ2 = v1
v2 =c/n1
c/n2 = n2
n1
(17.6)
This gives: n1λ1=n2λ2 (17.7)
If medium 1 is vacuum (or air), then n1=1 andλ1≡λ.In addition, if n is the index of refraction of medium 2, andλnis its refracted wavelength, then we find that:
n= λ λn
= Wavelength of the incident light in vacuum
Wavelength of refracted light in the medium (17.8) Since Eq.17.3leads to the ratiov1/v2=n2/n1,then the law of refraction given by Eq.17.2can be written as:
n1sinθ1=n2sinθ2 (Snell’s Law) (17.9) This form of the law of refraction is known as Snell’s law of refraction, and we will use this form in tackling most of our examples.
To compare the refractive angleθ2with the incident angleθ1and the relative ratio n1/n2for a light beam propagating from medium 1 to medium 2, we present the following results:
• If n2=n1, then θ2=θ1. In other words, the light beam will not be deflected (refracted) as it changes media, as in Fig.17.4a.
• If n2>n1, thenθ2< θ1. In other words, the light beam will refract and bend toward the normal as in Fig.17.4b.
• If n2<n1,thenθ2> θ1.In other words, the light beam will refract and bend away from the normal as in Fig.17.4c.
Normal
θ1
θ2
θ1
n1
n2
2 1
n >n
Normal
θ1
θ2
θ1
n1
n2
Normal
θ1
θ2
θ1
n1
n2
2 1
n =n n2< n1
(a) (b) (c)
Fig. 17.4 Light propagating from a medium of index of refraction n1into a medium of index of refraction n2.(a) When n2=n1,the beam does not bend. (b) When n2>n1,the beam bends toward the normal.
(c) When n2<n1,the beam bends away from the normal
Example 17.1
The wavelength of yellow light in vacuum is 600 nm.(a) What is the speed of this light in vacuum and water? (b) What is the frequency of this light in vacuum and water? (c) What is the wavelength of this light in water?
Solution: (a) The speed of the yellow light in vacuum (n=1) and water (n=1.333) can be obtained by using Eq.17.3as follows:
c=3×108m/s and v= c
n =3×108m/s
1.333 =2.25×108m/s
17.2 Reflection and Refraction of Light 567 (b) We use the equationv=λf to prove that the frequency of the yellow light in vacuum and water is the same, as follows:
f = c
λ = 3×108m/s
600×10−9m =5×1014Hz fn= v
λn = c/n λ/n = c
λ =5×1014Hz
(c) By using Eq.17.8, we can calculate the wavelength of the yellow light in water as follows:
λn= λ
n =600×10−9m
1.333 =4.501×10−7m=450.1 nm
Example 17.2
A beam of monochromatic light traveling through air strikes a slab of glass at an angleθ1=60◦to the normal, see Fig.17.5. The glass has a thickness t=1 cm and refractive index n=1.52. (a) Find the angle of refractionθ2.(b) Show that the emerging beam is parallel to the incident beam. (c) At what distance d does the beam shift from the original?
Fig. 17.5
θ1
θ2
θ1
n1
n2
Air
Glass
Air
θ2
θ3
d
3 1
n = n a
b t
Solution: (a) We apply Snell’s law at point a on the upper surface:
(1) n1sinθ1=n2sinθ2
sinθ2=n1
n2
sinθ1= 1
1.52sin 60◦=0.5698 ⇒ θ2=34.7◦ (b) Applying Snell’s law again at point b on the lower surface gives:
(2) n2sinθ2=n3sinθ3=n1sinθ3
Substituting n2sinθ2from Eq. (1) into (2) gives:
sinθ1=sinθ3
Therefore, θ3=θ1
Thus, the slab does not alter the direction of the emerging beam, it only shifts the beam laterally by an offset distance of magnitude d.
(c) From the geometry of the figure, we find that:
d=ab sin(θ1−θ2) and cosθ2=t/ab
Thus: d =sin(θ1−θ2)
cosθ2
t
Therefore, for a given incident angle θ1, the refracted angleθ2is solely deter- mined by n2, and the shift d is directly proportional to the thickness of the slab, t.
Substituting the values ofθ1,θ2, and t, into the above relation gives:
d =sin(60◦−34.7◦)
cos 34.7◦ ×1 cm=0.52 cm