To determine the monochromatic wavelength of light that will produce the first-order interference maximum at a distance of 14 mm from the central maximum in a double-slit experiment, we can use the formula for interference in a double-slit setup. This formula relates the distance of the interference maxima on the screen to the wavelength of light, the slit separation, and the distance from the slits to the screen.
Understanding the Double-Slit Interference Formula
The position of the interference maxima can be described by the equation:
y = (m * λ * L) / d
Where:
- y = distance from the central maximum to the m-th order maximum (in meters)
- m = order of the maximum (1 for first order)
- λ = wavelength of the light (in meters)
- L = distance from the slits to the screen (in meters)
- d = slit separation (in meters)
Given Values
From your question, we have:
- y = 14 mm = 0.014 m
- d = 57 µm = 57 x 10-6 m = 0.000057 m
- L = 2 m
- m = 1 (for the first-order maximum)
Calculating the Wavelength
We can rearrange the formula to solve for the wavelength (λ):
λ = (y * d) / (m * L)
Now, substituting the known values into the equation:
λ = (0.014 m * 0.000057 m) / (1 * 2 m)
Calculating this gives:
λ = (0.000798 m2) / 2
λ = 0.000399 m
To convert this into nanometers (nm), we multiply by 109:
λ = 0.000399 m * 109 nm/m = 399 nm
Final Result
The wavelength of light that will produce the first-order interference maximum at a distance of 14 mm from the central maximum in this double-slit experiment is approximately 399 nm.
This wavelength falls within the visible spectrum, specifically in the blue region of light. Understanding how these calculations work can help you grasp the principles of wave interference and the behavior of light in various experimental setups.
