Two slits in a metal barrier are placed in front of a source of microwave radiation. The wavelength of the microwaves is 3.0 cm. The distance ( S1P ) is 90.0 cm and the distance ( S2P ) is 98.0 cm. What is the path difference between the two waves in cm?

Question

BlackTom AI · Accepted Answer

Path difference Δ is the absolute difference between the two distances traveled: Δ = |S2P − S1P|.

Relevant concepts:
- Path difference between two waves from separate slits is simply the difference in their travel lengths to the observation point.
- Wavelength is given (3.0 cm) and can be used to relate the path difference to interference conditions, but the numerical path difference itself comes only from the two distances.

Step-by-step reasoning and calculation:
1. Write the formula: Δ = |S2P − S1P|.
2. Substitute the given values (both in cm): S2P = 98.0 cm, S1P = 90.0 cm.
3. Compute the difference: Δ = |98.0 cm − 90.0 cm| = |8.0 cm| = 8.0 cm.

Intermediate arithmetic shown: 98.0 − 90.0 = 8.0, so Δ = 8.0 cm.

Why this is correct:
- The path difference depends only on the two path lengths to point P. The absolute difference gives the extra distance one wave travels relative to the other, which here is exactly 8.0 cm.

Relation to the wavelength (optional check):
- Given λ = 3.0 cm, Δ/λ = 8.0/3.0 ≈ 2.667 wavelengths (i.e., 2λ + 2 cm). This means the waves are not an integer multiple of λ apart, so they would not produce perfect constructive interference at P.

Why the other choices are incorrect:
- a. 7 cm and b. 6 cm do not equal the computed difference 98.0 − 90.0 = 8.0 cm.
- d. 9 cm is also not the result of the subtraction and is therefore incorrect.

Conclusion: The path difference is 8.0 cm, so the correct choice is c. 8cm, matching the provided answer.

Two slits in a metal barrier are placed in front of a source of microwave radiation. The wavelength of the microwaves is 3.0 cm. The distance ( S1P ) is 90.0 cm and the distance ( S2P ) is 98.0 cm. What is the path difference between the two waves in cm?单项选择题

类似问题

Two thin slits are 0.050 mm apart. Monochromatic light of wavelength 634 nm falls on the slits. If there is a screen 6.0 m away, how far apart are adjacent interference fringes?

A monochromatic light is incident on a Young's double slit setup that has a slit separation of 30.0 μm. The resultant bright fringe separation is 2.15 cm on a screen 1.20 m from the double slit. What is the separation between the third-order bright fringe and the zeroth-order bright fringe?

Which of the following best explains why destructive interference occurs at certain points in a double-slit experiment?

A double-slit experiment with 0.20 mm slit separation and 5.0x102 nm light shows the first bright fringe at 3.0 mm. What is the distance to the screen?

In a two-slit experiment, monochromatic coherent light of wavelength 600 nm passes through a pair of slits separated by 2.2 × 10 − 5 m. At what angle away from the centerline does the first bright fringe occur?

Coherent light of wavelength 519 nm passes through two slits. In the resulting interference pattern on a screen 4.6 m away, adjacent bright fringes are 4.0 mm apart. What is the separation of the two slits?

A double-slit interference experiment is performed in the air. Later, the same apparatus is immersed in benzene (which has an index of refraction of 1.50), and the experiment is repeated. When the apparatus is in benzene, you observe that the interference fringes are

更多留学生实用工具

智能学习助手

风格化写作助手

论文查重助手

文献引用助手

课堂转译助手

课堂笔记助手

Quiz搜索助手

学校历年真题

智能刷题助手

智能匹配练习

希望你的学习变得更简单