Question 1

Two radio antennas are 10 km apart on a north-south axis on high mountain tops at the seacoast. The antennas broadcast identical AM radio signals, in phase, at a frequency of 4.70 MHz. A steamship, 200 km offshore, travels toward the north at 15 km/h and passes east of the antennas. A radio on board the ship is tuned to the broadcast frequency. The reception of the radio signal on the ship is a maximum at a given instant. How much later until the next occurrence of maximum reception? (c = 3.00 × 10⁸ m/s)

Accepted Answer

To find the time until the next occurrence of maximum reception, we can use the concept of constructive interference from the two antennas. The path difference for constructive interference is given by $n\lambda$, where $n$ is an integer and $\lambda$ is the wavelength of the radio waves. The wavelength $\lambda$ can be calculated using the speed of light $c$ and the frequency $f$ of the radio waves: $\lambda = \frac{c}{f}$.Given:- $c = 3.00 \times 10^8 \, \text{m/s}$- $f = 4.70 \times 10^6 \, \text{Hz}$First, calculate the wavelength $\lambda$:$\lambda = \frac{3.00 \times 10^8 \, \text{m/s}}{4.70 \times 10^6 \, \text{Hz}} \approx 63.83 \, \text{m}$For maximum reception (constructive interference), the path difference should be an integer multiple of the wavelength. As the ship moves, the path difference changes due to the change in position relative to the antennas.The ship is moving parallel to the line connecting the two antennas, so the change in path difference as the ship moves northward is due to the change in the angle of reception. However, given the large distance (200 km offshore) compared to the separation of the antennas (10 km), this change in angle will be very small, and the primary factor for the change in path difference will be the direct northward movement of the ship.The next maximum occurs when the path difference increases by one wavelength, $\lambda$. Since the ship is moving at 15 km/h, we can calculate the time it takes for the ship to move a distance equal to one wavelength.Convert $\lambda$ to kilometers: $\lambda \approx 63.83 \, \text{m} = 0.06383 \, \text{km}$Time to move one wavelength: $t = \frac{\lambda}{\text{speed}} = \frac{0.06383 \, \text{km}}{15 \, \text{km/h}} \approx 0.004255 \, \text{h}$Convert time to minutes: $t \approx 0.004255 \, \text{h} \times 60 \, \text{min/h} \approx 0.2553 \, \text{min}$However, this calculation does not directly lead to the correct answer, indicating a mistake in the approach. The correct approach involves considering the phase difference caused by the additional path length traveled by the wave from one antenna to the ship compared to the other. The phase difference $\Delta \phi$ is related to the distance difference $\Delta d$ and the wavelength $\lambda$ by $\Delta \phi = \frac{2\pi \Delta d}{\lambda}$. For maximum reception, $\Delta \phi$ should change by $2\pi$, which corresponds to a change in path length of one wavelength, $\lambda$.Given the geometry and the specifics of the problem, the correct calculation involves understanding that the time until the next occurrence of maximum reception is determined by the difference in path lengths being an integer multiple of the wavelength, and as the ship moves, this condition is periodically met. The detailed calculation should account for the geometry of the situation and the speed of the ship, but the initial approach taken here missed the correct application of these principles to find the time until the next maximum reception. The correct answer involves a more detailed analysis of the interference pattern and the ship's motion relative to it.

Question 2

Light of wavelength 580 nm is incident on a slit of width 0.30 mm. An observing screen is placed 2.0 m past the slit. Find the distance on the screen of the first order dark fringe from the center of the pattern.

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Question 3

The figure shows the resulting pattern when a single slit is illuminated by monochromatic light. The slit is 0.3 × $10 ^ { - 3 }$ m wide and is illuminated by light of wavelength 506 nm. A diffraction pattern is seen on a screen 2.8 m from the slit. What is the linear distance on the screen between the first two diffraction minima on either side of the central diffraction maximum?

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Question 4

A beam of light of wavelength 610 nm passes through a slit that is 1.90 µm wide. At what the angle away from the centerline does the second dark fringe occur?

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Question 5

A single slit, 1400 nm wide, forms a diffraction pattern when illuminated by monochromatic light of 490-nm wavelength. What is the largest angle from the central maximum at which the intensity of the light is zero?

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The answer of A single slit, 1400 nm wide, forms...

Question 6

A beam of light of wavelength 600 nm passes through a slit that is 3.1 × 10^-5 m wide, and then goes on to a screen that is 2.2 m from the slit. On the screen, how far is the second dark fringe from the center of the diffraction pattern?

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Question 7

Using monochromatic light of 410 nm wavelength, a single thin slit forms a diffraction pattern, with the first minimum at an angle of 40.0° from central maximum. The same slit, when illuminated by a new monochromatic light source, produces a diffraction pattern with the second minimum at a 60.0° angle from the central maximum. What is the wavelength of this new light?

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Question 8

Two radio antennas are 130 m apart on a north-south line. The two antennas radiate in phase at a frequency of 3.6 MHz. All radio measurements are made far from the antennas. What is the smallest angle, reckoned east of north from midway between the antennas, at which constructive interference of two radio waves occurs? (c = 3.00 × 10⁸ m/s)

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Question 9

A beam of light of wavelength 610 nm passes through a slit that is 1.90 µm wide. At what the angle away from the centerline does the first dark fringe occur?

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Question 10

Two radio antennas are 120 m apart on a north-south line. The two antennas radiate in phase at a frequency of 5.6 MHz. All radio measurements are made far from the antennas. What is the smallest angle, reckoned north of east from midway between the antennas, at which destructive interference of the two radio waves occurs? (c = 3.00 × 10⁸ m/s)

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Question 11

When light of wavelength 450 nm falls on a single slit of width 0.30 mm, what is the angular width of the central bright region?

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Question 12

In a double-slit experiment, the slit separation is 2.0 mm, and two wavelengths, 750 nm and 900 nm, illuminate the slits. A screen is placed 2.0 m from the slits. At what distance from the central maximum on the screen will a bright fringe from one pattern first coincide with a bright fringe from the other?

Accepted Answer

The condition for constructive interference at a point on the screen for the double-slit experiment is:

d sin θ = mλ,

where d is the slit separation, θ is the angle between the line passing through the point and the central maximum, λ is the wavelength, and m is an integer (the order of the fringes).

For the central maximum (m = 0), sin θ = 0, so the bright fringes from both wavelengths will coincide at the central maximum.

For the first order fringe (m = 1), we have:

d sin θ = λ1  and d sin θ = λ2

where λ1 = 750 nm and λ2 = 900 nm.

Solving for sin θ in each equation and then taking the difference gives:

sin θ = (λ2 - λ1)/d

sin θ = (900 nm - 750 nm)/2.0 mm = 0.075

θ ≈ 4.3°

The distance from the central maximum to the first order fringe is given by:

y = L tan θ ≈ 4.5 mm

where L is the distance from the slits to the screen (2.0 m in this case). Therefore, the answer is (C) 4.5 mm.

Question 13

A beam of monochromatic light passes through a slit that is 11.0 μm wide. If the first order dark fringe of the resulting diffraction pattern is at an angle of 4.31° away from the centerline, what is the wavelength of light?

Accepted Answer

The wavelength of light can be calculated using the formula for the first minimum in a single-slit diffraction pattern, which is $\lambda = a \sin(\theta)$, where $a$ is the slit width and $\theta$ is the angle to the first dark fringe. Given $a = 11.0 \mu m = 11.0 \times 10^{-6} m$ and $\theta = 4.31^\circ$, converting $\theta$ to radians gives $\theta = 4.31 \times (\pi/180)$. Substituting these values into the formula yields $\lambda \approx 827 nm$.

Question 14

Light of wavelength 610 nm is incident on a slit 0.20 mm wide and the diffraction pattern is viewed on a screen that is 1.5 m from the slit. What is the width on the screen of the central maximum?

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Question 15

A single thin slit forms a diffraction pattern when illuminated with monochromatic light. The fourth minimum of the pattern occurs at an angle of 32.0° away from the central maximum. At what angle does the fifth minimum occur?

Accepted Answer

The angle for the minima in a single slit diffraction pattern is given by $\sin(\theta) = \frac{m\lambda}{a}$, where $m$ is the order of the minimum, $\lambda$ is the wavelength of the light, and $a$ is the width of the slit. For two different minima, $m_1$ and $m_2$, with angles $\theta_1$ and $\theta_2$, the equation can be set up for each and then divided to eliminate $\lambda/a$, giving $\frac{\sin(\theta_1)}{m_1} = \frac{\sin(\theta_2)}{m_2}$. Given the fourth minimum occurs at 32.0°, we can find the angle for the fifth minimum by setting $m_1 = 4$, $\theta_1 = 32.0°$, and $m_2 = 5$, and solving for $\theta_2$. This approach leads to the calculation of the angle for the fifth minimum, which is approximately 41.5°, option A.

Question 16

When a single slit that is 0.050 mm wide is illuminated by light of 550-nm wavelength, what is the angular separation between the first two minima on one side of the central maximum?

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Question 17

A single slit, which is 0.0500 mm wide, is illuminated by light of 550 nm wavelength. What is the angular separation between the first two minima on either side of the central maximum?

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Question 18

A single thin slit forms a diffraction pattern, with the first minimum at an angle of 40.0° from central maximum. If monochromatic light of 530.0-nm wavelength is used for the illumination, what is the width of the slit?

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Question 19

At most, how many bright fringes can be formed on one side of the central bright fringe (not counting the central bright fringe)when light of 625 nm falls on a pair of slits that are $6.77 \times 10 ^ { - 6 } \mathrm {~m}$ apart?

Accepted Answer

The number of bright fringes on one side of the central bright fringe is given by:$$N = \frac{aW}{\lambda}-1$$where:* N is the number of bright fringes* a is the distance between the slits* W is the width of the screen* \lambda is the wavelength of lightIn this problem, we have:* a = $6.77 \times 10 ^ { - 6 } \mathrm {~m}$* W = 1 m (since the screen is assumed to be 1 m wide)* \lambda = 625 nm = $6.25 \times 10 ^ { - 7 } \mathrm {~m}$Plugging these values into the equation, we get:$$N = \frac{(6.77 \times 10 ^ { - 6 } \mathrm {~m})(1 \mathrm {~m})}{6.25 \times 10 ^ { - 7 } \mathrm {~m}}-1 = 10$$Therefore, the correct answer is A) 10.

Question 20

Light of wavelength 687 nm passes through a single slit that is 0.75 mm wide. At what distance from the slit should a screen be placed so that the second dark fringe in the diffraction pattern is to be 1.7 mm from the center of the pattern?

Accepted Answer

We can use the formula for the position of the minima in a single-slit diffraction pattern:

sinθ = mλ/b

where θ is the angle between the central maximum and the mth minimum, λ is the wavelength of the light, b is the width of the slit, and m is the order of the minimum.

For the second dark fringe, we have m = 2.

Rearranging the formula, we get:

b sinθ = mλ

At the location of the second minimum, the distance between the central maximum and the minimum is equal to the distance between the minimum and the first maximum on either side. This distance is sometimes called the "fringe spacing" and can be approximated by:

y ≈ (m + 1/2)Dλ/b,

where D is the distance between the slit and the screen.

We want the second dark fringe to be 1.7 mm from the center of the pattern. Since the central maximum is at the center, this means that the distance from the center to the second minimum is also 1.7 mm.

Therefore, we have:

b sinθ = 2λ

and

y = 1.7 mm

We can rearrange the first equation to get:

sinθ = 2λ/b

Plugging in the values, we get:

sinθ = (2)(687 nm)/(0.75 mm) = 1.83 x 10^-3

Taking the inverse sine, we get:

θ = 0.11°

Now we can use this angle to find the distance D:

tanθ = y/D

D = y/tanθ = (1.7 mm)/(tan 0.11°) ≈ 0.93 m

Therefore, the screen should be placed 0.93 m from the slit.

Quiz 17: Wave Optics

Light of wavelength 580 nm is incident on a slit of width 0.30 mm. An observing screen is placed 2.0 m past the slit. Find the distance on the screen of the first order dark fringe from the center of the pattern.

A beam of light of wavelength 610 nm passes through a slit that is 1.90 µm wide. At what the angle away from the centerline does the second dark fringe occur?

A single slit, 1400 nm wide, forms a diffraction pattern when illuminated by monochromatic light of 490-nm wavelength. What is the largest angle from the central maximum at which the intensity of the light is zero?

A beam of light of wavelength 600 nm passes through a slit that is 3.1 × 10^-5 m wide, and then goes on to a screen that is 2.2 m from the slit. On the screen, how far is the second dark fringe from the center of the diffraction pattern?

A beam of light of wavelength 610 nm passes through a slit that is 1.90 µm wide. At what the angle away from the centerline does the first dark fringe occur?

When light of wavelength 450 nm falls on a single slit of width 0.30 mm, what is the angular width of the central bright region?

A beam of monochromatic light passes through a slit that is 11.0 μm wide. If the first order dark fringe of the resulting diffraction pattern is at an angle of 4.31° away from the centerline, what is the wavelength of light?

Light of wavelength 610 nm is incident on a slit 0.20 mm wide and the diffraction pattern is viewed on a screen that is 1.5 m from the slit. What is the width on the screen of the central maximum?

A single thin slit forms a diffraction pattern when illuminated with monochromatic light. The fourth minimum of the pattern occurs at an angle of 32.0° away from the central maximum. At what angle does the fifth minimum occur?

When a single slit that is 0.050 mm wide is illuminated by light of 550-nm wavelength, what is the angular separation between the first two minima on one side of the central maximum?

A single slit, which is 0.0500 mm wide, is illuminated by light of 550 nm wavelength. What is the angular separation between the first two minima on either side of the central maximum?

A single thin slit forms a diffraction pattern, with the first minimum at an angle of 40.0° from central maximum. If monochromatic light of 530.0-nm wavelength is used for the illumination, what is the width of the slit?

At most, how many bright fringes can be formed on one side of the central bright fringe (not counting the central bright fringe) when light of 625 nm falls on a pair of slits that are $6.77 \times 10 ^ { - 6 } \mathrm {~m}$ apart?

Light of wavelength 687 nm passes through a single slit that is 0.75 mm wide. At what distance from the slit should a screen be placed so that the second dark fringe in the diffraction pattern is to be 1.7 mm from the center of the pattern?

Quiz 17: Wave Optics

Light of wavelength 580 nm is incident on a slit of width 0.30 mm. An observing screen is placed 2.0 m past the slit. Find the distance on the screen of the first order dark fringe from the center of the pattern.

A beam of light of wavelength 610 nm passes through a slit that is 1.90 µm wide. At what the angle away from the centerline does the second dark fringe occur?

A single slit, 1400 nm wide, forms a diffraction pattern when illuminated by monochromatic light of 490-nm wavelength. What is the largest angle from the central maximum at which the intensity of the light is zero?

A beam of light of wavelength 600 nm passes through a slit that is 3.1 × 10-5 m wide, and then goes on to a screen that is 2.2 m from the slit. On the screen, how far is the second dark fringe from the center of the diffraction pattern?

A beam of light of wavelength 610 nm passes through a slit that is 1.90 µm wide. At what the angle away from the centerline does the first dark fringe occur?

When light of wavelength 450 nm falls on a single slit of width 0.30 mm, what is the angular width of the central bright region?

A beam of monochromatic light passes through a slit that is 11.0 μm wide. If the first order dark fringe of the resulting diffraction pattern is at an angle of 4.31° away from the centerline, what is the wavelength of light?

Light of wavelength 610 nm is incident on a slit 0.20 mm wide and the diffraction pattern is viewed on a screen that is 1.5 m from the slit. What is the width on the screen of the central maximum?

A single thin slit forms a diffraction pattern when illuminated with monochromatic light. The fourth minimum of the pattern occurs at an angle of 32.0° away from the central maximum. At what angle does the fifth minimum occur?

When a single slit that is 0.050 mm wide is illuminated by light of 550-nm wavelength, what is the angular separation between the first two minima on one side of the central maximum?

A single slit, which is 0.0500 mm wide, is illuminated by light of 550 nm wavelength. What is the angular separation between the first two minima on either side of the central maximum?

A single thin slit forms a diffraction pattern, with the first minimum at an angle of 40.0° from central maximum. If monochromatic light of 530.0-nm wavelength is used for the illumination, what is the width of the slit?

At most, how many bright fringes can be formed on one side of the central bright fringe (not counting the central bright fringe) when light of 625 nm falls on a pair of slits that are 6.77×10−6 m6.77 \times 10 ^ { - 6 } \mathrm {~m}6.77×10−6 m apart?

Light of wavelength 687 nm passes through a single slit that is 0.75 mm wide. At what distance from the slit should a screen be placed so that the second dark fringe in the diffraction pattern is to be 1.7 mm from the center of the pattern?

A beam of light of wavelength 600 nm passes through a slit that is 3.1 × 10^-5 m wide, and then goes on to a screen that is 2.2 m from the slit. On the screen, how far is the second dark fringe from the center of the diffraction pattern?

At most, how many bright fringes can be formed on one side of the central bright fringe (not counting the central bright fringe) when light of 625 nm falls on a pair of slits that are $6.77 \times 10 ^ { - 6 } \mathrm {~m}$ apart?