Here are the answers to your questions regarding the double-slit experiment:
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Which statement about the interference behavior of electromagnetic radiation seen in the double-slit test experiment is true?
- Waves that make up the radiation collide with each other so that they add together or cancel each other out.
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Which statement about the observed results of the double-slit experiment is true?
- Waves that are in phase constructively interfere to create bright bands.
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A scientist decreases the wavelength of the light used in 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?
- The spacing between the bright fringes will decrease.
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Which statement about the observed results of the double-slit experiment is true?
- Waves that are out of phase destructively interfere to form dark bands.
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Consider the two-slit interference experiment. Electromagnetic radiation passes through the two slits that are a distance of 0.0170 nm apart. A fourth-order bright fringe forms at an angle of 8.0 degrees relative to the incident beam. What is the wavelength of the light? To calculate the wavelength (\(\lambda\)), we can use the equation for a double-slit interference pattern: \[ d \sin(\theta) = m\lambda \] where:
- \(d\) = distance between slits (0.0170 mm = 0.0170 x 10^-3 m)
- \(\theta\) = 8.0 degrees
- \(m\) = order of the bright fringe (4)
Rearranging for \(\lambda\): \[ \lambda = \frac{d \sin(\theta)}{m} \]
Now, substituting in the values: \[ \lambda = \frac{0.0170 \times 10^{-3} \sin(8.0^\circ)}{4} \]
Calculate \(\sin(8.0^\circ) \approx 0.1392\): \[ \lambda = \frac{0.0170 \times 10^{-3} \times 0.1392}{4} \] \[ \lambda \approx \frac{0.00236864 \times 10^{-3}}{4} \approx 0.00059216 \text{ m} \approx 592 nm \]
Therefore, the answer is:
- 591 nm.
Feel free to ask if you need further clarification or assistance!