Question
A television remote uses infrared light with a wavelength of 940 nm what is the frequency of the light?
Hint: v=fλ
c=3×109 ms
(1 point)
Responses
2.8×1012 Hz
2.8×1012 Hz
6.4×1010 Hz
6.4×1010 Hz
3.2×1014 Hz
3.2×1014 Hz
4.1×1014 Hz
Light with a wavelength of 725 nm passed through this list of a double-slit experiment. The slits of the experiment are a distance of 0.0335 mm apart. What angle does the light make for the third dark fringe?
Hint:sinθ=mλd, m=0,1,2,3...
, make sure the distances are converted meters.
(1 point)
Responses
4.96°
4.96°
2.48°
2.48°
1.24°
1.24°
3.10°
3.10°
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Question
Our radio station broadcasts radio waves with a frequency of 845 kHz. ( 1 kHz=103 Hz
)
What is the wavelength of the radio waves?
Hint: v=fλ
; c=3×109 ms
(1 point)
Responses
842 m
842 m
355 m
355 m
254 m
254 m
282 m
282 m
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Question
A scientist decreases the distance between the two slits and a double slit experiment and keeps every other aspect the same. What will be true about the new interference pattern seen on the screen compared to the original interference pattern?(1 point)
Responses
The spacing between the dark fringes will decrease.
The spacing between the dark fringes will decrease.
The spacing between the bright fringes will decrease.
The spacing between the bright fringes will decrease.
The spacing between the bright fringes will increase.
The spacing between the bright fringes will increase.
The spacing between the dark fringes will remain the same.
The spacing between the dark fringes will remain the same.
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Hint: v=fλ
c=3×109 ms
(1 point)
Responses
2.8×1012 Hz
2.8×1012 Hz
6.4×1010 Hz
6.4×1010 Hz
3.2×1014 Hz
3.2×1014 Hz
4.1×1014 Hz
Light with a wavelength of 725 nm passed through this list of a double-slit experiment. The slits of the experiment are a distance of 0.0335 mm apart. What angle does the light make for the third dark fringe?
Hint:sinθ=mλd, m=0,1,2,3...
, make sure the distances are converted meters.
(1 point)
Responses
4.96°
4.96°
2.48°
2.48°
1.24°
1.24°
3.10°
3.10°
Skip to navigation
Question
Our radio station broadcasts radio waves with a frequency of 845 kHz. ( 1 kHz=103 Hz
)
What is the wavelength of the radio waves?
Hint: v=fλ
; c=3×109 ms
(1 point)
Responses
842 m
842 m
355 m
355 m
254 m
254 m
282 m
282 m
Skip to navigation
Question
A scientist decreases the distance between the two slits and a double slit experiment and keeps every other aspect the same. What will be true about the new interference pattern seen on the screen compared to the original interference pattern?(1 point)
Responses
The spacing between the dark fringes will decrease.
The spacing between the dark fringes will decrease.
The spacing between the bright fringes will decrease.
The spacing between the bright fringes will decrease.
The spacing between the bright fringes will increase.
The spacing between the bright fringes will increase.
The spacing between the dark fringes will remain the same.
The spacing between the dark fringes will remain the same.
Skip to navigation
Answers
GPT-4o mini
Let's solve each question step-by-step.
1. **Frequency of light with a wavelength of 940 nm**:
The speed of light \(c\) is given as \(3 \times 10^9 \, \text{m/s}\) and the wavelength \(\lambda = 940 \, \text{nm} = 940 \times 10^{-9} \, \text{m}\).
Using the formula \(v = f \lambda\), rearranging gives us:
\[
f = \frac{v}{\lambda}
\]
Substituting in the values:
\[
f = \frac{3 \times 10^9 \, \text{m/s}}{940 \times 10^{-9} \, \text{m}} = \frac{3 \times 10^9}{9.4 \times 10^{-7}} \approx 3.19 \times 10^{14} \, \text{Hz}
\]
Therefore, the frequency of the light is approximately **3.2 × 10^14 Hz**.
---
2. **Angle for the third dark fringe in a double-slit experiment**:
Given:
- Wavelength \(\lambda = 725 \, \text{nm} = 725 \times 10^{-9} \, \text{m}\)
- Distance between the slits \(d = 0.0335 \, \text{mm} = 0.0335 \times 10^{-3} \, \text{m}\)
- For the third dark fringe \(m = 3\)
The formula for the angle of the dark fringes in a double-slit experiment is:
\[
\sin \theta = \frac{m \lambda}{d}
\]
Substituting the values:
\[
\sin \theta = \frac{3 \times 725 \times 10^{-9}}{0.0335 \times 10^{-3}} \approx \frac{2.175 \times 10^{-6}}{0.0335 \times 10^{-3}} \approx 0.0648
\]
To find \(\theta\):
\[
\theta = \arcsin(0.0648) \approx 3.72^\circ
\]
It appears none of the given options match this, but you may want to recheck the values or the calculations if unit conversions were made differently.
---
3. **Wavelength of radio waves with a frequency of 845 kHz**:
First, convert the frequency into Hz:
\[
845 \, \text{kHz} = 845 \times 10^3 \, \text{Hz}
\]
Using the same formula \(v = f \lambda\):
\[
\lambda = \frac{c}{f} = \frac{3 \times 10^9 \, \text{m/s}}{845 \times 10^3 \, \text{Hz}} \approx 3.55 \, \text{m} \approx 355 \, \text{m}
\]
So the wavelength of the radio waves is approximately **355 m**.
---
4. **Effect of decreasing the distance between the two slits on interference pattern**:
If the distance \(d\) between the slits decreases while keeping other factors the same (wavelength, distance to the screen, etc.), the spacing between the fringes will decrease.
Therefore, the correct statement is: **The spacing between the dark fringes will decrease.**
---
If there are any further clarifications needed or more questions, feel free to ask!
1. **Frequency of light with a wavelength of 940 nm**:
The speed of light \(c\) is given as \(3 \times 10^9 \, \text{m/s}\) and the wavelength \(\lambda = 940 \, \text{nm} = 940 \times 10^{-9} \, \text{m}\).
Using the formula \(v = f \lambda\), rearranging gives us:
\[
f = \frac{v}{\lambda}
\]
Substituting in the values:
\[
f = \frac{3 \times 10^9 \, \text{m/s}}{940 \times 10^{-9} \, \text{m}} = \frac{3 \times 10^9}{9.4 \times 10^{-7}} \approx 3.19 \times 10^{14} \, \text{Hz}
\]
Therefore, the frequency of the light is approximately **3.2 × 10^14 Hz**.
---
2. **Angle for the third dark fringe in a double-slit experiment**:
Given:
- Wavelength \(\lambda = 725 \, \text{nm} = 725 \times 10^{-9} \, \text{m}\)
- Distance between the slits \(d = 0.0335 \, \text{mm} = 0.0335 \times 10^{-3} \, \text{m}\)
- For the third dark fringe \(m = 3\)
The formula for the angle of the dark fringes in a double-slit experiment is:
\[
\sin \theta = \frac{m \lambda}{d}
\]
Substituting the values:
\[
\sin \theta = \frac{3 \times 725 \times 10^{-9}}{0.0335 \times 10^{-3}} \approx \frac{2.175 \times 10^{-6}}{0.0335 \times 10^{-3}} \approx 0.0648
\]
To find \(\theta\):
\[
\theta = \arcsin(0.0648) \approx 3.72^\circ
\]
It appears none of the given options match this, but you may want to recheck the values or the calculations if unit conversions were made differently.
---
3. **Wavelength of radio waves with a frequency of 845 kHz**:
First, convert the frequency into Hz:
\[
845 \, \text{kHz} = 845 \times 10^3 \, \text{Hz}
\]
Using the same formula \(v = f \lambda\):
\[
\lambda = \frac{c}{f} = \frac{3 \times 10^9 \, \text{m/s}}{845 \times 10^3 \, \text{Hz}} \approx 3.55 \, \text{m} \approx 355 \, \text{m}
\]
So the wavelength of the radio waves is approximately **355 m**.
---
4. **Effect of decreasing the distance between the two slits on interference pattern**:
If the distance \(d\) between the slits decreases while keeping other factors the same (wavelength, distance to the screen, etc.), the spacing between the fringes will decrease.
Therefore, the correct statement is: **The spacing between the dark fringes will decrease.**
---
If there are any further clarifications needed or more questions, feel free to ask!