reactance X Ω ) The proportionality constant between
3. What is the difference between a real image and a virtual image?
4. In each situation in Figure 33.37, draw the three rays emanating from the top of the object and reflecting or refracting from the optical element shown. Show the image, and state whether it is real or virtual.
answers:
1. On a calm day, the lake surface is smooth, and specular reflection is like that of a mirror. On a windy day, the surface is rough, which makes the reflection diffuse and prevents the formation of an image.
2. (a) The wavelength decreases when the wave travels more slowly in the second medium (i and iii) and in- creases when the wave travels faster in the second medium (ii and iv). (b) See Figure 33.38.
3. Real image: All rays actually pass through the location of the image, and the image can be seen on a screen placed at the image location. Virtual image: All rays do not pass through the location of the image (only the extensions of the rays do), and the image cannot be seen on a screen placed at the image location.
4. See Figure 33.39. For the lenses, the three principal rays can be used to locate the image; for the mirror, any rays and the law of reflection can be used to locate the image. The images are (a) real, (b) virtual, (c) virtual.
self-quiz
Figure 33.36
fast slow
slow fast
fast slow
slow
(i) (ii) (iii) fast (iv)
Figure 33.37
(a) (b) (c)
object object object
mirror
Figure 33.38
Figure 33.39
33.5 snel’s law 909
Quantitative tools
33.5 snel’s law
In the first part of this chapter, we saw that light refracts when it travels from one medium into another because the speed of light depends on the medium. The speed of light in a medium is specified by the index of refraction:
n K c0
c , (33.1)
where c is the speed of light in the medium and c0 is the speed of light in vac- uum. (By definition, nvacuum= 1; in air nair≈ 1.) If a light wave of frequency f travels from one medium into another, the frequency doesn’t change because the source determines the frequency (see also Checkpoint 16.10). The wavelength, however, does change; it is greater in the medium in which wave speed is greater.
The wavelength l of the light is related to the wave speed and frequency, in the same manner that these quantities are related for harmonic waves (Eq. 16.10).
In vacuum, for example,
l = c0
f (vacuum). (33.2)
In a medium in which a wave has speed c1, the wavelength l1 is given by l1= c1
f = c0>n1
f = 1
n1l , (33.3)
where l is the wavelength of the wave in vacuum and n1 is the index of refrac- tion of the medium. Thus, the wavelength decreases as the index of refraction increases. As discussed in Section 33.3, the amount of refraction a light wave undergoes varies somewhat with wavelength (see Figure 33.22) because different wavelengths of light travel at different speeds. Therefore the index of refraction depends on the wavelength. table 33.1 lists the indices of refraction for some common transparent materials at a wavelength of 589 nm.
Let us now work out the quantitative relationship between the angle of in- cidence and the angle of refraction. Figure 33.40 shows wavefronts and one ray for a beam of light incident on the interface between medium 1 and medium 2 at angle u1 from the normal. The angle of refraction is u2. Using right triangles ABD and ACD, we can express angles u1 and u2 in terms of the wavelengths l1 and l2:
sin u1= BD AD = l1
AD (33.4)
and sin u2= AC
AD = l2
AD . (33.5)
table 33.1 Indices of refraction for common transparent materials
Material n (for L=589 nm)
Air (at standard temperature and pressure) 1.00029
Liquid water 1.33
Sugar solution (30%) 1.38
Sugar solution (80%) 1.49
Microscope cover slip glass 1.52
Sodium chloride (table salt) 1.54
Flint glass 1.65
Diamond 2.42
Figure 33.40 Relationship between angle of incidence u1 and angle of refraction u2.
n1 n2 l2
l1 B
D A
C
u1 u1
u2
u2
Quantitative tools
Combining these equations to eliminate AD and substituting l1= l >n1 (Eq. 33.3), we get
sin u1 sin u2= l1
l2= l >n1
l >n2= n2
n1, (33.6)
which can be written as
n1 sin u1= n2 sin u2. (33.7) This relationship between the indices of refraction and the angles of incidence and refraction is called Snel’s law, after the Dutch astronomer and mathemati- cian Willebrord Snel van Royen (1580–1626).
Earlier in this chapter, we found that for rays traveling from a denser medium to a less dense medium, we can define a critical angle of incidence uc such that the angle of refraction u1 is equal to 90° (Figure 33.42); beyond this critical angle uc, total internal reflection occurs. We can calculate the critical angle uc for an interface between two media with indices of refraction n1 and n2 (n27 n1) by applying Snel’s law (Eq. 33.7) and setting u1= 90°:
sin u1 sin u2= 1
sin uc= n2
n1. (33.8)
Solving for uc gives uc= sin-1 a n1
n2b . (33.9)
n1 n2 uc
u1 = 90°
Figure 33.42 Critical angle for a ray traveling from a denser medium (n2) to a less dense medium (n16n2).
example 33.6 Bending 90°
A ray traveling through a medium for which the index of refraction is n1 is incident on a medium for which the index of refraction is n2. At what angle of incidence u1, expressed in terms of n1 and n2, must the ray strike the interface between the two media for the reflected and transmitted rays to be at right angles to each other?
❶ GettinG started This problem involves both reflection and refraction at an interface between two media. To visualize the problem, I draw the incident, reflected, and refracted rays and indicate that the reflected and refracted rays are 90° apart (Figure 33.41).
reflected and refracted rays. Therefore I need to use those re- lationships to obtain an expression that tells me the value of u1 that produces reflected and refracted rays oriented 90° to each other. To obtain u1 in terms of n1 and n2, I need to eliminate u2 from Eq. 33.7. To do so, I use the fact that the angles on the right side of the normal to the interface must add to 180°. Thus, with reflected and refracted rays forming a 90° angle, I can say 180° =u1+90° +u2. Solving this expression for u2 gives u2=90° −u1, which I can substitute into Eq. 33.7.
❸ execute plan Substituting u2=90° −u1 into Eq. 33.7, I get
n1 sin u1=n2 sin (90° −u1)=n2 cos u1, and isolating the terms that contain u1 gives
sin u1
cos u1=tan u1=n2 n1
u1=tan-1 an2 n1b. ✔
❹ evaluate result My result says that u1 increases as n2 in- creases. This makes sense because as n2 increases, the refracted ray bends more, meaning that u2 becomes smaller. To keep the reflected and refracted rays perpendicular to each other, the angle of reflection must increase, and so u1 must also increase.
Figure 33.41
❷ devise plan Snel’s law (Eq. 33.7), the law of reflection, and the indices of refraction determine the paths taken by the
33.5 snel’s law 911
Quantitative tools
example 33.7 Fermat’s principle
For a light ray that crosses the interface between medium 1 hav- ing index of refraction n1 and medium 2 having index of re- fraction n2, what relationship between u1 and u2 follows from Fermat’s principle (page 902)?
❶ GettinG started I begin with a diagram that shows the two media and a ray traveling from an arbitrary point A in the n1 me- dium to an arbitrary point C in the n2 medium (Figure 33.43).
Fermat’s principle states that the path the ray takes from A to C is the path for which the time interval needed for the motion is a minimum. Therefore this ray must cross the interface at a point B that makes the time interval a minimum.