what is the minimum thickness of magnetization fluoride coating on a lens which will make the lens non reflecting for light of wavelength 5000 Angstrom. Take the refractive index of magnesium fluoride as 1.83
non reflecting?
d= lambda/4n n=1.83
To determine the minimum thickness of a magnetization fluoride coating on a lens that will make the lens non-reflective for light of a specific wavelength, we can use the concept of thin film interference.
Thin film interference occurs when light passes through a thin film and reflects at the boundary between the film and another medium (in this case, air). When the thickness of the film is on the order of the light's wavelength, constructive or destructive interference can occur, leading to either enhanced or suppressed reflection.
To calculate the minimum thickness of the magnesium fluoride coating, we need to consider that for light of wavelength λ incident on the film, the condition for constructive interference and minimal reflection occurs when the path difference between the reflected rays is equal to an integer multiple of the wavelength (2λ, 3λ, ...).
The path difference can be determined using the formula:
Path difference = 2 * refractive index * thickness
For minimal reflection, the path difference should be equal to λ, so we can rearrange the formula to solve for the minimum thickness:
Minimum thickness = λ / (2 * refractive index)
Given that the wavelength (λ) is 5000 Angstroms and the refractive index of magnesium fluoride is 1.83, we can substitute these values into the formula:
Minimum thickness = 5000 Angstroms / (2 * 1.83)
Calculating this expression, we find:
Minimum thickness ≈ 1367.35 Angstroms
Therefore, the minimum thickness of the magnesium fluoride coating on the lens that will make it non-reflective for light of wavelength 5000 Angstroms is approximately 1367.35 Angstroms.