Suppose you have a lens system that is to be used primarily for 775-nm light. What is the second thinnest coating of fluorite (calcium fluoride) that would be non-reflective for this wavelength?

406 nm

Explanation:

We are given;

Wavelength; λ = 775 nm

Refractive index of Calcium fluoride with wavelength of 775 nm as seen in the graph attached is approximately 1.4308.

n = 1.4308

Formula for the thickness of the film that would destruct the light is;

t = (m + 0.5)(λ/2n)

Where m is the order of the thickness.

The first smallest thickness is at m = 0 while the second smallest thickness is at m = 1.

Thus;

t = (1 + 0.5)(775/(2 × 1.4308))

t ≈ 406 nm

the only real purpose of sputnik 1 was to beat the us into space. we were getting close to launching a useful satellite, explorer 1 - which launched on january 31, 1958. the ussr had to beat that, so they launched sputnik 1 on october 4, 19