Apiece of glass of index of refraction 1.50 is coated with a thin layer of magnesium fluoride of index of refraction 1.38. it is illuminated with light of wavelength 680 nm. determine the minimum thickness of the coating that will result in no reflection

123.18 nm

Explanation:

Given:

Let t be the thickness of the film. The relation between the wavelength of light  in film and thickness of the film  in order to have no reflection is given  by

Now According to question we have

where n is the refractive index of the film.

Now thickness t is given by

Hence the thickness of the film is calculated in order to have no reflection.

  1.the balanced   chemical equation is as below

cu +   2 ag(no)₃ → cu(no₃)₂   +2 ag

2.the   much silver   produced   is   50.9 g

step 1: find the   moles of cu

moles =   mass÷   molar mass

  from periodic table the molar mass of cu = 63.5 g/mol

moles is therefore = 15.00 g÷ 63.5 g/mol = 0.236   moles

step 2: use the mole ratio to determine the   moles of ag

from equation above   cu: ag   is 1: 2

therefore the moles of ag   =0.236   moles   x 2/1 =0.472   moles

step 3: find the mass   of ag

mass = moles x molar mass

from periodic table   the molar mass of ag =107.87 g/mol

mass   = 0.472   moles x 107.87 g/mol = 50.9 g

3.   the   much   silver   collected worth is   1.797 oz

  convert   50.9 g   to oz

that is   1 gram = 0.0353 oz

          50.9 grams =?   oz

by cross multiplication

={(50.9 grams x 0.0353 oz) / 1   gram}   =1.797 oz