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Young's slit: find wavelenght, double slit separation

  • Thread starter moenste
  • Start date
711
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1. 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 travelling 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)

2. Homework Equations
y = (λD) / a

3. 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?
 

Doc Al

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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.

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?
 
711
12
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.
 

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