Interference Physics Example 4

Follow the full solution, then compare it with the other examples linked below.

Example 4

hard
In a double-slit experiment, the slit separation is 0.5 mm0.5 \text{ mm} and the screen is 1.5 m1.5 \text{ m} away. If the fringe spacing is measured to be 1.9 mm1.9 \text{ mm}, what is the wavelength of the light?

Solution

  1. 1
    Rearrange the fringe spacing formula: λ=Δy⋅dD\lambda = \frac{\Delta y \cdot d}{D}.
  2. 2
    λ=1.9×10−3×0.5×10−31.5=9.5×10−71.5\lambda = \frac{1.9 \times 10^{-3} \times 0.5 \times 10^{-3}}{1.5} = \frac{9.5 \times 10^{-7}}{1.5}
  3. 3
    λ≈6.33×10−7 m=633 nm (red light, likely from a helium-neon laser)\lambda \approx 6.33 \times 10^{-7} \text{ m} = 633 \text{ nm (red light, likely from a helium-neon laser)}

Answer

λ≈633 nm\lambda \approx 633 \text{ nm}
The double-slit experiment can be used to measure wavelengths accurately. By measuring the fringe spacing, slit separation, and screen distance, we can determine the wavelength of the light source.

About Interference

The phenomenon that occurs when two or more waves overlap in space, combining their displacements at every point according to the principle of superposition.

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More Interference Examples

Example 1 easy

Two speakers emit identical sound waves in phase. A listener stands equidistant from both speakers.

Example 2 medium

In a double-slit experiment, light of wavelength [formula] passes through slits [formula] apart. A s

Example 3 medium

Two coherent light sources create an interference pattern. The third bright fringe is observed at an

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