nelson physics 12

Copyright 2003 Nelson Unit 4 Are You Ready? 519

Unit 4 The Wave Nature of Light

ARE YOU READY? (Pages 440441)

Knowledge and Understanding 1.

The rays are reflected so that if normals are drawn in, the angle of incidence equals the angle of reflection for each ray. 2. Rays of light travelling from air into glass at an oblique angle will slow down and bend toward the normal. 3.

4. (a) i = 60 R = 21 n = ?

i

R

sin sin

sin 60 sin 21

2.42

n

n

n

=

=

=

The index of refraction of the diamond is 2.42. (b) v1 = 3.00 108 m/s v2 = ?

1

2

12

8

82

3.00 10 m/s2.42

1.24 10 m/s

vnvvvn

v

=

=

=

=

The speed of light in diamond is 1.24 108 m/s.

520 Unit 4 The Wave Nature of Light Copyright 2003 Nelson

(c) Since the angle of incidence in diamond (60) is greater than the critical angle, we expect that total reflection would occur.

( )( )

( )( )

2 2

1 1

2 11

2

1

sinsin

sinsin

sin 60 2.421

sin 2.10

nn

nn

=

=

=

=

Since sin 1 > 1, is undefined, there is no refracted ray. If no light is reflected, then total reflection occurs. 5.

6.

7. In longitudinal waves, the particles in the medium vibrate in the same direction as the motion of the waves. In transverse

waves, the particles in the medium vibrate at right angles to the direction of motion of the waves. 8. (a) A crest; B wavelength; C amplitude; D trough (b), (c)

(d) number of cycles = 10 cycles t = 2.0 s T = ? First calculate the frequency:

number of cycles

10 cycles2.0 s

5.0 Hz

ft

f

=

=

=

Copyright 2003 Nelson Unit 4 Are You Ready? 521

We can now calculate the period:

1

15.0 Hz 0.20 s

Tf

T

=

=

=

The period of the wave is 0.20 s. 9. f = 3.0 Hz v = 5.0 m/s = ?

5.0 m/s3.0 Hz

1.7 m

v fvf

=

=

=

=

The distance between adjacent troughs is 1.7 m, which is one wavelength. 10.

Points

Transmitted pulses are never inverted. When travelling from a fast medium to a slow medium, the reflected pulse is inverted. When travelling in a slower medium, the wavelength is smaller. The reflected pulse has the same wavelength as the incident pulse. The amplitude of the transmitted and reflected pulses is smaller than the amplitude of the incident pulse.


11. The three conditions necessary for two pulses to interfere destructively are 1. The two waves have the same wavelength (frequency). 2. The waves are in the same phase (i.e., crest on trough, trough on crest). 3. The waves have approximately the same amplitude.

12. (a) In a standing wave, the distance between adjacent nodes is 12 . From the diagram, 3 nodes = 6.0 m = 3

2 .

Therefore,

3 6.0 m2

4.0 m

=

=

The wavelength of the wave is 4.0 m. (b) number of cycles = 90 vibrations t = 1.0 min = 60 s v = ? Determine the frequency of the waves:

number of cycles

90 vibrations60 s

1.5 Hz

ft

f

=

=

=

We can now calculate the speed of the waves:

( )( )1.5 Hz 4.0 m6.0 m/s

v f

v

==

=

The speed of the waves is 6.0 m/s.

Inquiry and Communication 13. The observations suggest that as the wavelength decreases and travels through an opening that remains constant in size,

the amount of diffraction decreases.

14. Destructive interference will occur on a nodal line, when the path difference is: 1 3 5, , , etc.2 2 2

Since = 1.2 cm. the path differences are: 0.6 cm, 1.8 cm, 3.0 cm, etc.

Math Skills 15. 1

2nx n L

d

=

12

nx d

n L =

16. n = 3 = 632 nm = 6.32 106 m d = 153 m = 153 106 m = 1.53 104 m n = ?

6

3 4

1

3

1sin 2

1 6.32 10 msin 32 1.53 10 m

1.630 105.93

n n d

=

=

=

=

To the correct number of significant digits, the value of is 5.93.

Copyright 2003 Nelson Chapter 9 Waves and Light 523

Technical Skills and Safety 17. The principle precaution when working with bright sources of light is not to look directly into the beam of light. The light

can affect vision either temporarily or permanently. Although some filters can shield the eyes to a certain degree from welding torches or lasers, there are no filters that are absolutely dependable when looking directly at the Sun.

CHAPTER 9 WAVES AND LIGHT

Reflect on Your Learning (Page 442)

1. The photo is the interference of white light when passing through double slits (Young's experiment). Although the student may guess that it is some form of dispersion, it is interference.

2. The wavelengths (frequencies) in the source of light determine the colours seen. 3. Nodal lines would appear as dark lines or areas where destructive interference has occurred. (Dark is the absence of light.) 4. The dark vertical lines represent complete destructive interference.

Try This Activity: Diffraction of Light (Page 443)

2. Holding the double slits vertically in front of one eye, the student should see an equally spaced vertical pattern of alternating red and black lines or areas.

3. With a green filter covering the lamp, the same pattern should be seen as in step 2, except the bright lines are green. Also, the green lines and dark lines are closer together.

4. (a) A smaller slit width produced a wider pattern (b) Green light produced a wider pattern than the red light. (c) We know that diffraction has occurred because the light spreads out after passing through the slit. We know that

interference of light has occurred because there are bright lines and black lines, which represent areas of constructive and destructive interference.

9.1 WAVES IN TWO DIMENSIONS

PRACTICE (Pages 445, 450)

Understanding Concepts 1. Using the subscript 1 for shallow water and the subscript 2 for deep water, 2 = 2.0 cm v1 = 10.0 cm/s v2 = 18.0 cm/s 1 = ?

1 1

2 2

1 22

2

2

(10.0 cm/s)(2.0 cm)18.0 cm/s

1.1 cm

vv

vv

=

=

=

=

The wavelength of the wave in shallow water is 1.1 cm. 2. Using the subscript 1 for shallow water and the subscript 2 for deep water, v1 = 0.75v2 1 = 2.7 cm 2 = ?


2 22

1

2

2

2

(2.7 cm)0.75

3.6 cm

vv

vv

=

=

=

The wavelength of the wave in deep water is 3.6 cm. 3. Using subscript 1 for deep water and the subscript 2 for shallow water, v1 = 18.0 cm/s 1 = 2.0 cm v2 = 10.0 cm/s 2 = 1.1 cm f = ? For deep water:

1 1

1

1

18.0 cm/s2.0 cm

9.0 Hz

v fvf

f

=

=

=

=

For shallow water:

2

2

10.0 cm/s1.1 cm

9.0 Hz

vf

f

=

=

=

The frequency of the wave in both deep water and shallow water is 9.0 Hz. 4. 1 = 60 2 = 45 n = ?

1

2

sinsinsin 60sin 451.225, or 1.2

n

n

=

=

=

(a), (b) To calculate the ratios in the two media of the wavelengths, and the velocities, we can use the following ratio:

1 1 12 2 2

sinsin

vv

= =

Since 12

sinsin

= 1.2, the ratio for both the wavelengths and the velocities is 1.2.

(c) Since the frequency remains the same in the two media, the ratio is 1.0. 5. Using the subscript 1 for shallow water and the subscript 2 for deeper water, v1 = 28 cm/s 1 = 40 2 = 46 v2 = ?


1 1

2 2

2

2

sinsin

sin 40 28 cm/ssin 46

31 cm/s

vv

vv

=

=

=

The speed in the deeper water is 31 cm/s. 6. Using the subscript 1 for deep water and the subscript 2 for shallow water, f = 10.0 Hz v1 = 38.0 cm/s v2 = 28.0 cm/s 1 = 30 (a) n = ?

2 1

1 2

38.0 cm/s28.0 cm/s1.36

n vnn v

n

= =

=

=

The index of refraction is 1.36. (b) 1 = ? 2 = ? For the deep water:

1 1 1

11

1

1

38.0 cm/s10.0 Hz

3.8 cm

v fvf

=

=

=

=

For the shallow water:

( )( )

1 1

2 2

2 12

1

2

28.0 cm/s 3.8 cm38.0 cm/s

2.8 cm

vv

vv

=

=

=

=

The wavelengths in the deeper water and the shallow water are 3.8 cm and 2.8 cm, respectively. (c) 2 = ?

1

2

12

2

sinsin

sinsin

sin 301.36

21.6

n

n

=

=

=

=

The angle of refraction in shallow water is 21.6. 7. (a) Using the subscript 1 for region A and the subscript 2 for region B, 1 = 30 2 = 20 n = ?


1

2

sinsin

sin 30sin 20

1.46

n

n

n

=

=

=

The refractive index is 1.46. (b) 1 = 2.0 cm (by measurement) 2 = 1.35 cm (by measurement); 1.36 cm (by calculation) f = 6.0 Hz v1 = ? v2 = ?

For region A:

( )( )1 1 1

1

6.0 Hz 2.0 cm12 cm/s

v f

v

==

=

For region B:

( )( )2 2 2

2

6.0 Hz 1.36 cm8.2 cm/s

v f

v

==

=

The speed in region A is 12 cm/s. The speed in region B is 8.2 cm/s. 8. Using the subscript 1 for crown glass and the subscript 2 for air, sin 2 = 60 n1 = 1.52 n2 = 1.00 1 = ?

1 1 2 2

1

1

sin sin1.52 sin 1.00 sin 60

34.7

n n

=

=

=

The angle of incidence in crown glass is 34.7. 9. Using the subscript 1 for air and the subscript 2 for diamond, n1 = 1.33 n2 = 2.42 1 = 60 2 = ?

1 1 2 2

1 12

2

2

2

sin sinsin

sin

1.33 sin 60 2.42 sin28.0

n nn

n

=

=

=

=

The angle of refraction in diamond will be 28.0.

Section 9.1 Questions (Page 452)

Understanding Concepts 1. Using the subscript 1 for deep water and the subscript 2 for shallow water, v1 = 24 cm/s f1 = 4.0 v2 = 15 cm/s 1 = 40 2 = ?


(a) 1 12 2

sinsin

vv

=

( )( )

1 22

1

2

sinsin

sin 40 15 cm/s24 cm/s

24

vv

=

=

=

The refracted wave front makes a 24 angle with the boundary. (b) 2 = ? Calculate the wavelength in the deep water:

1 1

11

1

24 cm/s4.0 Hz

6.0 cm

v fvf

=

=

=

=

We can now calculate the wavelength in the shallow water:

( )( )

1 1

2 2

1 22

1

2

6.0 cm 15 cm/s24 cm/s

3.75 cm, or 3.8 cm

vv

vv

=

=

=

=

The wavelength in the shallow water is 3.8 cm. 2. Using the subscript 1 for deep water and the subscript 2 for shallow water, number of cycles = 10 waves t = 5.0 s 21 = 24.0 cm 1 = 24.0 cm

2 = 12.0 cm

22 = 18.0 cm 2 = 18.0 cm

2 = 9.0 cm

(a) v1 = ? v2 = ? In order to determine the speed of each wave, we must first calculate the frequency.

number of cycles

10 waves5.0 s

2.0 Hz

ft

f

=

=

=

For the speed of the wave in deep water:

( )( )1 1

1

2.0 Hz 12.0 cm24 cm/s

v f

v

==

=


For the speed of the wave in shallow water:

( )( )2 2

2

2.0 Hz 9.0 cm18 cm/s

v f

v

==

=

The speeds of the wave in deep water and in shallow water are 24 cm/s and 18 cm/s, respectively. (b) n = ?

1

2

24 cm/s18 cm/s1.33, or 1.3

vnv

n

=

=

=

The index of refraction is 1.3. 3. Using the subscript 1 for deep water and the subscript 2 for shallow water, f = 5.0 Hz v1 = 30 cm/s v2 = 27 cm/s 1 = 50 2 = ? (a) 1 1

2 2

sinsin

vv

=

22

sin 50 30 cm/ssin 27 cm/s

46.9 , or 44

=

=

The angle of refraction in the shallow water is 44. (b) n = ?

1

2

30 cm/s27 cm/s1.1

vnv

n

=

=

=

The index of refraction is 1.1. (c) = ?

27 cm/s5.0 Hz

5.4 cm

v fvf

=

=

=

=

The wavelength in shallow water is 5.4 cm. 4. Using the subscript 1 for deep water and the subscript 2 for shallow water: 1 = 2.0 cm f = 11 Hz 1 = 30 2 = 60 v1 = ? v2 = ? For the wave in deep water:

( )( )1 1

1

2.0 cm 11 Hz22 cm/s

v f

v

==

=


For the wave in shallow water, we must determine the angle of incidence and the angle of refraction. The angle between the boundary and the wave front is 60. Therefore, the angle of incidence (1) is 60. Similarly, the angle of refraction (2) is 30.

1 1

2 2

2

2

sinsin

sin 60 22 cm/ssin 30

12.7 cm/s, or 13 cm/s

vv

vv

=

=

=

The speeds of the wave in deep water and in shallow water are 22 cm/s and 13 cm/s, respectively. 5. Using the subscript 1 for cold air and the subscript 2 for warm air, v1 = 320 m/s v2 = 354 m/s 1 = 30 2 = ?

1 1

2 2

2

2

sinsin

sin 30 320 m/ssin 354 m/s

33.58 , or 34

vv

=

=

=

The angle of refraction in warm air is 34. 6. v1 = 7.75 km/s v2 = 7.72 km/s 1 = 20 2 = ?

1 1

2 2

2

2

sinsin

sin 20 7.75 km/ssin 7.72 km/s

19.9

vv

=

=

=

The angle of refraction is 19.9. 7. The two conditions for wave rays in water and light rays to exhibit total internal reflection are: 1. The energy (wave or light) must be travelling from a denser medium to a less dense medium. 2. The angle of incidence must be greater than the critical angle. 8. n1 = 1.30 n2 = 1.00 2 = 45 1 = ?

1 1

2 2

1

1

sin sin sin 1.30 sin 45 1.00

66.8 , or 67

nn

=

=

=

The angle of incidence is 67. 9. n1 = 1.33 n2 = 1.63 1 = 30 2 = ?


1 1

2 2

2

2

sin sin

sin 30 1.63 sin 1.33

24

nn

=

=

=

The angle of refraction is 24. 9.2 DIFFRACTION OF WATER WAVES


Understanding Concepts 1. The conditions required to maximize the diffraction of waves through a slit are: 1. long waves relative to the slit width, and/or 2. a small slit relative to the wavelength. 2. = 2.0 m w = 4.0 m

The condition for significant diffraction is w 1.

The ratio w = 2.0 m

4.0 m= 0.5, therefore, diffraction will not be noticeable.

3. = 6.3 104 m w = ?

The condition for significant diffraction is w 1.

4

4

4

6.3 10 m 1

6.3 10 m

6.3 10 m

ww

w

The maximum slit width that will produce noticeable diffraction is 6.3 104 m. 4. If there are larger slit widths (w > 6.3 104 m), there will only be slight diffraction at the edges of the slit, and the

majority of the wave will pass straight through without being diffracted. 9.3 INTERFERENCE OF WAVES IN TWO DIMENSIONS

PRACTICE (Page 459)

Understanding Concepts 1. = 1.98 m

1 2

1 2

1P S P S2112

1P S P S2

n n

n n

n

=

=

=

The smallest possible path difference is 12.


Since = 1.98,

1 0.99 m2 =

Therefore, the smallest corresponding path length difference is 0.99 m. 2. PS1 = 35.0 cm PS2 = 42.0 cm n = 3 f = 10.5 Hz = ? v = ? For the wavelength:

1 21P S P S2

35.0 cm - 42.0 cm132

2.80 cm

n n n

=

=

=

For the speed of the waves:

( )( )10.5 Hz 2.8 cm29.4 cm/s

v f

v

==

=

The wavelength of the sources is 2.80 cm. The speed of the waves is 29.4 cm/s. 3. PS1 = 29.5 cm PS2 = 25.0 cm n = 2 v = 7.5 cm/s For the wavelength:

n 1 n 2

n 1 n 2

1P S P S2

P S P S12

29.5 cm 25.0 cm122

3.0 cm

n

n

=

=

=

=

For the frequency:

7.5 cm= 3.0 cm

2.5 Hz

v fvf

f

=

=

=

The wavelength of the sources is 3.0 cm. The frequency of the sources is 2.5 Hz.



Understanding Concepts 1. For a two-point interference pattern to remain stable the two sources must: - be in-phase, - have the same frequency (wavelength), - have a fixed separation.

2. For two waves from identical sources to interfere destructively their path lengths must differ by 12.

3. Given sin n = 12n d

The maximum value of sin n is 1. Therefore,

1 12

112

nd

d n

Since n = 1,

1112

2

d

d

A ratio of less than 2 for d would produce no nodal line.

4. (a) If the distance between the sources is large, the distance between the nodal lines is also large, to the point where no nodal lines are observed.

(b) If the phase of the sources is constantly changing, the interference pattern is also constantly changing and probably cannot be observed.

5. xn = 10.0 cm L = 50.0 cm d = 5.0 cm n = 1 f = 6.0 Hz

To calculate the wavelength:

12

10.0 cm 5.0 cm150.0 cm 12

2.0 cm

nx dL n

=

=


To calculate the speed of the waves:

( )( )6.0 Hz 2.0 cm12 cm/s

v f

v

==

=

The wavelength is 2.0 cm and the speed of the waves is 12 cm/s. 6. Since the sources are in phase, an area of constructive interference is located along the right bisector. This passes through

the centre of the square whether the sources are adjacent or on opposite corners.

7. 3 = 3.00 m = 1.00 m d = 2.00 m L = 5.00 m n = 1

1

1

12

12

(5.00 m)( 1.00 m) 112.00 m 2

1.25 m

nx dL n

Lx nd

x

=

=

= =

A person would stand 1.25 m from the perpendicular bisector of the line between the openings. 8. n = 3 x3 = 35 cm L = 77 cm d = 6.0 cm 3 = 25 5 crests = 4.2 cm = ? Method 1:

5 crests = 44.2 cm

41.05 cm, or 1.0 cm

=

=


Method 2:

1sin 2

sin=12

(6.0 cm)(sin 25 )132

1.0 cm

n n dd

n

=

=

=

Method 3:

12

35 cm 6.0 cm177 cm 32

1.0 cm

nx dL n

=

=

=

The wavelength of the waves is 1.0 cm when calculated by all three methods. 9. = 2.00 m v = 338 m/s (a) f = ?

338 m/s2.00 m

169 Hz

vf

f

=

=

=

The frequency of the sound is 169 Hz. (b) Nodal lines occur when the path difference is:

1 1 3 5 or , , , etc.2 2 2 2

n Since = 2.00 m, the path differences could be 1.00 m, 3.00 m, and 5.00 m. Three possible distances from S2 are 7.00 m + 1.00 m = 8.00 m 7.00 m + 3.00 m = 10.00 m 7.00 m + 5.00 m = 12.00 m

(c) 2 11P S P S2n n

n =

112.0 m 5.00 m (2.00 m)2

7.00 m 12.00 m 24

n

n

n

=

= +

=

|

N is located on the 4th nodal line.

Applying Inquiry Skills 10. (a) Water waves are approximately transverse; however the water molecules move slightly back and forth as well as up

and down. In fact, an individual particle moves in a small oval path, as we see a cork move in water (see the figure below). This characteristic helps cause ocean waves to break if they become too large, or when they approach a beach where the depth becomes smaller and the waves slow down. Because small water ripples are very nearly


transverse, we can use the water in lakes and the water in ripple tanks in our study of waves. The ripples move slowly enough that we are able to study them directly.

(b) As the depth of the water decreases, the speed of the water and of the wavelength also decrease. The volume of the

displaced water in the wave is equal to the area under the wave. If the wavelength decreases, the amplitude increases to maintain the same volume of water (illustrated below). Also, water particles do not move straight up and down as in an ideal wave but move in small circles. As the water becomes shallow over the beach, there is bottom friction. The combination of these two effects causes the waves to break near the shore.

(c) Water waves are a good approximation of transverse waves because they can be observed more readily than sound,

light or electromagnetic waves. As long as the depth remains constant, water waves have a uniform speed and can be used to discover the properties of transverse waves. For a visual demonstration, see

http://www.infoline.ru/g23/5495/Physics/English/waves.htm.

Making Connections 11. d = 4.00 102 m f = 1.00 106 Hz v = 3.00 108 m/s (a) First we must calculate the wavelength of the radio signal:

8

63.00 10 m/s1.00 10 Hz3.00 m

vf

=

=

=

For interference maxima:

2

2

sin

3.00 10 m4.00 10 m

sin 0.750

n

n

nd

n

=

=

=

The only possible values for n are 0 and 1. Therefore, 0 = 0, or 1 = 49. For maximum intensity, the directions of the radio signal would be north, 49E of N and 49W of N.


9.4 LIGHT: WAVE OR PARTICLE?


Understanding Concepts 1.

2. If the speed of light changed when light was reflected, the wavelength would also change since the frequency is fixed by

the source. As a result, the colour of the reflected light might be different from the incident light, which would be difficult to observe. A better test would be to cause the reflected light to interfere with the incident light. If the wavelengths were different, interference would be impossible provided the light originated from the same source. Since we know that interference is possible, we conclude that the speed of light does not change when it is reflected.

3. Huygens principle is a method used to construct a succeeding wave based on the previous wave front. The procedure can be used in the same way for all waves, including sound and water waves.

4. The experimental evidence that indicates light could be a wave is: - both light and waves obey the laws of reflection - both light and waves obey Snells law - the speed of a wave slows down in a more dense medium, as does light - both waves and light can exhibit partial reflection-partial refraction and total internal reflection - waves exhibit dispersion, as does light 5. n = 1.50 va = 3.00 108 m/s vg = ? Using the particle theory:

( )( )

g

a

g a

8

8g

1.50 3.00 10 m/s

4.50 10 m/s

vn

vv nv

v

=

=

=

=

According to the particle theory of light, the speed of light in glass is 4.50 108 m/s. Note that this speed is greater than the speed of light in air, which is an impossible situation and will be discussed in Chapter 11.


Applying Inquiry Skills 6. Any reasonable mass moving at the speed of light has enormous momentum. When this mass is stopped by a black

absorbing surface, or reflected by a hand-held mirror, a large force would have to be applied to change this momentum. Since no force has to be applied at all (at least not one we can measure with normal instruments), we can conclude that the particles have an extremely small mass.

9.5 WAVE INTERFERENCE: YOUNGS DOUBLE-SLIT EXPERIMENT

PRACTICE (Page 473)

Understanding Concepts 1. 6x = 6.0 cm x = 6.0 cm

6 = 1.0 cm = 1.0 102 m

L = 3.0 m d = 220 m = 220 x 106 m = 2.2 104 m = ?

( ) 427

2.2 10 m1.0 10 m3.0 m

7.3 10 m

dxL

=

= =

The wavelength of the light is 7.3 107 m. 2. d = 0.042 mm = 4.2 105 m n = 5 5 = 3.8

( )5

7

1sin2

sin12

(4.2 10 m) sin 3.8152

6.2 10 m

n

n

nd

d

n

= =

=

=

The wavelength of the light is 6.2 107 m, or 6.2 102 nm. 3. = 6.3 107 m d = 43 m = 4.3 105 m L = 2.5 m x = ?

7

5

2

6.328 10 m2.5 m4.3 10 m

3.7 10 m

dxL

x Ld

x

=

=

=

=

The separation of adjacent nodal lines is 3.7 102 m, or 3.7 cm.


4. r = 6.0 102 nm = 6.0 107 m L = 1.5 m 10xr = 13.2 cm xr = 13.2 cm10 = 1.32 cm = 1.32 10

2 m

(a) d = ?

r

r7

2

5

(1.5 m)(6.0 10 m)1.32 10 m

6.8 10 m

Ldx

d

=

=

The separation of the slits is 6.8 105 m, or 68 m. (b) b = 4.5 102 nm xb = ?

x, therefore,

b b

r r

bb r

r2

2

b

4.5 10 nm (1.32 cm)6.0 10 nm0.99 cm, or 1.0 cm

xx

x x

x

=

=

=

=

The spacing between adjacent nodal lines for blue light is 1.0 cm. 5. = 656 nm = 6.56 107 m d = 0.050 mm = 5.0 105 m L = 2.6 m x = ?

7

4

2

(2.6 m)(6.56 10 m)5.0 10 m

3.4 10 m

dxL

Lxd

x

=

=

=

=

The fringes are 3.4 102 m, or 3.4 cm apart. 6. = 6.8 102 nm = 6.8 107 m n = 4 x = 48 mm = 4.8 102 m L = 1.5 m d = ?

7

2

5

12

12

1 (1.5 m)(6.8 10 m)42 4.8 10 m

7.4 10 m

n

n

x dL n

Ld nx

d

=

=

=

=

The separation of the two slits is 7.4 105 m, or 7.4 102 mm.


7. = 6.0 107 m L = 3.0 m 9x = 5.0 cm x = 5.0 cm

9 = 0.56 cm = 5.6 103 m

d = ?

7

3

4

(6.0 10 m)(3.0 m)5.6 10 m

3.2 10 m

dxL

Ldx

d

= =

=

=

The separation of the two slits is 3.2 104 m. 8. r = 6.0 107 m L = 1.5 m 10x = 2.0 cm x = 2.0 cm

10 = 0.20 cm = 2.0 103 m

(a) b = 4.5 107 m d = ?

b

7

4

3

(1.5 m)(4.5 10 m)4.5 10 m

1.5 10 m

Lx

d

x

=

=

=

The spacing between adjacent nodal lines is 1.5 103 m. (b) d = ?

r

r

7

3

4

(6.0 10 m)(1.5m)2.0 10 m

4.5 10 m

dxL

Ldx

d

= =

=

=

The separation of the slits is 4.5 104 m.


Understanding Concepts 1. Grimaldis work had shown that a beam of light passing through two successive narrow slits resulted in a beam of light

slightly larger than the width of the slits. He hypothesized that the beam bent slightly outward from the edges of the second slit, and named this diffraction. Diffraction was the important property Young used to demonstrate the interference of light. The sunlight fell on a card with two closely spaced pinholes, which allowed light to pass through onto a second card. These acted as point sources in phase. Young could not have made his discovery without the knowledge of diffraction provided by Grimaldi.

2. The observation of the double-slit interference pattern was more convincing evidence for the wave theory of light than the observation of diffraction. It was more convincing because destructive interferencea null resultis predicted and observed.

3. Since red > blue , there would be more nodal lines if other factors were kept constant.


4. Apart from the difficulty of making measurements in water, the spacing of the interference fringes would decrease by a factor of n (the index of refraction) of the water. For example, the first dark fringe would occur when the path difference was one-half the wavelength in water.

5. n = 5.4 d

= ?

1sin2

12

sin 122

sin 5.4

15.9 or 16

n

n

nd

nd

d

=

=

=

=

The ratio of the slit separation to the wavelength of the light is 16. 6. d = 0.15 mm = 1.5 104 m L = 2.0 m x = 0.56 cm = 5.6 103 m (a) = ?

3 4

7

(5.6 10 m)(1.5 10 m)2.0 m

4.2 10 m

dxL

=

=

=

The wavelength of the source is 4.2 107 m, or 4.2 102 nm. (b) = 6.0 102 nm = 6.0 107 m x = ? x, therefore,

1 1

2 2

11 2

27

37

31

(6.0 10 m) (5.6 10 m)(4.2 10 m)

8.0 10 m

xx

x x

x

=

=

=

=

The spacing of the dark bands would be 8.0 103 m, or 0.80 cm apart. 7. d = 0.80 mm = 8.0 104 m L = 49 cm = 4.9 101 m x = 0.33 mm = 3.3 104 m (a) = ?

4

41

7

8.0 10 m3.3 10 m4.9 10 m

5.4 10 m

dxL

=

=

=

The wavelength of the light is 5.4 107 m.


(b) d = 0.60 mm = 6.0 104 m x = ?

7 1

4

4

(5.4 10 m)(4.9 10 m)6.0 10 m

4.4 10 m

dxL

Lxd

x

=

=

=

=

The separation of the nodal lines would be 4.4 104 m. 8. d = 0.040 mm = 4.0 105 m L = 5.00 m x = 5.5 cm = 5.5 102 m = ?

( ) 527

4.0 10 m5.5 10 m5.0 m

4.4 10 m

dxL

=

= =

The wavelength of the light is 4.4 107 m. 9. = 6.3 102 nm = 6.3 107 m d = 3.3 105 m

1 = ? 2 = ? 3 = ?

1

7

5

1

sin (maxima)

(1)(6.3 10 m)=3.3 10 m

1.1

nd

=

=

Similarly, we find that 2 = 2.2 and 3 = 3.3. The angles, with respect to the slits, that locate the first-, second-, and third-order bright fringes on the screen, are 1.1, 2.2, and 3.3, respectively.

Applying Inquiry Skills 10. Since the sources are now 180 out-of-phase, the pattern would shift so that a nodal (dark) line along the right bisector

would occur, whereas previously there was an area of constructive interference (bright).


9.6 COLOUR AND WAVELENGTH

PRACTICE (Page 478)

Understanding Concepts

1. = 6.0 107 m f = ? v = c = 3.00 108 m/s

87

14

3.00 10 m/s6.0 10 m

5.0 10 Hz

c fcf

f

=

=

=

=

The frequency of the orange light is 5.0 1014 Hz. 2. f = 3.80 1014 Hz

= ?

8

14

7

3.00 10 m/s3.80 10 Hz7.89 10 m

cf

=

=

=

The wavelength is 7.89 107 m, or 789 nm. 3. Using the subscript 1 for air and the subscript 2 for alcohol,

1 = 4.4 107 m n1 = 1.00 n2 = 1.40 2 = ?

( )( )

2 1

1 2

1 12

2

7

72

1.00 4.4 10 m

1.403.1 10 m

nn

nn

=

=

=

=

The wavelength of the violet light in alcohol is 3.1 107 m. 4. Using the subscript 1 for air and the subscript 2 for turpentine,

n1 = 1.00 n2 = 1.47 1 = 6.5 107 m 1 = 40.0

(a) 2 = ?

( )( )

2 1

1 2

1 12

2

7

72

1.00 6.5 10 m

1.474.4 10 m

nn

nn

=

=

=

=

The wavelength of the red light in the turpentine is 4.4 107 m.


(b) 2 = ?

( )( )

1 1 2 2

1 12

2

2

sin sin sin

sin

1.00 sin 401.47

26

n nn

n

=

=

=

=

The angle of refraction is 26. 5. d = 0.12 mm = 1.2 104 m

L = 0.80 m x = 9.0 mm = 9.0 103 m n = 3 = ?

3 4

7

(9.0 10 m)(1.2 10 m)(3)(0.80m)

4.5 10 m

x nLd

xdnL

= =

=

=

The wavelength of the light used was 4.5 107 m. The colour used was blue. 6. d = 0.20 mm = 2.0 104 m

L = 2.0 m 1 = 4.0 102 nm = 4.0 107 m 2 = 6.0 102 nm = 6.0 107 m x = ?

To find x, we must first find xv and xr. For maxima (bright) for the violet light:

v

7

4

3v

4.0 10 m(2.0 m)(1)2.0 10 m

4.0 10 m

nx dL n

x Lnd

x

=

=

=

=

For the second-order band of the red light:

( )( )

r

7

4

3r

6.0 10 m2.0 m 12.0 10 m

6.0 10 m

x Lnd

x

=

=

=

We can now calculate x:

r v

3 3

3

6.0 10 m 4.0 10 m

2.0 10 m

x x x

x

=

=

=

The distance separating the violet from the red light is 2.0 103 m.



Understanding Concepts 1. The index of refraction for violet light in glass is slightly larger than that for red light. As a result, the violet light will be

refracted slightly more than the red light, and the focal lengths will be different in each type of lens. 2. White light passing through a flat piece of window glass is not broken down into colours as it is by a prism. This is

because window glass has parallel sides and is thin relative to a prism. As a result, there is not sufficient refraction for dispersion to be visible.

3. Using the subscript 1 for air and the subscript 2 for alcohol, 1 = 7.50 107 m n1 = 1.00 n2 = 1.40

2 = ?

2 1

1 2

1 12

27

72

(1.00)(7.5 10 m)1.40

5.4 10 m

nn

nn

=

=

=

=

The wavelength of the red light in alcohol is 5.4 107 m. 4. r = 4.00 102 nm = 4.00 107 m

v = 7.50 102 nm = 7.50 107 m fr = ? fv = ?

To determine the range of frequencies, we first calculate the frequency of red light:

rr

8

7

14r

3.00 10 m/s4.00 10 m7.50 10 Hz

cf

f

=

=

=

Calculate the frequency of the violet light:

v

8

7

14v

3.00 10 m/s7.50 10 m4.00 10 Hz

v

cf

f

=

=

=

The range of frequencies of visible light is from 4.00 1014 Hz to 7.50 1014 Hz. 5. = 5.8 102 nm = 5.8 107 m

n = 3 d = 0.10 mm = 1.0 104 m n = 3 = ?

7

4

3

sin

5.8 10 m(3)1.0 10 m

1.0

n n d

=

=

=

The angle for the third-order maximum is 1.0.


6. = 6.10 102 nm = 6.10 107 m n = 1 n = 3.0 d = ?

7

5

sin

sin

(1)(6.10 10 m)sin 3.0

1.2 10 m

n

n

nd

nd

d

= =

=

=

The separation between the two slits is 1.2 105 m. 7. In questions 5 and 6, it is difficult to observe and analyze the interference pattern produced by a double slit. The angle

predicting the direction of the fringes is so small, and therefore is difficult to measure. This problem will be remedied in the next chapter when we study the diffraction grating.

Making Connections 8. To produce a rainbow, all of the colours in white light must be present for reflection, refraction, and dispersion to occur

inside suspended raindrops in the atmosphere. Violet and red rays intersect inside the raindrop, and when the rays leave the raindrop, we see violet at the top, red at the bottom, and the other colours of the spectrum in between. This is known as a primary rainbow. Just after sunrise and just before sunset, the sunlight must pass through more of the Earth's atmosphere and the green-blue end of the spectrum is removed, primarily by scattering. Since all the colours are not present, a rainbow cannot be formed.

CHAPTER 9 LAB ACTIVITIES

Investigation 9.1.1: Transmission, Reflection and Refraction of Water Waves in a Ripple Tank (Pages 480481)

Observations Part 1: Transmission 23. When the surface of the water was lightly touched, the wave moved out in an expanding circle as illustrated. The arrows

indicate the direction of the motion of the wave front.


4. Rolling the dowel produced straight waves as illustrated. The arrows indicate the direction of the wave movement.

Part 2: Reflection 7. Reflection straight on

8. Reflection at an angle


Part 3: Refraction 11. Straight waves moving from deep to shallow water with an angle of incidence of 0.

14. Straight waves moving from deep to shallow water with an angle of incidence > 0.

15. Trial 1:

1

2

sinsinsin 45sin 301.41

n

n

=

=

=

16. Trial 1:

1

2

sinsinsin 40sin 281.38

n

n

=

=

=

Analysis (a) From a point source the wave front is circular, indicating that each part of the wave front has travelled the same distance

in the same time interval. Therefore, we can conclude that the speed of the wave is the same in all directions.


(b) An increase in the frequency results in a decrease in the wavelength. Any change in the frequency does not affect the speed of the wave in the ripple tank.

(c) As the waves pass from deep to shallow water the wavelength decreases and the wave ray is bent towards the normal.

(d) deepshallow

6.0 cm4.3 cm

=

deepshallow

1.39

=

The Snell's law value calculated in step 15 was 1.41. The values are compatible, within experimental error.

(e) To determine the ratio deepshallow

vv

, we must first calculate the frequency of the generator.

number of cycles

10 cycles5.0 s

2.0 Hz

ft

f

=

=

=

Using this value, we can calculate the speed of the wave in deep water:

( )( )deep

deep

2.0 Hz 6.0 cm12 cm/s

v f

v

==

=

We then calculate the speed of the wave in shallow water:

( )( )shallow

shallow

2.0 Hz 4.3 cm8.6 cm/s

v f

v

==

=

Using these values, we can determine the ratio:

deep

shallow

deep

shallow

12 cm/s8.6 cm/s

1.39

vvv

v

=

=

The ratio of deepshallow

vv

is 1.39. This compares favourably with the ratio of the wavelengths, as one would expect.

(f) The appraisal of the Prediction will vary for each student, depending on the prediction that was made.

Evaluation (g)

In the investigation we used the shadow wave, not the actual wave. Through appropriate scaling, the actual wavelengths could be used in the measurements and calculations.

More care could be taken when measuring the frequency of the generator by measuring the number of cycles over a longer time period or by using the strobe to 'stop' the waves and measuring the strobe frequency instead. In either case, a more accurate value for the frequency and the calculated speed could result.

The speeds of a wave front in deep and shallow water could be measured directly on the screen by recording the time interval for a wave front to travel a fixed distance (e.g., 30 cm). A more accurate value for the speeds would give a

more accurate value for the ratio of the speeds, deepshallow

vv

.

(h) Both the waves are stopped by the stroboscope because the frequency of the wave is the same in both deep and shallow water.


Synthesis (i) (i) Waves passing from shallow to deep water would be refracted away from the normal. (ii) Snell's law would predict their behaviour as follows

1

2

1

2

sin 1sin

11.38

sin0.72

sin

n

=

=

=

(iii) Partial internal reflection should be observed.

Investigation 9.2.1: Diffraction of Water Waves (Page 482)

Procedure (typical example) Diffraction around an Obstacle 1. Fill the tank with water to a depth of 1 cm. 2. Place a full wax block in the tank approximately 10 cm from the straight wave generator. The block should be standing on

its end, and parallel to the wave generator. Position the light source directly above the block. Using the wave generator, generate a straight periodic wave with a long wavelength. Make a sketch showing a series of wave fronts on both sides of the wax block.

3. Gradually reduce the wavelength of the wave by increasing the frequency.

Diffraction around an Edge 4. Place a wax block, with one side bevelled, at the centre of the wave tank, about 10 cm away from the wave generator.

Line up the other wax blocks so they prevent wave fronts from passing by the other side of the first wax block. 5. Generate a straight periodic wave with a long wavelength, and observe how the wave fronts pass by the bevelled edge of

the block. Record your observations in a sketch showing a series of wave fronts on both sides of the block. 6. Slowly decrease the wavelength of the wave, noting any changes in the amount of diffraction. Draw a sketch to illustrate

your answer.

Diffraction through an Opening 7. Using two bevelled pieces of wax, create a barrier about 10 cm from the wave generator, with an opening of

approximately 4 cm. Generate a wave with a long wavelength, observing the amount of bending that occurs. Increase the frequency gradually. Record your observations in the form of sketches.

8. While generating a wave with a constant frequency, slowly decrease the size of the opening by moving one of the blocks, noting how the size of the opening affects the diffraction. Observe the relationship between the size of the opening and the wavelength to achieve significant diffraction.

Observations Diffraction around an Obstacle 2. Waves were observed moving behind the edges of the obstacle as illustrated. Longer wavelength waves were diffracted

more than shorter wavelength waves.


Diffraction around an Edge 56. Waves were observed moving behind the edge of the obstacle as illustrated. Again, waves with longer wavelengths were

diffracted more than waves with shorter wavelengths.

Diffraction through an Opening 7. Waves were observed moving behind the edges of the opening, as illustrated. Increasing the frequency decreased the

wavelength and the amount of diffraction.

8. For a fixed wavelength, if the size of the opening was reduced, the amount of diffraction increased. It appears that to

maximize diffraction, the size of the opening must be smaller than the wavelength.

Analysis (b) There is more diffraction behind the obstacle or opening if the wavelength is longer. (c) Diffraction around an edge increases when the wavelength increases. (d) To keep diffraction minimal, the aperture width should be larger than the wavelength. (e) For maximum diffraction the wavelength should be large and the aperture should be smaller than the wavelength.

(f) When is approximately the same as the width w, diffraction is noticeable. In other words, 1w= .

If > w, noticeable diffraction will be observed. (g) The accuracy of the prediction will depend on the prediction made.

Evaluation (h) The evaluation of the design will depend on the design used in the Procedure.


Investigation 9.3.1: Interference of Water Waves in Two Dimensions (Pages 482484)

Observations, Data, and Calculations 2. As the two circular waves pass through one another neither is affected. At the point where they interfere for an instant, the

line is darker on the screen. 4. The nodal pattern is symmetrical on either side of the right bisector of the line connecting the two sources.

5. The frequency was determined for a fixed number of cycles of the two-point generator as follows:

number of cycles

20 cycles5.0 s

4.0 Hz

ft

f

=

=

=

The frequency was 4.0 Hz. 6. As the separation of the sources increased, the number of nodal lines increased and the spacing between the lines was

smaller. 7. As the phase of the sources was changed, the nodal pattern shifted. At 180, when the two sources were out of phase, a

nodal line ran down the right bisector to the line joining the two sources. No matter how the phase changed, the number of nodal lines remained constant, as long as the frequency of the sources was kept constant.

10. Points P1, P2, and P3 are on the first nodal line (n = 1) to the left side of the right bisector. Method 1:

12

12

x nL d

x dL n

=

=


Method 2:

1sin2

sin12

nd

d

n

= =

n = 1, d = 16.8 cm (This is the shadow distance on the screen, actual = 6.0 cm)

Method 1 Method 2

L (cm) x (cm) (cm) (cm) 18.0 4.0 13 7.5 7.6 38.0 8.0 12 7.1 6.0 45.0 23.5 14 7.8 8.1

11. Points P1, P2, and P3 are on the second nodal line (n = 2) to the right side of the right bisector. n = 2, d = 16.8 cm

Method 1 Method 2

L (cm) x (cm) (cm) (cm) 21.0 13.5 40 6.8 7.2 35.0 23.5 42 7.7 7.5 45.0 29.0 40 7.2 7.2

12. For the first prediction:

( )

avg

avg

67.5 7.1 7.8 7.6 6.0 8.1 cm

644.1 cm

67.4 cm

=

+ + + + +=

=

=

The average predicted wavelength is 7.4 cm. For the second prediction:

( )

avg

avg

66.8 7.7 7.2 7.2 7.5 7.2 cm

643.6 cm

67.3 cm

=

+ + + + +=

=

=

The average predicted wavelength is 7.3 cm. 13. These values were obtained by measuring the wavelength directly on the screen.

4 45.5 cm30.5 cm

47.6 cm

=

=

=

The wavelength was measured to be 7.6 cm. Both predictions were acceptable, within experimental error.


14. When the waves passed through the two openings, each opening acted as a point source and a two-point interference pattern was created on the other side of the paraffin block.

15.

Analysis (c) When the frequency of the source increases, the wavelength of the waves decreases and more nodal line are created in the

interference pattern. (d) When the sources are separated, the number of nodal lines on each side of the right bisector increases. (e) When the phase changes, the number of nodal lines remains constant and the pattern shifts position.

Evaluation (f) The accuracy of the prediction will depend on the prediction made. (g) The values for the mathematical predictions for the wavelength are close to those measured directly.

Synthesis (h) The measurements on the screen are measurements made on the shadows of the waves on the ripple tank surface. Since all

measurements are scaled up by the same factor, we are justified in using these values to test the mathematical relationships.

(i) If the relative phase of the identical sources is constantly changing phase, the interference pattern will be constantly shifting. Therefore, the pattern will be unstable, difficult to observe, and difficult to use to make measurements.

(j) The pattern created by diffraction of the waves from a single source through two slits is more stable than the pattern produced by two sources. This occurs because the waves from the two slits are always in phase since they originate from a single source. With two separate sources, there will always be some variations in the phase, which will shift the pattern every time the phase changes. We observed this in the experiment.


Investigation 9.5.1: Young's Double-Slit Experiment (Pages 484485)

Observations and Calculations 1. The pattern on the screen is made up of a series of equally spaced, narrow, vertical red bars with dark areas between. 2. The distance between the first and the eighth bright line was 4.2 cm

8 bright lines = 7 4.2 cm4.2 cm

70.60 cm

x

x

x

=

=

=

3. d = 3.20 104 m (on the slit plate)

4 2

7

(3.20 10 m)(0.60 10 m)3.00 m

6.4 10 m

xL d

d xL

=

=

=

=

The first predicted wavelength of the helium-neon laser light is 6.4 107 m. 4. First pair of slits: 7x = 7.0 cm x = 7.0 cm

7 = 1.00 cm

d = 2.00 104 m

7 2

7

(2.00 10 m)(1.00 10 m)3.00 m

6.67 10 m

xL d

d xL

=

=

=

=

The second predicted wavelength of the helium-neon laser light is 6.67 107 m. Second pair of slits: 7x = 13.3 cm x = 13.3 cm

7 = 1.90 cm

d = 1.00 104 m

4 2

7

(1.00 10 m)( 1.90 10 m)3.00 m

6.33 10 m

xL d

d xL

=

=

=

=

The third predicted wavelength of the helium-neon laser light is 6.33 107 m.


5. Accepted value for the wavelength of a helium neon laser is 633 nm, or 6.33 107 m The average value of the calculated wavelengths is:

avg

7

7avg

3(6.40 + 6.67 + 6.33) 10 m

36.46 10 m

=

=

=

Therefore, the average wavelength is 6.46 107 m.

Analysis (a) The average value of the wavelength that was calculated is correct to two significant digits. Since the accepted value is

6.33 107 m, we will use 6.46 107 m for the calculation of experimental error.

7 7

7

accepted value experimental value percent difference 100%

accepted value

6.33 10 m 6.46 10 m100%

6.33 10 mpercent difference 2.05%

=

=

=

This is an acceptable value for experimental error.

Evaluation (b) Factors that contribute to errors in measuring the wavelength of the light would include the given value of d and the

measurements of L and x. (c) To obtain a more accurate value for the wavelength of the laser light we would have concentrated on the measured values

of x, since d and L are fixed. More accurate values could be achieved by measuring the distance between a larger number of bright lines. Also, a magnifier may assist in achieving more accurate measurements.

Synthesis (d) As the value for d decreased, the interference pattern of bright and dark lines became more spread out as indicated by the

values for x. In other words as d decreased, x increased, as indicated by the proportionality statement 1xd

.

(e) This relationship 1xd

was exactly the same observation for water wave interference in the ripple tank. (f) This investigation strongly supports the wave theory of light. The predicted wavelength of light, based on the two-point

interference pattern, compares well with the predicted wavelength of light for water wave interference. Also, the mathematical relationships hold for both interferences, and can be used when making a prediction for the wavelength of light.

Investigation 9.6.1: Wavelengths of Visible Light (Pages 485486)

Prediction (a) Red light has a longer wavelength than green light.

Observations, Data, and Calculations 1. A pattern of nodal and bright spectral lines is observed on either side of the light filament. 2. The red light interference pattern is more spread out than that for green. In other words, xred > xgreen. 3. With the red filter covering the filament and the paper sliders 25 cm apart, 16 nodal lines were observed. 4. Prediction: The green light should have fewer nodal lines than the red light. 5. With the green filter covering the filament and the paper sliders 25 cm apart, 12 nodal lines were observed. 6. The ratio of the number of nodal lines can be found using the following relationship:

redgreen

16 1.3312

= =


7. Red light: Number of nodal lines between the paper sliders was 9. Sliders positioned at 25.0 cm and 45.0 cm. Distance between sliders is 45.0 cm 25.0 cm = 19.0 cm.

8 19.0 cm19.0 cm

82.38 cm

x

x

x

=

=

=

The average separation x of adjacent nodal lines for red light is 2.38 cm, or 2.38 102 m. 8. x = 2.38 102 m L = 1.00 m

d = 2.61 105 m

red

2 5

7red

(2.38 10 m)(2.61 10 m)1.00 m

6.21 10 m

dxL

=

=

=

The predicted wavelength of red light is 6.21 107 m. 9. Green light: (Step 6) Number of nodal lines between the paper sliders was 7.

Sliders positioned at 25.0 cm and 32.0 cm. Distance between sliders is 36.0 cm 32.0 cm = 12.0 cm.

6 12.0 cm12.0 cm

62.00 cm

x

x

x

=

=

=

The average separation x of adjacent nodal lines for green light is 2.00 cm, or 2.0 102 m.

(Step 7) L = 1.00 m d = 2.61 105 m

green

2 5

7green

(2.00 10 m)(2.61 10 m)1.00 m

5.22 10 m

dxL

=

=

=

The predicted wavelength of green light is 5.22 107 m. 10. The comparison of results from other groups will vary. 11. To find the wavelength of each source, we use a green helium-neon laser and the same procedure as followed in

Investigation 9.5.1. d = 1.0 104 m L = 3.00 m 7x = 11.3 cm x = 11.3 cm

7 = 1.61 cm = 1.61 102 m

4 2

7

(1.0 10 m)( 1.61 10 m)3.0 m

5.37 10 m

d xL

=

=

=

The wavelength of the green helium-neon laser is 5.37 107 m, or 537 nm. This value is comparable to 543 nm, the value provided with the laser specifications.


Analysis (b) The spectral colours are observed when white light is viewed through a double slit. This occurs because the various

colours of the visible spectrum have unique wavelengths and therefore interfere constructively at different positions in the interference pattern.

(c) The interference pattern for green light is more closely spaced than the interference pattern for red light because green light has a shorter wavelength.

(d) The ratio redgreen

16 1.3312

= = indicates that red light has a longer wavelength than green light. Since v = f, red has a lower

frequency than green light. (e) This agrees with the prediction that the wavelength for red light is longer than the wavelength for green light.

(f) The ratio from steps 8 and 9 is 7

red7

green

6.21 10 m 1.195.22 10 m

= =

. This compares favourably with the ratio 1.33 from step 6.

Evaluation (g) difference in measurementspercent difference 100%

average measurement=

1.33 1.19

1.26percent difference 11.1%

=

=

The percent difference is 11.1%. (h) The values from group to group will vary because of precision in measuring the number of nodal lines. Values will also

vary because the red and green filters will produce a range of wavelengths for that sector of the visible spectrum.

Synthesis (i) LEDs and lasers emit monochromatic (one wavelength) light, not a range of wavelengths. Since there would not be as

much variation from group to group, the observations and calculations should be more accurate.

CHAPTER 9 SUMMARY (Page 487)

Making a Summary

Property of Light Particle Theory Wave Theory transmission in a straight line

very strong weaker waves tend to spread out

reflection strong also supported by mechanics strong with both straight and circular waves refraction weak obeys Snell's law, but light must

speed up to bend towards the normal strong obeys Snell's law, wave rays bend towards the normal as the waves slow down

partial reflection/ refraction

very weak some particles reflect, some refract at the interface

strong can be demonstrated with water waves

dispersion very weak some particles bend more than others

strong demonstrated with waves of differing wavelengths

diffraction very weak edge affects direction of particles

strong can be demonstrated for obstacles and openings with water waves

interference does not explain very strong interference of waves predict interference of light

CHAPTER 9 SELF QUIZ (Pages 488489)

True/False 1. F The wave equation, v = f, can be applied to all waves. 2. T 3. F Waves with long wavelengths experience more diffraction than waves with shorter wavelengths.


4. F For a given slit, the amount of diffraction depends on the ratio w . For observable diffraction, 1

w .

5. F In a two-point, in phase interference pattern, increasing the wavelength of the two sources decreases the number of nodal lines.

6. F Decreasing the separation of the two-point interference pattern sources decreases the number of nodal lines. 7. T 8. F Early attempts to demonstrate the interference of light were unsuccessful because the two sources were too far apart and

out of phase, and the frequency of light is very large. 9. F Dispersion occurs because the refractive index of light is slightly higher for violet light than it is for red light. 10. F Youngs experiment validated the wave theory of light by showing the interferences of light waves.

Multiple Choice 11. (b) 2

1

vnv

=

8

83.00 10 m/s2.13 10 m/s1.41n

=

=

12. (d) Option (ii) is not valid because decreasing the depth of the water decreases the wavelength, which therefore decreases the amount of diffraction.

13. (a) (a) increasing the frequency of the source, decreases the wavelength and decreases the diffraction (b) increasing the amplitude of the waves has no affect on diffraction (c) decreasing the width of the slit increases the diffraction (d) decreasing the distance between the wave generator and the slit produces a wider area of diffraction (e) using a longer wavelength increases the amount of diffraction 14. (d) = 0.024 m = 2.4 cm n = 2

1path difference212 2.4 cm2

path difference 3.6 cm

n =

= =

15. (b) d = 4.5 cm n = 5

1sin2n

nd =

The maximum value for sin is 1. Therefore,

( )

( )

1 12

112

1 4.5 cm152

1.0 cm

nd

d

n

16. (e) For a point to be on a nodal line in a two-point interference pattern, where the sources are in phase, the condition is:

sin = xL

= 12

nd


In the diagram, the point is on the second nodal line, so the expression reduces to

122

23

xL d

dxL

= =

17. (d) 12

x nL d

=

12

dxnL

= +

The value of n can be increased by increasing either d or the separation of the sources, or by decreasing . In turn, can be decreased by increasing the frequency of the waves. A phase change would change the position of the nodal lines but would not increase their number.

18. (c) See the arguments made in 17. 19. (c) According to the model that Newton proposed, the particle theory cannot explain interference. 20. (c) All the statements are true except c. A nodal line only will be found down the centre line if the two point sources of

light are 180 out of phase 21. (a) x1 = 0.30 cm

L1 = 1.50 m = 1.5 102 cm L2 = 1.00 m = 1.00 102 cm x2 = ?

( )( )

1 1

2 2

1 22

1

2

2

2

0.30 cm 1.0 10 cm

1.50 10 cm0.20 cm

xL dx Lx L

x LxL

x

=

=

=

=

=

22. (d) = 340 nm = 3.40 107 m = 3.40 105 cm L = 2.0 m = 2.0 102 cm x = 3.4 cm

5 2

3

(3.4 10 cm)(2.0 10 cm)3.4 cm

2.0 10 cm, or 0.002 cm

xL d

Ldx

d

=

=

=

=

23. (e) The pattern has doubled or the value of x is half of its original value. Given that x

L =

d

(i) If the frequency of the source was decreased, the wavelength would increase and x would increase. (ii) If the frequency of the source was increased, the wavelength would decrease and x would decrease. (iii) If the width of each slit was increased, it would no affect on x. (iv) If the separation of the slits was increased, x would decrease. (v) If the separation of the slits was decreased, x would increase.


CHAPTER 9 REVIEW (Pages 490491)

Understanding Concepts 1. Since v = f, v , provided the frequency remains constant. Thus, if v decreases, also decreases. 2. Similarities between refraction and diffraction

both bend or change the direction of a ray of light Differences between refraction and diffraction

there is a change in speed with refraction; there is no change in speed with diffraction there is a wavelength change with refraction; there is no wavelength change with diffraction diffraction requires a slit or obstacle; refraction does not require a slit or obstacle

3. ng = n1 = 1.52 nw = n2 = 1.33

1 2

2 1

1

2

1.331.52

0.88

nn

=

=

=

To accommodate the same number of wavelengths, the water would have to be thicker than the glass since the wavelengths in water are longer. Thus, the thickness ratio (glass to water) is 0.88.

4. The measurements of the speeds of light in various media were not available to Newton. These measurements would have shown him that light slows down when it bends toward the normal, and does not speed up as he predicted.

5. Experimentally, reflection, refraction, partial-reflection, total internal reflection/refraction, interference, and dispersion can all be demonstrated with waves.

6. The light from the two headlights of a car does not produce an interference pattern because the headlights are too far apart, their light is not in-phase, and the sources are not point sources.

7. Youngs experiment is a pivotal event in the history of science because the experiment produced a null result (total destructive interference) providing the most important validation of the wave theory of light. Once established, the wave concept could be applied to a host of applications and concepts including electromagnetic waves.

8. 1 = 6.33 107 m 2 = 3.30 107 m nz = 1.92 nd = 2.42

1 1

2 2

1

27

7

6.33 x 10 m3.30 x 10 m1.92

vnv

n

n

= =

=

=

=

Therefore, the material is zircon. 9. 1 = 10.0 3

4n = 1.00

n = ( )4 1.003

= 0.75


1

2

12

2

sinsin

sinsin

sin 10.00.75

13.4

n

n

=

=

=

=

According to a particle theorist, the angle in water would be 13.4. 10. d = 0.50 mm = 5.0 104 m x = 7.7 mm = 7.7 103 m

L = 6.50 m = ?

( ) 437

5.0 10 m7.7 10 m6.50 m

5.9 10 m

dxL

=

= =

The wavelength of the light is 5.9 107 m. 11. n = 2.0

d = 0.038 mm = 3.8 105 m n = 1 = ?

( )1

5

7

sin (maxima)

sin

(3.8 10 m) sin 2.01

6.6 10 m

n n dd

n

= =

=

=

The wavelength of the light is 6.6 107 m. 12. d = 0.15 mm = 1.5 104 m = 482 nm = 4.82 107 m L = 2.0 m

( )7

1 4

31

(minima)12

12

1 4.82 10 m2.0 m 12 1.5 10 m

3.8 10 m

n

n

x dL n

x L nd

x

x

=

=

=

=

The second-order dark bands are 3.8 103 m, or 3.8 mm away from the central axis on either side. 13. = 5.50 102 nm = 5.50 107 m

n = 7 L = 2.00 m d = 0.10 mm = 1.0 107 m x7 = ?


( )( )7

7 4

27

(maxima)

5.50 10 m2.00 m 71.0 10 m

7.7 10 m

n

n

x dL n

x Lnd

x

x

=

=

=

=

The distance between the two 7th bright lines on either side of the right bisector would be 2 7.7 cm = 15.4 cm. 14. From the relationship sin n = 12n d

we can see that the maximum value of sin n is 1. Since n is the number of the

fringe, we can write d 1, or d. For this to be true, cannot be greater than d.

15. = 632.8 nm = 6.328 107 m L = 2.00 m x = 1.0 cm = 1.0 102 m

(a) d = ?

7

2

4 4

(6.328 10 m)(2.00 m)1.0 10 m

1.27 10 m, or 1.3 10 m

LxdLdx

d

=

=

=

=

The separation of the slits is 1.3 104 m. (b) n = 1 = ?

7

4

1sin2

1 6.328 10 m12 1.27 10 m

0.43

n

n

nd

=

=

=

The angle of the first-order dark fringes is 0.43. 16. n = 2.0

d = 3.8 105 m = ?

5

7

sin

sin

(3.8 10 m)sin2

6.6 10 m

n

n

n

nd

dn

= =

=

=

The wavelength of the light is 6.6 107 m. 17. The maximum is determined by the relationship sin n = nd

. The largest value for sin n is 1. Thus, nd 1.

Re-arranging the relationship we get: d 61.20 10 m A wavelength less than 1.20 106 m, or 12 nm, is invisible in the deep ultraviolet section of the spectrum.


18. f = 4.75 1014 Hz = ?

8

14

7

3.00 10 m/s4.75 10 Hz6.32 10 m

v fvf

=

=

=

=

The wavelength is 6.32 107 m, or 632 nm. By referring to Table 1 in Section 9.6, we see that this radiation is found in

the red part of the electromagnetic spectrum. 19. = 4.60 102 nm = 4.60 107 m

n = 2 min = ?

7

2

sin (maxima)

4.66 10 msin 2

n n d

d

=

=

2

1sin (minima)2

1sin 2 22

n n d

d

=

=

Since in the same position,

7

7

4.66 10 m 12 22

6.21 10 m

d d

= =

The wavelength of visible light that would have a minimum at P is 6.21 107 m, or 621 nm. 20. d = 0.158 mm = 1.58 104 m

r = 665 nm = 6.65 107 m yg = 565 nm = 5.65 107 m L = 2.2 m x = ?

We must first calculate the maxima for red light:

( )( )

r

7

4

2r

6.65 10 m2 2.2 m1.58 10 m

1.85 10 m

x nLd

x

=

=

=

We can then calculate the maxima for yellowgreen light:

yg

7

4

2yg

5.65 x 10 m(2)(2.2 m)1.58 x 10 m

1.57 10 m

x nLd

x

=

= =


Finally, we calculate the distance between the two fringes:

r yg

2 2

2

3

1.85 10 m 1.57 10 m

0.28 10 m

2.8 10 m

x x x

x

=

=

=

=

The distance between the third-order red fringe and the third-order yellow-green fringe is 2.8 103 m, or 28 mm. 21. Using the subscript 1 for air and the subscript 2 for water,

1 = 4.00 107 m d = 5.00 105 m L = 40.0 cm = 0.40 m n2 = 1.33 n1 = 1.00 x2 = ?

We must first calculate the wavelength in water:

( )( )

2 1

1 2

1 12

2

7

72

4.00 10 m 1.00

1.333.00 10 m

nn

nn

=

=

=

=

Using this value, we can calculate the distance between the fringes:

22

7

5

32

(3.00 10 m)(0.40 m)5.00 10 m)

2.4 10 m

Lxd

x

=

=

=

The fringes are 2.4 103 m, or 2.4 mm apart.

Applying Inquiry Skills 22. L = 1.00 m

d = 6.87 104 m r = ? b = ? Measurements on the photograph for red light (Figure 1, Section 9.6): 14xr = 5.1 cm

xr = 5.1 cm14 = 3.64 101 cm = 3.64 103 m

actual xr = 31 (3.64 10 m)4 = 6.94 104 m

rr

4 4

7r

(6.87 10 m)(9.10 10 m)1.00 m

6.25 10 m

d xL

=

=

=


Measurements on the photograph for blue light (Figure 1, Section 9.6): 9xb = 2.5 cm xb = 2.5 cm9 = 2.78 10

1 cm = 2.78 103 m

actual xb = 31 (2.78 10 m)4 = 6.94 104 m

bb

4 4

7b

(6.87 10 m)(6.94 10 m)1.00 m

4.77 10 m

d xL

=

=

=

The wavelengths of the red and blue light are 6.2 107 m and 4.77 107 m, respectively. 23. Measured quantities: d = 2.2 cm

PS1 = 4.9 cm PS2 = 4.6 cm Given: n = 3

2 1

2 1

1PS PS2

|PS PS |12

4.9 mm 4.6 mm132

1.2 mm

n

n

=

=

=

=

Since the scale is 1.0 mm = 1.0 cm, the true wavelength is 1.2 cm.

Making Connections 24. In their research the students will find that:

The storms with the large waves occur off the east coasts of Canada and the United States They usually occur in the fall. The largest waves and rogue waves are the result, not of one storm but two or more storms, often coming from

different directions. Large rogue waves can occur in relative calm water, sometimes a large distance for any storm. Constructive interference is the probably cause. Research is centering on past storm information which is being put into a computer model to simulate and predict

storms in the future. 25. d1 = 7.00 km = 7.00 103 m

d2 = 8.12 km = 8.12 103 m f = 536 kHz = 5.36 105 Hz To calculate wavelength:

8

5

2

3.00 10 m/s5.36 10 Hz5.6 10 m

cf

=

=

=


To calculate the difference:

2 1

3 3

3

8.12 10 m 7.00 10 m

1.12 10 m

d d d

d

= =

=

The difference is 1.12 103 m, which is equivalent to 2 (5.6 103 m), or 2. Since constructive interference occurs when the path difference is n, there is constructive interference at your house.

26. d = 7.00 m f = 85 Hz v = 346 m/s n = ?

First, we must calculate the wavelength of the bass tone from the speakers:

346 m/s842 Hz4.07 m

vf

=

=

=

We can then calculate the angle:

1

1

1sin21 4.07 msin 12 7.00 m

17

n n d

=

= =

The smallest angle where the audience has trouble hearing the bass tone is 17. 27. (a) Since the speakers are 180 out-of-phase, there will be a nodal line running down the right bisector. Therefore, you are

sitting at a low intensity point in the interference pattern. (b) d = 4.00 m

L = 5.0 m f = 842 Hz

83.00 10 m/s

842 Hz0.41 m

cf

=

=

=

Since the sources are 180 out-of-phase, the pattern shifts so that an intensity peak is now where a null area would have been and visa versa. Thus, the first loud intensity peak will be located using the following relationship for n = 1:

1

1

121 0.41m1 (5.0 m)2 4.00m

0.26 m

nx n L d

x

x

=

= =

You should move 0.26 m, or 26 cm due east along the opposite wall. 28. f = 1.0 MHz = 1.0 106 Hz

d = 585 m L = 19 km = 1.9 103 m xn = ?


First, we must calculate the wavelength:

8

6

2

c

3.00 10 m/s1.0 10 Hz3.0 10 m

=

=

=

Using the relationship (maxima)nx dL n

= :

2

3

2

3.0 10 m(1)(1.9 10 m)585 m

9.7 10 m

n

n

x nLd

x

=

= =

The receiver should be moved 9.7 102 m, or 9.7 km north. 29. = 488 nm = 4.88 107 m 2 1 = 1.0

1

71

1

sin

sin

1(4.88 10 m)sin

n

n

nd

nd

d

=

=

=

and 7

12

2(4.88 10 m)sind

=

Given 2 1 = 1.0

7 71 1

7 7

7

5

2(4.88 10 m) 1(4.88 10 m)sin sin 1.0

2(4.88 10 m) 1(4.88 10 m) sin1.0

4.88 10 msin1.0

2.8 10 m

d d

d d

d

d

=

=

=

=

The slit separation is 2.8 105 m. 30. 1 = 4.80 102 nm = 4.80 107 m

2 = 632 nm = 6.32 107 m d = 0.52 mm = 5.2 104 m L = 1.6 m n = 2 xn = ? Using the equation xn =

12

n Ld

, we can calculate the value for x480:

7

480 4

3480

1 4.80 10 m2 (1.6 m)2 5.2 10 m

2.2 10 m

x

x

=

=


Similarly, we can also calculate the value for x632:

7

632 4

3632

1 6.32 10 m2 (1.6 m)2 5.2 10 m

2.9 10 m

x

x

=

=

Using these values, we can calculate the distance between the second-order fringes:

632 480

3 3

2

2.9 10 m 2.2 10 m

7.0 10 m

x x x

x

=

=

=

The second-order fringes are 7.0 102 m, or 7 cm apart. 31. d = 2.0 m

f = 3.0 109 Hz L = 1.0 102 m t = 0.20 s v = ?

First we must calculate the value of the wavelength of the radio waves:

8

9

1

3.00 10 m/s3.0 10 Hz

1.0 10 m

cf

=

=

=

We can now calculate the distance between the point sources:

1

2 1.0 10 m(1.0 10 m)2.0 m

5.0 m

dxL

x Ld

x

=

=

= =

Finally, we calculate the speed of the car:

5.0 m0.20 s25 m/s

dvt

v

=

=

=

The car is moving at a speed of 25 m/s. 32. The radio will respond to the direct signal and to the reflected signal. If they are in-phase, constructive interference will

occur, and if out-of-phase, destructive interference will occur. When the girl moves a distance of 9.0 m, the intensity goes

from a maximum to a minimum. Since the 9.0 m move produces a path difference of 18 m, 12 = 18 m and = 36 m.

8

6

3.00 10 m/s=36 m

= 8.3 10 Hz

vf

f

=

The frequency of the radio transmitters is 8.3 MHz.

Copyright 2003 Nelson Chapter 10 Wave Effects of Light 569

CHAPTER 10 WAVE EFFECTS OF LIGHT

Reflect on Your Learning (Page 492) 1. Reflected light is polarized in the horizontal plane on some flat surfaces. Polaroid glasses polarize light in the vertical

plane. Since little light polarized in the vertical plane is received from the surface, the reflected light (glare) is significantly reduced.

2. The pits and bumps on the surface of the CD act as lines and the clear spaces between the adjacent lines reflected light. Together they act as a reflective diffraction grating producing the spectral colours since different wavelengths are diffracted constructively different amounts.

3. A transparent thin coating is placed on the lens to reduce unwanted internal reflection in the lens system. These losses by reflection reduce the transmitted light. Light directed at the thin surface coating constructively reflects light of a specific wavelength, in this case in the blue violet region of the spectrum.

4. As in the colours produced by soap films, some of the surface feathers have thin films that selectively produce reflective constructive interference. The thickness of the film determines how interference occurs constructively in the blue and green segments of the electromagnetic spectrum. Light soap bubbles, the thickness of the films is not uniform causing changes in the colours, depending on the angle of the incident light.

Try This Activity: Thin Film on Water (Page 493) (a) Dark areas in the oil film represent destructive interference. (b) The pattern is caused because the reflected light from the surface of the oil and from the oil-water interface interfere. The

thickness of the oil film determines whether constructive or destructive interference occurs for each colour, since each colour has a different wavelength.

(c) The pattern changes from broad bands of spectral colours to single bright and dark areas because the different filters change the light from white light to monochromatic light. Since a small range of wavelengths are striking the surface of the oil and water, the reflected light can be either diminished (black) or increased (bright) depending on the thickness of the oil film.

10.1 POLARIZATION OF LIGHT

Try This Activity: Polaroid Sheets (Page 494)

Observations 1. No change should be observed when the Polaroid is rotated through 180. 2. As the Polaroid is rotated, the intensity of light should diminish almost to zero, and then increase again as the rotation

continues. 3. As the Polaroid is rotated, the glare from the disk diminishes. 4. When rotating the Polaroid film and looking at various locations in the sky, some darkening should be observed

depending on the direction. This may bring clouds into more prominence.

Try This Activity: Polaroid Sunglasses (Page 496) You should see less reflection from the points where the light is reflected from a flat surface. Some types of autoglass may

show patterns of dark lines. In this case, the safety glass is under tension, causing polarization.



Understanding Concepts 1. Double refraction: certain crystals doubly refract the light creating two rays from one. Each ray is polarized in a different

direction. Polarizing filter: polarizing filters, such as Polaroid, polarize the light in one plane. Reflection: light striking a flat surface is polarized in a plane parallel to the surface. Scattering: small particles in the atmosphere scatter the light causing the transmitted light to be polarized.

2. Using two Polaroid filters, the intensity can be changed from maximum to zero, by rotating one of the Polaroids through 90.

3. Since there is no absorption and polarization from a mirrored surface, the Polaroid will not reduce the intensity of the light. Also, since the Sun is low in the sky, even glare will not be reduced as much, because of the large angle of incidence.

Applying Inquiry Skills 4. (i) Hold two pairs of sunglasses together. Holding them both up to the light so they are overlapping, keep one pair in a

fixed position and rotate the other one 90. If they are polarized, when they cross the light intensity should be zero (black).

(ii) Wear the sunglasses outside on a sunny day. Look at the glare coming off some flat surface, such as a car hood. If the glasses are Polaroid, the glare will be reduced.

(iii) Look at various points in a blue sky. If the glasses are Polaroid, you should see some darkening. 5. If you rotate the filter, the two As will disappear and reappear since each A is polarized in a different plane.

Making Connections 6. Students may list some of the following points:

liquid crystals are affected by electric current nematic liquid crystal, called twisted nematics, (TN), is naturally twisted; applying an electric current to these liquid

crystals untwists them to varying degrees, depending on the current's voltage LCDs use these liquid crystals because they react predictably to electric current a special polymer is added to two pieces of polarized glass that creates microscopic grooves in the surface; grooves are

on the side of the glass that does not have the polarizing film on it; a coating of nematic liquid crystals is added in to the grooves

a second piece of glass with the polarizing film and a grooved coating is lined up at a right angle to the first piece light that strikes the first filter is polarized and guided to the next layer; the liquid crystal layers change the light's plane

of vibration to match their own angle light reaches the far side of the liquid crystal substance and vibrates at the same angle as the final layer of molecules; if

the final layer is matched up with the second polarized glass filter, the light will pass through when liquid crystal molecules straighten out, they change the angle of the light passing through them so that it no

longer matches the angle of the top polarizing filter no light passes through that area of the LCD, making that area darker than the surrounding areas; controlling the

current in various parts of the film, black and clear areas are created that can be combined to create images (i.e., the numbers on a calculator)

small and inexpensive LCDs usually reflect light from external light sources; an LCD watch displays numbers where small electrodes charge the liquid crystals and make the layers untwist so that light is not transmitting through the polarized film.

computer displays have built-in fluorescent tubes above, beside, and sometimes behind the LCD; a white diffusion panel behind the LCD redirects and scatters the light evenly to ensure a uniform display

colour displays can be passive or active 7. (a) Until 1980, sugar beet farmers were paid by weight. The quality of the product (sugar level) was not a factor. The

polarimeter was called a saccharimeter. This was an important innovation for farmers because higher prices could be obtained for smaller bushels of higher quality sugar beets.

(b) Industries that widely use polarimetry to determine the active purity of raw materials such as vitamins, steroids, and antibiotics. The most common application is for sugar content in such products as chocolate, wine, jellies, flour, and lactose in milk.


10.2 DIFFRACTION OF LIGHT THROUGH A SINGLE SLIT

PRACTICE (Page 504) Understanding Concepts 1. = 7.50 102 nm = 7.50 107 m n = 2 w = 2.0 m = 2.0 108 m = ?

( )( )7

2 8

2

sin (minima)

2 7.50 10 msin

2.0 10 m49

nnw

=

=

=

The light produces a second minimum at an angle of 49. 2. = 15 = 580 nm = 2.80 107 m w = ?

1

17

6

sin (minima)

sin

5.80 10 msin 15

2.2 10 m

nwnw

w

=

=

=

=

The width of the slit is 2.2 106 m, or 2.2 m. 3. r > b

From the equation = w yL , we know that y .

Therefore, spacing for red light will be larger than spacing for blue light. 4. = 6.328 107 m w = 43 m = 4.3 105 m L = 3.0 m y = ?

( )( )7

5

2

3.0 m 6.328 10 m

4.3 10 m4.4 10 m

Lyw

y

=

=

=

Other than the central maxima, the separation of adjacent minima is 4.4 102 m, or 4.4 cm. 5. w = 3.00 106 m = 25.0 = ?


( )6

7

sin (minima)

sin

3.00 10 m sin12.5

16.49 10 m

n

n

nww

n

=

=

=

=

The wavelength is 6.49 107 m. 6. w = 1.5 105 m = 694.3 nm = 6.943 107 m

m = 2 = ?

( )7

5

12sin (maxima)

12 6.943 10 m21.5 10 m

sin 6.6

m

m

m

w

+

=

+

=

=

The angular position of the second maximum is 6.6. 7. a = 56 n = 1

b = 34 a

b

ww

= ?

Since sin (minima),nnw = wa =

asinn

and wb = bsin

n

.

Therefore, ab

ww

= a

b

sin

sin

. Cancelling out the , we get:

a b

b a

a

b

sinsinsin 34sin 56

0.67

ww

ww

=

=

=

The ratio ab

ww

of the two slits is 0.67.

Try This Activity: Resolution (Page 505) As the size of the hole increased, it was easier to see the two filaments separately. When viewing through the smallest hole,

the two filaments appeared as one. The larger the opening, the better the resolution.



Understanding Concepts 1. From the equation = w y

nelson physics 12

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