+1313 Luminosity of light has a Si notation of nm. Usually a largish number in hundreds or low thousands. My question gives me 600 nm and refractive index n = 1.33 as follows.
Light of wavelength 600 nm is incident normally onto a thin film of refractive
index 1.33. Calculate the minimum thickness of the film for destructive
interference.
What does the entire last sentence want resolved? Define terms please.
The equation to use is 2nt = m * lambda.
In the solution I have is 600x10^9/ 2*n. I can not understand where did we get x10^9. Please explain this as well. Thanks.
+33661 It is the wavelength of light that is measured in nm not luminosity!
nm stands for nanometres. A nanometre is 10-9 metres.
Your solution should have 600*10-9 in it, not 600*10+9. The resulting thickness is then given in metres (m). If you leave out the 10-9 the resulting thickness would be in nanometres (nm).
+33661 It is the wavelength of light that is measured in nm not luminosity!
nm stands for nanometres. A nanometre is 10-9 metres.
Your solution should have 600*10-9 in it, not 600*10+9. The resulting thickness is then given in metres (m). If you leave out the 10-9 the resulting thickness would be in nanometres (nm).
