CHAPTER 26 SOLUTIONS5. REASONING Since the light will travel in glass at a constant speed v, the time it takes to

pass perpendicularly through the glass is given by , where d is the thickness of the glass. The speed v is related to the vacuum value c by Equation 26.1:

SOLUTION Substituting for v from Equation 26.1 and substituting values, we obtain

7. REASONING The refractive index n is defined by Equation 26.1 as n = c/v, where c is the speed of light in a vacuum and v is the speed of light in a material medium. The speed in a vacuum or in the liquid is the distance traveled divided by the time of travel. Thus, in the definition of the refractive index, we can express the speeds c and v in terms of the distances and the time. This will allow us to calculate the refractive index.

SOLUTION According to Equation 26.1, the refractive index is

Using dvacuum and dliquid to represent the distances traveled in a time t, we find the speeds to

be

Substituting these expressions into the definition of the refractive index shows that

9. REASONING AND SOLUTION

a. We know from the law of reflection (Section 25.2), that the angle of reflection is equal to

the angle of incidence, so the reflected ray is reflected at .

b. Snell’s law of refraction (Equation 26.2: can be used to find the angle of refraction. Table 26.1 indicates that the index of refraction of water is 1.333. Solving for 2 and substituting values, we find that

12. REASONING AND SOLUTION The angle of incidence is found from the drawing to be

Snell's law gives the angle of refraction to be

sin θ2 = (n1/n2) sin θ1 = (1.000/1.333) sin 73° = 0.72 or θ2 = 46°

The distance d is found from the drawing to be

d = 8.0 m + (4.0 m) tan θ2 =

13. REASONING We will use the geometry of the situation to determine the angle of incidence. Once the angle of incidence is known, we can use Snell's law to find the index of refraction of the unknown liquid. The speed of light v in the liquid can then be determined.

SOLUTION From the drawing in the text, we see that the angle of incidence at the liquid-air interface is

The drawing also shows that the angle of refraction is 90.0°. Thus, according to Snell's law (Equation 26.2: ), the index of refraction of the unknown liquid is

From Equation 26.1 ( ), we find that the speed of light in the unknown liquid is

17. REASONING When the incident light is in a vacuum, Snell’s law, Equation 26.2, can be used to express the relation between the angle of incidence (35.0), the (unknown) index of refraction n2 of the glass and the (unknown) angle θ2 of refraction for the light entering the

slab: . When the incident light is in the liquid, we can again use

Snell’s law to express the relation between the index of refraction n1 of the liquid, the angle

of incidence (20.3), the index of refraction n2 of the glass, and the (unknown) angle of

refraction θ 2: . By equating these two equations, we can determine

the index of refraction of the liquid.

SOLUTION Setting the two equations above equal to each other and solving for the index of refraction of the liquid gives

21. The drawing at the right shows the geometry of the situation using the same notation as that in Figure 26.6. In addition to the text's notation, let t represent the thickness of the pane, let L represent the length of the ray in the pane, let x (shown twice in the figure) equal the displacement of the ray, and let the difference in angles 1 – 2 be given by .

We wish to find the amount x by which the emergent ray is displaced relative to the incident ray. This can be done by applying Snell's law at each interface, and then making use of the geometric and trigonometric relations in the drawing.

SOLUTION If we apply Snell's law (see Equation 26.2) to the bottom interface we obtain . Similarly, if we apply Snell's law at the top interface where the ray

emerges, we have . Comparing this with Snell's law at the bottom face, we see that , from which we can conclude that 3 = 1.

Therefore, the emerging ray is parallel to the incident ray.

From the geometry of the ray and thickness of the pane, we see that , from

which it follows that . Furthermore, we see that .

Substituting for L, we find

Before we can use this expression to determine a numerical value for x, we must find the value of 2. Solving the expression for Snell's law at the bottom interface for 2, we have

Therefore, the amount by which the emergent ray is displaced relative to the incident ray is

27. REASONING AND SOLUTION According to Equation 26.4, the critical angle is related to the refractive indices n1 and n2 by , where n1 > n2. Solving for n1, we

find

29. REASONING The refractive index nLiquid of the liquid can be less than the refractive

index of the glass nGlass. However, we must consider the phenomenon of total internal

reflection. Some of the light will enter the liquid as long as the angle of incidence is less than or equal to the critical angle. At incident angles greater than the critical angle, total internal reflection occurs, and no light enters the liquid. Since the angle of incidence is 75.0º, the critical angle cannot be allowed to fall below 75.0º. The critical angle θc is

determined according to Equation 26.4:

As nLiquid decreases, the critical angle decreases. Therefore, nLiquid cannot be less than

the value calculated from this equation, in which θc = 75.0º and nGlass = 1.56.

SOLUTION Using Equation 26.4, we find that

34. REASONING In the ratio nB/nC each refractive index can be related to a critical angle for

total internal reflection according to Equation 26.4. By applying this expression to the A-B interface and again to the A-C interface, we will obtain expressions for nB and nC in terms

of the given critical angles. By substituting these expressions into the ratio, we will be able to obtain a result from which the ratio can be calculated

SOLUTION Applying Equation 26.4 to the A-B interface, we obtain

Applying Equation 26.4 to the A-C interface gives

With these two results, the desired ratio can now be calculated:

39. REASONING Brewster's law (Equation 26.5: ) relates the angle of

incidence B at which the reflected ray is completely polarized parallel to the surface to the

indices of refraction n1 and n2 of the two media forming the interface. We can use

Brewster's law for light incident from above to find the ratio of the refractive indices n2/n1.

This ratio can then be used to find the Brewster angle for light incident from below on the same interface.

SOLUTION The index of refraction for the medium in which the incident ray occurs is designated by n1. For the light striking from above .

The same equation can be used when the light strikes from below if the indices of refraction are interchanged

47. REASONING We can use Snell's law (Equation 26.2: ) at each face of the prism. At the first interface where the ray enters the prism, n1 = 1.000 for air and

n2 = ng for glass. Thus, Snell's law gives

(1)

We will represent the angles of incidence and refraction at the second interface as and

, respectively. Since the triangle is an equilateral triangle, the angle of incidence at the

second interface, where the ray emerges back into air, is . Therefore, at the second interface, where n1 = ng and n2 = 1.000, Snell’s law becomes

(2)

We can now use Equations (1) and (2) to determine the angles of refraction at which the red and violet rays emerge into the air from the prism.

SOLUTION

The index of refraction of flint glass at the wavelength of red light is

ng = 1.662. Therefore, using Equation (1), we can find the angle of refraction for the red

ray as it enters the prism:

Substituting this value for 2 into Equation (2), we can find the angle of refraction at which

the red ray emerges from the prism:

For violet light, the index of refraction for glass is ng = 1.698. Again using

Equation (1), we find

Using Equation (2), we find

49. REASONING The ray diagram is constructed by drawing the paths of two rays from a point on the object. For convenience, we will choose the top of the object. The ray that is parallel to the principal axis will be refracted by the lens so that it passes through the focal point on the right of the lens. The ray that passes through the center of the lens passes through undeflected. The image is formed at the intersection of these two rays. In this case, the rays do not intersect on the right of the lens. However, if they are extended backwards they intersect on the left of the lens, locating a virtual, upright, and enlarged image.

SOLUTION

a. The ray-diagram, drawn to scale, is shown below.

Object

Scale:

20 cm

FF

Image

20 cm

From the diagram, we see that the image distance is and the magnification is

. The negative image distance indicates that the image is virtual. The positive magnification indicates that the image is larger than the object.

b. From the thin-lens equation [Equation 26.6: ], we obtain

The magnification equation (Equation 26.7) gives the magnification to be

m –dido

––75.0 cm30.0 cm

+2.50

52. REASONING The height of the mountain’s image is given by the magnification equation as hi = –hodi/do. To use this expression, however, we will need to know the image distance

di, which can be determined using the thin-lens equation. Knowing the image distance, we

can apply the expression for the image height directly to calculate the desired ratio.

SOLUTION According to the thin-lens equation, we have

(1)

For both pictures, the object distance do is very large compared to the focal length f.

Therefore, 1/do is negligible compared to 1/f, and the thin-lens equation indicates that di = f.

As a result, the magnification equation indicates that the image height is given by

(2)

Applying Equation (2) for the two pictures and noting that in each case the object height ho

and the focal length f are the same, we find

61. REASONING The magnification equation (Equation 26.7) relates the object and image distances and , respectively, to the relative size of the of the image and object:

. We consider two cases: in case 1, the object is placed 18 cm in front of a diverging lens. The magnification for this case is given by m1. In case 2, the object is

moved so that the magnification is reduced by a factor of 2 compared to that in case 1.

In other words, we have . Using Equation 26.7, we can write this as

(1)

This expression can be solved for do2. First, however, we must find a numerical value for

di1, and we must eliminate the variable di2.

SOLUTION

The image distance for case 1 can be found from the thin-lens equation [Equation 26.6: ]. The problem statement gives the focal length as .

Since the object is 18 cm in front of the diverging lens, . Solving for di1, we

find

where the minus sign indicates that the image is virtual. Solving Equation (1) for do2, we

have

(2)

To eliminate di2 from this result, we note that the thin-lens equation applied to case 2 gives

Substituting this expression for di2 into Equation (2), we have

Solving for do2, we find

69. REASONING The problem can be solved using the thin-lens equation [Equation 26.6: ] twice in succession. We begin by using the thin lens-equation to

find the location of the image produced by the converging lens; this image becomes the object for the diverging lens.

SOLUTION

a. The image distance for the converging lens is determined as follows:

This image acts as the object for the diverging lens. Therefore,

Thus, the final image is located .

b. The magnification equation (Equation 26.7: ) gives

Therefore, the overall magnification is given by the product .

c. Since the final image distance is negative, we can conclude that the image is .

d. Since the overall magnification of the image is negative, the image is .

e. The magnitude of the overall magnification is less than one; therefore, the final image is

.

75. REASONING We will apply the thin-lens equation to solve this problem. In doing so, we must be careful to take into account the fact that the lenses of the glasses are worn at a distance of 2.2 or 3.3 cm from her eyes.

SOLUTION

a. The object distance is 25.0 cm – 2.2 cm, since it is measured relative to the lenses, which are worn 2.2 cm from the eyes. As discussed in the text, the lenses form a virtual image located at the near point. The image distance must be negative for a virtual image, but the value is not –67.0 cm, because the glasses are worn 2.2 cm from the eyes. Instead, the image distance is –67.0 cm + 2.2 cm. Using the thin-lens equation, we can find the focal length as follows:

b. Similarly, we find

83. REASONING The angular size of a distant object in radians is approximately equal to the diameter of the object divided by the distance from the eye. We will use this definition to calculate the angular size of the quarter, and then, calculate the angular size of the sun; we

can then form the ratio .

SOLUTION The angular sizes are

Therefore, the ratio of the angular sizes is

quarter

sun

0.034 rad 3 70.0093 rad

.

91. REASONING The angular magnification of a compound microscope is given by Equation 26.11:

where is the focal length of the objective, is the focal length of the eyepiece, and L is the separation between the two lenses. This expression can be solved for , the focal length of the objective.

SOLUTION Solving for , we find that the focal length of the objective is

100. REASONING AND SOLUTION a. The lens with the largest focal length should be used for the objective of the telescope. Since the refractive power is the reciprocal of the focal length (in meters), the lens with the smallest refractive power is chosen as the objective, namely, the .b. According to Equation 26.8, the refractive power is related to the focal length f by

. Since we know the refractive powers of the two lenses, we can solve Equation 26.8 for the focal lengths of the objective and the eyepiece. We find that . Similarly, for the eyepiece,

. Therefore, the distance between the lenses should be

c. The angular magnification of the telescope is given by Equation 26.12 as

120. REASONING The angular magnification of a magnifying glass is given by Equation

26.10: , where N is the distance from the eye to the near-point. For

maximum magnification, the closest to the eye that the image can be is at the near point, with (where the minus sign indicates that the image lies to the left of the lens and is virtual). In this case, Equation 26.10 becomes . At minimum magnification, the image is as far from the eye as it can be ( ); this occurs when the object is placed at the focal point of the lens. In this case, Equation 26.10 simplifies to

.

Since the woman observes that for clear vision, the maximum angular magnification is 1.25 times larger than the minimum angular magnification, we have . This

equation can be written in terms of N and f using the above expressions, and then solved for f.

SOLUTION We have

Solving for f, we find that