f 7.5 x 1014 Hz FIGURE 24-12 The spectrum of visible light, showing the range of frequencies and wavelengths in air for the various colors. Many colors, such as brown, do not appear in the spectrum; they are made from a mixture of wavelengths. 6 x 1014 Hz 5 x 1014 Hz 4 x 1014 Hz UV IR 400 nm 500 nm 600 nm 700 nm EXAMPLE 24–3 Wavelengths from double-slit interference. White light passes through two slits 0.50 mm apart, and an interference pattern is observed on a screen 2.5 m away. The first-order fringe resembles a rainbow with violet and red light at opposite ends. The violet light is about 2.0 mm and the red 3.5 mm from the center of the central white fringe (Fig. 24–11). Estimate the wavelengths for the violet and red light. APPROACH We find the angles for violet and red light from the distances given and the diagram of Fig. 24–10. Then we use Eq. 24–2a to obtain the wavelengths. Because 3.5 mm is much less than 2.5 m, we can use the small-angle approximation. SOLUTION We use Eq. 24–2a (d sin 0 = m)) with m = 1, d = 5.0×10¬ª m, and sin 0 = tan 0 = 0. Also 0 ~ x/l (Fig. 24–10), so for violet light, x = 2.0 mm, and 5.0 x 10-4 m d sin 0 de d x 2.0 × 10-³ m = 4.0 × 10-7 m, m m m l 1 2.5 m or 400 nm. For red light, x = 3.5 mm, so d x 5.0 × 10¬ª m 3.5 × 10-³ m = 7.0 × 10-7 m = 700 nm. | 1 2.5 m EXERCISE B For the setup in Example 24–3, how far from the central white fringe is the first-order fringe for green light A 500 nm?
f 7.5 x 1014 Hz FIGURE 24-12 The spectrum of visible light, showing the range of frequencies and wavelengths in air for the various colors. Many colors, such as brown, do not appear in the spectrum; they are made from a mixture of wavelengths. 6 x 1014 Hz 5 x 1014 Hz 4 x 1014 Hz UV IR 400 nm 500 nm 600 nm 700 nm EXAMPLE 24–3 Wavelengths from double-slit interference. White light passes through two slits 0.50 mm apart, and an interference pattern is observed on a screen 2.5 m away. The first-order fringe resembles a rainbow with violet and red light at opposite ends. The violet light is about 2.0 mm and the red 3.5 mm from the center of the central white fringe (Fig. 24–11). Estimate the wavelengths for the violet and red light. APPROACH We find the angles for violet and red light from the distances given and the diagram of Fig. 24–10. Then we use Eq. 24–2a to obtain the wavelengths. Because 3.5 mm is much less than 2.5 m, we can use the small-angle approximation. SOLUTION We use Eq. 24–2a (d sin 0 = m)) with m = 1, d = 5.0×10¬ª m, and sin 0 = tan 0 = 0. Also 0 ~ x/l (Fig. 24–10), so for violet light, x = 2.0 mm, and 5.0 x 10-4 m d sin 0 de d x 2.0 × 10-³ m = 4.0 × 10-7 m, m m m l 1 2.5 m or 400 nm. For red light, x = 3.5 mm, so d x 5.0 × 10¬ª m 3.5 × 10-³ m = 7.0 × 10-7 m = 700 nm. | 1 2.5 m EXERCISE B For the setup in Example 24–3, how far from the central white fringe is the first-order fringe for green light A 500 nm?
Related questions
Question
Assume that light of a single color, rather than white
light, passes through the two-slit setup described in
Example 24–3. If the distance from the central fringe to
a first-order fringe is measured to be 2.9 mm on the screen,
determine the light’s wavelength (in nm) and color (see
Fig. 24–12).
Transcribed Image Text:f
7.5 x 1014 Hz
FIGURE 24-12 The spectrum of visible
light, showing the range of frequencies
and wavelengths in air for the various
colors. Many colors, such as brown, do
not appear in the spectrum; they are
made from a mixture of wavelengths.
6 x 1014 Hz
5 x 1014 Hz
4 x 1014 Hz
UV
IR
400 nm
500 nm
600 nm
700 nm
Transcribed Image Text:EXAMPLE 24–3 Wavelengths from double-slit interference. White light
passes through two slits 0.50 mm apart, and an interference pattern is observed
on a screen 2.5 m away. The first-order fringe resembles a rainbow with violet
and red light at opposite ends. The violet light is about 2.0 mm and the red 3.5 mm
from the center of the central white fringe (Fig. 24–11). Estimate the wavelengths
for the violet and red light.
APPROACH We find the angles for violet and red light from the distances given
and the diagram of Fig. 24–10. Then we use Eq. 24–2a to obtain the wavelengths.
Because 3.5 mm is much less than 2.5 m, we can use the small-angle approximation.
SOLUTION We use Eq. 24–2a (d sin 0 = m)) with m = 1, d = 5.0×10¬ª m, and
sin 0 = tan 0 = 0. Also 0 ~ x/l (Fig. 24–10), so for violet light, x = 2.0 mm, and
5.0 x 10-4 m
d sin 0
de
d x
2.0 × 10-³ m
= 4.0 × 10-7 m,
m
m
m l
1
2.5 m
or 400 nm. For red light, x = 3.5 mm, so
d x
5.0 × 10¬ª m
3.5 × 10-³ m
= 7.0 × 10-7 m = 700 nm.
|
1
2.5 m
EXERCISE B For the setup in Example 24–3, how far from the central white fringe is
the first-order fringe for green light A
500 nm?
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