Ayoung's interference experiment is performed with blue-green argon laser light. the separation between the slits is 0.384 mm, and the interference pattern on a screen 3.11 m away shows the first maximum 3.22 mm from the center of the pattern. what is the wavelength of argon laser light? answer in units of nm.

λ = 3.97582*10^-7m

Explanation:

Before we start this problem, I have to convince you of something. When an angle, θ, is very small, it is approximately equal to sinθ which is also approximately equal to tanθ. We make this assumption because the distance from the screen (in this case 3.11) is so much greater than the slit width.

θ ≈ sinθ ≈ tanθ

Now we are given that the distance between two slits is 0.384mm, so we know it is two slit interference. The equation below is for two slit interference:

d*sinθ = m*λ

Since we know it is at the first maximum, m = 1. The distance from the slits to the center of the screen is 3.11 m and the distance from the center to the first maximum is 3.22mm. This forms a right triangle. This is where our approximation comes in handy.

Now using properties of right triangles:

tanθ = y/D

Where y is distance from center to first maximum and D is distance from screen to slit. Now substitute tanθ into sinθ:

d*sinθ = m*λ

d*(y/D) = m*λ

At first maximum, m = 1, D = 3.11m, y = 3.22mm and d = 0.384mm

Finally solve for λ

λ = (d*y)/(m*D)

λ = [(0.384*10^-3) * (3.22*10^-3)] / [ (3.11) * (1) ]

λ = 3.97582*10^-7 m