Physics 1402 Homework Solutions - Walker, Chapter 26 Conceptual ...

Physics 1402 Homework Solutions - Walker, Chapter 26 Conceptual Questions 11. When the mug is empty there is just air in it; therefore as light bounces off the bottom surface of the inside of the mug, it is not refracted when it reaches the top of the mug. When you fill the mug with water, light bouncing off the bottom surface of the inside of the cup reaches the surface of the water, where it passes from a region of high index of refraction (the water) to a region of lower index of refraction (the air). Whenever light passes from a region of high n to a region of lower n, the light is bent away from the normal. This means that it is possible for light rays to be bent far enough to reach your eye when the mug is filled with water, but not reach your eye when the mug is empty. Conceptual Exercises 3. The back side of the spoon acts like a convex spherical mirror. Therefore the image is right-side up, smaller than the object, and virtual. 9.

The satellite that the antenna receives signals from is far enough away that the rays coming into the antenna are essentially parallel. In order for parallel rays coming into the antenna to be focused onto the receiver, the receiver should be placed at the focal point, which is at a distance f = R / 2 in front of the dish.

Problems 3. The reflected ray is rotated through an angle 2θ . Consider the following figure showing the mirror in its original orientation (horizontal) and rotated clockwise through an angle θ . Before the mirror is rotated, the incident ray makes a certain angle with the normal that I will call (θi )old . In the statement of the problem, it says (θi )old is 37°. By the law of reflection, the reflected ray makes an angle (θ r )old = (θi )old on the other

side of the normal, as shown in the figure.

(θi )new

(θi )new θ

(θi )old

(θi )old

φ

When the mirror is rotated through any angle θ , nothing happens to the incident ray. But, as shown in the figure, the normal is rotated through the angle θ . If the mirror is rotated clockwise, as shown above, this increases the angle of incidence by θ , so that the new angle of incidence is (1) (θi )new = (θi )old + θ , as can be seen from the figure. Therefore, by the law of reflection, the angle between the new reflected ray and the new normal must be (θ r )new = (θi )new , as shown in the figure.

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Physics 1402 Homework Solutions - Walker, Chapter 26 Now consider the angle through which the reflected ray got rotated. This angle is shown as φ in the figure. From the figure, we see that φ = (θi )new − ⎡⎣(θi )old − θ ⎤⎦ φ = (θi )new − (θi )old + θ But, from (1), (θi )new = (θi )old + θ , so

φ = ⎡⎣(θi )old + θ ⎤⎦ − (θi )old + θ φ = 2θ So the reflected ray gets rotated through an angle that is twice the angle through which the mirror was rotated.

5.

Consider the following picture.

∆y

y2 y1 42° 32°

32°

2.0 m The purple reflected ray is the reflected ray before the mirror is tilted. It follows from the law of reflection that because the incident ray makes an angle of 32° with the horizontal, this reflected ray will also make an angle of 32° with the horizontal, as indicated in the figure. When the mirror is tilted upward through an angle of 5°, the reflected ray gets rotated through an angle of 10°, as discussed in Problem 3. This means that the new reflected ray (shown as blue in the figure) makes an angle 32D + 10D = 42D with the horizontal, as indicated in the figure. Now consider the purple and blue right triangles shown in the figure. From these two right triangles, it follows that y2 = ( 2.0 m ) tan 42D

and

Therefore

y1 = ( 2.0 m ) tan 32D ∆y = y2 − y1 = ( 2.0 m ) tan 42D − ( 2.0 m ) tan 32D = 0.55 m = 55 cm

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Physics 1402 Homework Solutions - Walker, Chapter 26 11. The situation is as shown below. The figure shows the observer’s eye 1.8 m above the floor. The mirror starts at the floor and has height H.

1.8 m h

θ θ H

0.80 m 1.5 m 3.0 m The observer just barely sees the table in the mirror when a ray from the top of the table hits the top edge of the mirror and enters the observer’s eye, as shown in the figure. Suppose a ray leaves the top of the table and makes an angle θ with the normal to the top edge of the mirror, as shown in the figure. Then by the law of reflection, this ray must reflect out to the person’s eye in such a way that the reflected ray makes the same angle θ with the normal, as shown in the figure. From the figure, the height of the mirror is (1) H = h + 0.80 m But also from the figure, h = (1.5 m ) tan θ (2) and

tan θ =

1.8 m − H 3.0 m

(3)

Using (2) and (3), (1) becomes ⎛ 1.8 m − H H = (1.5 m ) ⎜ ⎝ 3.0 m Solving this for H (a little algebra), I get H = 1.1 m

13. (a.)

⎞ ⎟ + 0.80 m ⎠

Consider the figure shown below. This figure shows the observer’s eye positioned a distance of 0.50 m away from the mirror. The height H of the building that the observer sees in the mirror is determined by rays reflected off the building that just barely hit the mirror, at the top and bottom edges of the mirror, as shown in the figure. Suppose the top incident ray in the figure makes an angle θ with the normal to the mirror, as indicated in the figure. Then by the law of reflection, this ray reflects out at the same angle θ below the mirror and enters the observer’s eye, as shown in the figure. Now notice the midline of the mirror, shown in blue in the figure. This midline and the normal at the top edge of the mirror constitute a pair of parallel lines being cut by a transversal (the ray reflected off the top of the mirror and entering the observer’s eye). Therefore, if the angle between the normal at the top edge of the mirror and the reflected ray is called θ , then the angle between this ray and the midline is also θ , as indicated in the figure.

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Physics 1402 Homework Solutions - Walker, Chapter 26

h

H 2

0.16 m

θ θ θ θ

θ H

θ 0.32 m

95 m

95.5 m

By symmetry, similar geometrical relationships exist for all the rays below the midline. Therefore, all the angles indicated as θ in the figure are, in fact, equal. Now consider the distance shown as H 2 in the figure. From the figure, this distance is given by H = h + 0.16 m (1) 2 Now let’s think about the right triangle formed by the top incident ray, the normal to the top edge of the mirror, and the distance h. From this right triangle, I get h tan θ = , 95.5 m from which it follows that h = ( 95.5 m ) tan θ (2) But also from the figure, I find tan θ =

0.16 m , 0.50 m

(3)

and putting (1), (2) and (3) together then gives H ⎛ 0.16 m ⎞ = ( 95.5 m ) ⎜ ⎟ + 0.16 m 2 ⎝ 0.50 m ⎠

⎡ ⎤ ⎛ 0.16 m ⎞ H = 2 ⎢( 95.5 m ) ⎜ ⎟ + 0.16 m ⎥ ⎝ 0.50 m ⎠ ⎣ ⎦ H = 61 m (b.)

The distance from the mirror to your eyes is the 0.50 m from Part (a.). Let’s call this distance d.

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Physics 1402 Homework Solutions - Walker, Chapter 26 Then the result we just got says

⎡ ⎤ ⎛ 0.16 m ⎞ H = 2 ⎢( 95.5 m ) ⎜ ⎟ + 0.16 m ⎥ ⎝ d ⎠ ⎣ ⎦ If the distance d decreases, H gets larger in this equation. Therefore, our answer to Part (a) would increase.

18.

If sunlight converges to a point 15 cm from the glass, then the focal length of the glass, viewed as a concave spherical mirror, is f = 15 cm . For a concave mirror, f is related to the radius of curvature R by f =

Therefore

R 2

R = 2 f = 2 (15 cm ) = 30 cm

19. We’re given that do = 30.0 cm and R = 40.0 cm . We want di . The mirror equation says 1 1 1 + = di do f For a concave spherical mirror, f = R / 2 = 20.0 cm , so 1 1 1 1 1 = − = − = 0.0167 cm -1 di f d o 20.0 cm 30.0 cm

di = 21.

1 0.0167 cm −1

= 60.0 cm

(a.)

I’ll let you draw the ray diagram. See the slides to check your answer. If you do this on graph paper, you should be able to estimate the image distance di and the image height hi from your diagram. This is what they want you to do.

(b.)

It’s the case of a concave mirror with the object farther away than the center of curvature. (The center of curvature is at a distance R = 2 f = 2 ( 0.50 m ) = 1.0 m from the mirror and we’re told that the object is 2.0 m from the mirror.) For this case, when you draw your ray diagram, you should find that the image is inverted. See the slides to see why and to check your ray diagram.

22. We know that ho = 42 cm , do = 2.0 m , and f = 0.50 m . From the mirror equation, we have 1 1 1 do − f = − = , di f do f do in which I got a common denominator on the right-hand side. Taking the reciprocal of both sides gives di : di = di =

f do do − f

( 0.50 m )( 2.0 m ) 2.0 m − 0.50 m di = 0.67 m

The magnification is m= m=−

hi d =− i ho do

0.67 m = −0.33 2.0 m

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Physics 1402 Homework Solutions - Walker, Chapter 26 37. The index of refraction is given by

c , v in which v is the speed of the light in the material with index of refraction n. We’re told that in this material, the light travels 0.960 m in 4.00 ns, so 0.960 m v= 4.00 × 10−9 s 3.00 × 108 m/s n= = 1.25 ⎛ 0.960 m ⎞ ⎜ ⎟ ⎝ 4.00 × 10−9 s ⎠ n=

46. Consider the figure shown below.

5.00 cm

θ2 θ1

20.0 cm As the incident ray comes in, it passes through the air-glass boundary at the left edge of the semicircular disk. At this boundary, the normal is as shown by the dashed black line in the figure. Therefore, the angle labeled as θ1 in the figure is the angle of incidence. (This is the angle called θ in the figure in the book.) By Snell’s law nair sin θ1 = nglass sin θ 2 , in which the angle θ 2 means the angle the refracted ray makes with the normal. (This angle is shown as θ 2 in the figure above.) So we have (1) sin θ1 = (1.52 ) sin θ 2 and we want θ1 . But what is θ 2 ? Well, consider the glass-to-air boundary the refracted ray crosses as it leaves the semicircular disk. The normal at this boundary is shown as the bold blue dashed line in the figure above. Notice that this normal passes through the center of the disk! This is a property of any circle: a line drawn perpendicular to the circle at any point on its circumference passes through the center of the circle. Therefore, because the refracted ray

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Physics 1402 Homework Solutions - Walker, Chapter 26 leaves from the center of the circle, this refracted ray comes into the glass-to-air boundary along the normal to this boundary. This means that the angle of incidence for the glass-to-air boundary is 0°! Therefore, by Snell’s law, the angle of refraction is also 0°, meaning that the ray is not refracted upon leaving the glass! Suppose, for example, that we call the angle between the refracted ray leaving the glass and the normal “ θ3 .” Then Snell’s law applied to this boundary says nglass sin 0D = nair sin θ3 . But sin 0D = 0 , from which it follows that θ3 = 0 , which means that this ray leaves along the direction of the normal. This is shown in the figure above. From the figure, the angle θ 2 is given by ⎛ 5.00 ⎞ D ⎟ = 14.0 20.0 ⎝ ⎠

θ 2 = tan −1 ⎜ Plugging this back into (1) gives

(

)

sin θ1 = (1.52 ) sin 14.0D = 0.369 So

θ1 = sin −1 ( 0.369 ) = 21.6D

47. Consider the figure shown in Part (a) below. Let’s call the angle that the observer’s line of sight makes with the horizontal θ 2 , as I’ve done. If the observer can just barely see the center of the bottom of the glass when it is filled with water, then the light reaching his eye from the center of the bottom of the glass must follow the path shown in Part (b). The light ray coming up from the midpoint of the bottom of the glass makes an angle I’ve called θ1 with the vertical, as shown in Part (b).

90D − θ 2

θ2

θ2 H 2 +W 2

H

H 2 +W 2

θ1

θ2

⎛W ⎞ H2 +⎜ ⎟ ⎝ 2⎠

2

θ2 W 2

W

(a)

(b)

Applying Snell’s law at the water-air boundary at the top of the glass, it must be true that nwater sin θ1 = nair sin ( 902 − θ 2 )

1.33sin θ1 = sin ( 90D − θ 2 )

(1)

But

sin ( 90D − θ 2 ) = sin 90D cosθ 2 − cos90D sin θ 2 (trig identity… see inside front cover of book)

So sin ( 90D − θ 2 ) = cos θ 2 and (1) becomes

1.33sin θ1 = cosθ 2

(2)

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Physics 1402 Homework Solutions - Walker, Chapter 26

Looking at Part (b.) of the figure below, I find that

( W2 ) 2 H 2 + ( W2 )

sin θ1 = and cos θ 2 =

Substituting these into (2), I get

W H +W 2 2

( W2 ) 2 H 2 + ( W2 )

(1.33)

W

=

H +W 2 2

,

which simplifies to 1.33

1

=

4H + W H +W 2 Now the task is to solve this for H. Squaring and cross-multiplying, I get 2

(1.33)

2

(H

2

2

2

+ W 2 ) = ( 4H 2 + W 2 )

⎡(1.33)2 − 1⎤ W 2 = ⎡ 4 − (1.33)2 ⎤ H 2 ⎣ ⎦ ⎣ ⎦

(1.33) − 1 2 4 − (1.33) 2

H =W Substituting in 6.2 cm for W, I get

57. (a.)

H = 3.6 cm

A concave lens (like the double-concave one we talked about in class) is the same as a diverging lens. I’ll let you draw the ray diagram. See the slides to check your answer.

(b.)

For a concave lens, you should find that the image is upright. See the slides to check your diagram.

(c.)

The image is virtual because there is no light really emanating from the location of the image. (See slides.)

59. (a.)

A convex lens (like the double-convex one we talked about in class) is the same as a converging lens. I’ll let you draw the ray diagram with the object being between the focal point and the lens. See the slides to check your answer.

(b.)

You should find that the image is upright. (See slides.)

(c.)

The image is virtual because there is no light really emanating from the location of the image. The refracted rays just appear to diverge from the location of the image. (See slides.)

Calculating Image Location ( di ) We know that do = f / 2 . So the thin-lens equation becomes 1 1 1 + = di ( f / 2 ) f

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Physics 1402 Homework Solutions - Walker, Chapter 26 1 1 1 1 2 1 = − = − =− di f ( f / 2) f f f Therefore di = − f This means that the image is farther away from the lens than the object is. (The object was at a distance f/2 from the lens.) The minus sign means that the image is on the same side of the lens as the object, so the image is virtual. If we wanted to calculate the magnification, we could do so from the equation h d m= i =− i ho do m=−

(− f ) = +2 ( f / 2)

The fact that the magnification is positive means that the image is upright. All of these facts are in agreement with the ray diagram shown in the slides for the diverging lens with the object between the focal point and the lens. (See the slides.)

74. A convex lens is a converging lens. We’re told that do = 1.2 cm . The fact that insect appears twice its actual size means that the magnification is either m = +2 (if the image is right-side up) or m = −2 (if the image is upside-down). They don’t say so, but they mean for you to assume that the image is right-side up. Therefore m=−

di =2 do

di = −2d o So the thin-lens equation becomes 1 1 1 + = f −2do d o 1 1 = f 2d o

f = 2d o = 2 (1.2 cm ) = 2.4 cm

93. (a.) m=

hi d =− i , ho do

and we’re told that do = 75.0 cm and f = 30.0 cm . But what’s di ? Well, we get this from the thinlens equation: 1 1 1 + = di do f 1 1 1 do − f = − = di f do f do di = di =

f do do − f

( 30.0 cm )( 75.0 cm ) 75.0 cm − 30.0 cm

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Physics 1402 Homework Solutions - Walker, Chapter 26 di = 50.0 cm So the magnification is m= (b.)

hi d 50.0 cm 2 =− i =− =− ho do 75.0 cm 3

Think of the left end of the arrow as being a dot. This dot is at an object distance ( do )left end = 76.0 cm . From the thin-lens equation, the image of this end of the arrow will be formed

at an image distance ( di )left end given by 1

+

1

( di )left end ( do )left end from which I get

( di )left and = ( di )left and =

=

1 , f

f ( d o )left end

( do )left end − f ( 30.0 cm )( 76.0 cm )

76.0 cm − 30.0 cm ( di )left end = 49.565 cm

Now think of the right end of the arrow as being a dot. This dot is at an object distance ( do )right end = 74.0 cm . From the thin-lens equation, the image of this end of the arrow will be formed at an image distance ( di )right end given by 1

+

1

( di )right end ( do )right end

=

1 f

Proceeding in a way exactly similar to what I did above, I get ( di )right end = 50.455 cm The length of the image is the difference between the image distance for the right end and the image distance for the left end: Li = ( di )right end − ( di )left end

Li = 0.889 cm We were told that the length of the object was Lo = 2.00 cm , so the longitudinal magnification is

Li 0.889 cm = = 0.44 2.00 cm Lo

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