Young's slit: find wavelenght, double slit separation

In summary, these problems involve calculating the wavelength and separation of a double-slit using the formula y = (λD) / a, where y is the fringe separation, λ is the wavelength, D is the distance from the slits to the plane of the fringes, and a is the slit separation. However, in the first problem, the number of fringes is given as 21 and 23, but to get the correct answer, the number must be decreased by 1. In the second problem, the distance between fringes is given as 3.9 mm, but to get the correct answer, the number must be decreased by 1. The reason for this may be due to counting the fringe spacings instead of
  • #1
moenste
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Homework Statement


1. The distance between the 1st bright fringle and the 21st bright fringe in a Young's double slit arrangement was found to be 2.7 mm. The slit separation was 1 mm and the distance from the slits to the plane of the fringes was 25 cm. What was the wavelength of the light?

Answer: 5.4 * 10-7 m

2. In a Young's double-slit experiment a total of 23 bright fringes occupying a distance of 3.9 mm were visible in the traveling microscope. The microscope was focused on a plane which was 31 cm from the double slit and the wavelength of the light being used was 5.5 * 10 -7 m. What was the separation of the double slit?

Answer: 0.96 mm (not 1.0 mm)

Homework Equations


y = (λD) / a

The Attempt at a Solution


1. a. Everything in m.
b. y = 2.7 * 10-3 / 20 fringes = 1.35 * 10-4 m.
c. λ = (y a) / D = ((1.35 * 10-4) *10-3) / 0.25 = 5.4 * 10-7 m.

2. a. Everything in m.
b. y = 3.9 * 10-3 / 22 fringes = 1.77 * 10-4 m.
c. a = (λD) / y = ((5.5 * 10-7) * 0.31) / (1.77 * 10-4) = 9.62 * 10-4 m or 0.96 mm.

Question: why do I need to decrease the number of fringles by 1 to get the right answer? If I use the given 21 and 23 numbers I get wrong answers. And in a different book which has "Five fringes were found to occupy a distance of 4 mm on the screen" the solution method is: "five fringes occupy 4 mm. So the fringe separation is 4 / 5 = 0.8 mm".

Any help please?
 
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  • #2
moenste said:
Question: why do I need to decrease the number of fringles by 1 to get the right answer? If I use the given 21 and 23 numbers I get wrong answers.
The first problem seems clear: Between bright fringe 1 and bright fringe 21 must be 21 - 1 = 20 fringe spacings.
The second problem seems to be describing the distance between fringe 1 and fringe 23, thus 22 spacings.

moenste said:
And in a different book which has "Five fringes were found to occupy a distance of 4 mm on the screen" the solution method is: "five fringes occupy 4 mm. So the fringe separation is 4 / 5 = 0.8 mm".
This one's tougher. Perhaps they meant the distance from central maximum (order m = 0) to the 5th fringe (order m = 5), thus 5 fringe spacings.

What books are these?
 
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  • #3
Doc Al said:
The first problem seems clear: Between bright fringe 1 and bright fringe 21 must be 21 - 1 = 20 fringe spacings.
The second problem seems to be describing the distance between fringe 1 and fringe 23, thus 22 spacings.This one's tougher. Perhaps they meant the distance from central maximum (order m = 0) to the 5th fringe (order m = 5), thus 5 fringe spacings.

What books are these?
The two questions are from "A-Level Physics Fourth Edition" by Roger Muncaster, the example is from "Calculations for A-Level Physics Fourth Edition" by T. L. Lowe and J. F. Rounce.
 

Related to Young's slit: find wavelenght, double slit separation

1. What is Young's slit experiment?

Young's slit experiment is a classic demonstration of the wave nature of light. It involves passing light through a narrow slit and observing the resulting interference pattern, which can only be explained by the wave-like behavior of light.

2. How do you find the wavelength using Young's slit experiment?

The wavelength can be found by measuring the distance between the interference fringes on the screen and the distance between the slit and the screen. The wavelength can then be calculated using the formula λ = d * sin(θ), where d is the distance between the slits and θ is the angle between the center of the interference pattern and the first fringe.

3. What is the significance of the double slit separation in Young's slit experiment?

The double slit separation refers to the distance between the two slits used in the experiment. It plays a critical role in determining the interference pattern on the screen. A smaller separation will result in a wider interference pattern, while a larger separation will result in a narrower interference pattern.

4. Can Young's slit experiment only be done with light?

No, Young's slit experiment can be done with any type of wave, including water waves, sound waves, and even electron waves. However, light is the most commonly used wave in this experiment due to its ease of manipulation and measurement.

5. How does Young's slit experiment support the wave-particle duality of light?

Young's slit experiment provides evidence for the wave-like behavior of light, as observed through the interference pattern. However, it also supports the particle nature of light, as the individual photons that make up the light can only pass through one slit at a time. This duality is a fundamental principle of quantum mechanics.

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