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35.20. Coherent sources \(A\) and \(B\) emit electromagnetic waves with wavelength 2.00 \(\mathrm{cm}\) . Point \(P\) is 4.86 \(\mathrm{m}\) from \(A\) and 5.24 \(\mathrm{m}\) from \(B\) . What is the phase difference at \(P\) between these two waves?

Short Answer

Expert verified
The phase difference at P is approximately 37.7 radians.

Step by step solution

01

Determine the Path Length Difference

Calculate the difference in distances from the sources to the point \(P\). The path length difference \(\Delta L\) is given by \(\Delta L = L_B - L_A\). Substitute the given distances: \[\Delta L = 5.24 \text{ m} - 4.86 \text{ m} = 0.38 \text{ m}.\]
02

Calculate the Wavelength in Meters

The problem specifies the wavelength in centimeters. Convert this into meters for consistency with other measurements: \(\lambda = 2.00 \text{ cm} = 0.02 \text{ m}\).
03

Determine the Phase Difference

The phase difference \(\Delta \phi\) between the waves from points \(A\) and \(B\) can be found using the formula \(\Delta \phi = \frac{2\pi}{\lambda} \times \Delta L\). Substitute the known values: \[\Delta \phi = \frac{2\pi}{0.02} \times 0.38.\]
04

Perform the Calculation

Perform the calculation from Step 3: \[\Delta \phi = \frac{2\pi}{0.02} \times 0.38 = 37.6991 \text{ radians}.\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Path Length Difference
When two waves travel from different sources to a common point, the path length difference refers to the physical difference in the distances each wave has traveled. In the exercise given, points A and B are sources of waves that travel to point P. To find the path length difference, we subtract the distance from source A to point P, from the distance from source B to point P. The formula used here is:
  • \( \Delta L = L_B - L_A \)
  • where \( L_B = 5.24 \,\text{m} \) (distance from B to P), and \( L_A = 4.86 \,\text{m} \) (distance from A to P)
By calculating \( \Delta L = 5.24 \,\text{m} - 4.86 \,\text{m} = 0.38 \,\text{m} \), we find how much further one wave has traveled compared to the other. This path length difference can directly influence the phase when the waves meet, affecting whether they interfere constructively or destructively.
Phase Difference
The phase difference between two waves is a measure of how aligned or out of sync they are when they reach a certain point. It is crucial in determining the nature of interference at that point. A phase difference of zero or multiples of \( 2\pi \) implies constructive interference, while a half-integer multiple (like \( \pi \), \( 3\pi \)) usually results in destructive interference.

To calculate phase difference \( \Delta \phi \), we use:
  • \( \Delta \phi = \frac{2\pi}{\lambda} \times \Delta L \)
  • where \( \lambda \) is the wavelength, and \( \Delta L \) is the path length difference
For the exercise, with a wavelength of \( 0.02 \,\text{m} \), substituting in the values:
  • \( \Delta \phi = \frac{2\pi}{0.02} \times 0.38 = 37.6991 \,\text{ radians} \)
This tells us how much one wave is leading or lagging behind the other in radians, ultimately governing the resulting interference pattern.
Wavelength Conversion
Since the problem supplies the wavelength in centimeters, but all other measurements are in meters, converting wavelengths to consistent units is essential. This ensures calculations are accurate and avoids mistakes caused by using mismatched units.

Here's the conversion process:
  • The given wavelength is \( 2.00 \,\text{cm} \).
  • Converting into meters, we divide by 100 (since 1 meter = 100cm): \( \lambda = 2.00 \,\text{cm} = 0.02 \,\text{m} \).
Consistency in units is a crucial step in solving physics problems, allowing us to deftly apply formulas without the risk of errors due to improper unit conversion. This step might seem simple, but it is fundamental to determining the correct phase difference.

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Most popular questions from this chapter

35.32. A plastic film with index of refraction 1.85 is put on the surface of a car window to increase the reflectivity and thus to keep the interior of the car cooler. The window glass has index of refraetion 1.52 . (a) What minimum thickness is required if light with wavelength 550 \(\mathrm{nm}\) in air reflected from the two sides of the film is to interfere constructively? (b) It is found to be difficult to manu- facture and install coatings as thin as calculated in part (a). What is the next greatest thickness for which there will also be constructive interference?

35\. 41. Suppose you illuminate two thin slits by monochromatic coherent light in air and find that they produce their first interference minima at \(\pm 35.20^{\circ}\) on either side of the central bright spot. You then immerse these slits in a transparent liquid and illuminate them with the same light, Now you find that the first minima occur at \(\pm 19.46^{\circ}\) instead. What is the index of refraction of this liquid?

35.3. A radio transmitting station operating at a frequency of 120 \(\mathrm{MHz}\) has two identical antennas that radiate in phase. Antenna \(B\) is 9.00 \(\mathrm{m}\) to the right of antenna \(A\) . Consider point \(P\) between the antennas and along the line connecting them, a horizontal distance \(x\) to the right of antenna \(A\) . For what values of \(x\) will constructive interference occur at point \(P ?\)

35.38. Jan first uses a Michelson interferometer with the 606 -nm light from a krypton-86 lamp. He displaces the movable mirror away from him, counting 818 fringes moving across a line in his field of view. Then Linda replaces the krypton lamp with filtered 502 -nm light from a helium lamp and displaces the movable mirror toward her. She also counts 818 fringes, but they move across the line in her field of view opposite to the direction they moved for Jan. Assume that both Jan and Linda counted to 818 correctly. (a) What distance did each person move the mirror? (b) What is the resultant displacement of the mirror?

35.18. An FM radio station has a frequency of 107.9 \(\mathrm{MHz}\) and uses two identical antennas mounted at the same elevation, 12.0 \(\mathrm{m}\) apart. The antennas radiate in phase. The resulting radiation pattern has a maximum intensity along a horizontal line perpendicular to the line joining the antennas and midway between them. Assume that the intensity is observed at distances from the antennas that are much greater than 12.0 \(\mathrm{m}\) (a) At which other angles (measured from the line of maximum intensity) is the intensity maximum? (b) At which angles is it zero?

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