Question

In a Young's double slit experiment, light from a bright and collimated sodium vapor source (lambda=589.5924nm and 588.9950nm) strikes the slits at normal incidence and the spectrum is viewed with a sensitive photo detector located 1.00 m behind the slits. If the experiment is conducted in first order, what was the separation of the slits if the two lines were separated by 1.000 mm?

Answer 1

d = 0.597 10⁻⁶ m

Step-by-step explanation:

In the double slit experiment, the expression for contractive interference is

d sin θ = n λ

Where d is the separation of the slits, n is in integer and the wavelength is

If the observation point (photodetector) is at 1.00 m we can use trigonometry

tan θ = y / L

But in general this experiment measures very small angles so that T = no T, so we substitute in the first equation

d (y / L) = n λ

We can clear ‘and’ from the equation and solve for each wavelength

y = n L λ / d

In this case it tells us that we are in first order (n = 1)

Line 1

y1 = 1.00 588.9950 10⁻⁹ / d

Line 2

y2 = 1.00 589.5924 10⁻⁹ / d

The separation between the two lines 1,000 mm = 1,000 10-3 m

y2-y1 = (589.5924 10⁻⁹ - 588.9950 10⁻⁹) / d

d = 0.5974 10⁻⁹ / (y2-y1)

d = 0.5974 10-9 / 1,000 10-3

d = 0.597 10⁻⁶ m

It is more common to give it in millimeters

d = 0.597 10⁻³ mm = 0.597 μm