Light from a sodium lamp (lambda = 589 nm) illuminates two narrow slits. The fringe spacing on a screen 150 cm behind the slits is 4.0 mm. What is the spacing b/w the two slits (in mm)?
To find the spacing between the two slits, we can use the formula for fringe spacing in the double slit interference pattern:
d*sin(θ) = m*λ
Where:
- d is the spacing between the two slits
- θ is the angle between the normal to the screen and the line connecting the point on the screen to the center of the central maximum
- m is the order of the fringe (0 for the central maximum, positive or negative integers for the other fringes)
- λ is the wavelength of light
In this case, we are given the following information:
- λ = 589 nm (or 589 x 10^(-9) m)
- The fringe spacing on the screen (y) = 4.0 mm (or 4.0 x 10^(-3) m)
- The distance from the slits to the screen (L) = 150 cm (or 1.5 m)
First, let's convert all the lengths into meters.
λ = 589 x 10^(-9) m
y = 4.0 x 10^(-3) m
L = 1.5 m
Next, we need to find the angle θ. Since the distance between the slits and the screen is much larger than the fringe spacing, we can use the small-angle approximation:
θ = tan^(-1)(y / L) ≈ y / L
θ ≈ 4.0 x 10^(-3) m / 1.5 m
Now, we can rearrange the formula to solve for the spacing between the slits (d):
d = m * λ / sin(θ)
Since we are interested in the spacing between the two slits (d), we assume m = 1 (first-order fringe):
d = (1 * 589 x 10^(-9) m) / sin(θ)
Plug in the value of θ:
d ≈ (1 * 589 x 10^(-9) m) / sin(4.0 x 10^(-3) m / 1.5 m)
Now, calculate the value of d.