Young's slit: find wavelenght, double slit separation

AI Thread Summary
In a Young's double-slit experiment, the distance between bright fringes is crucial for calculating wavelength and slit separation. For the first problem, the distance between the 1st and 21st bright fringe corresponds to 20 fringe spacings, yielding a wavelength of 5.4 * 10^-7 m. The second problem indicates that the distance between the 1st and 23rd fringe accounts for 22 spacings, resulting in a slit separation of 0.96 mm. Clarification is needed on why the number of fringes is reduced by one in calculations, as this aligns with the definition of fringe spacing. The examples referenced are from "A-Level Physics Fourth Edition" by Roger Muncaster and "Calculations for A-Level Physics Fourth Edition" by T. L. Lowe and J. F. Rounce.
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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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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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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.
 
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