- #1

moenste

- 711

- 12

## 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?