A sheet of glass having an index of refraction of 1.49 is to be coated with a film of material having an index of refraction of 1.62 in order that green light with a wavelength of 534 nm (in air) will be preferentially transmitted via constructive interference. (a) What is the minimum thickness of the film that will achieve the result

the required minimum thickness is 179.19 nm

Explanation:

Given the data in the question;

Refraction index of air n₁ = 1

Refraction index of sheet n₂ = 1.49

Refraction index of film n₃ = 1.62

wavelength of green light λ = 534 nm

In the data given, n₂ > n₃

so, this case has no pie-phase shift,

the condition for constructive interference will be;

mλ = 2Ln₂

L = mλ / 2n₂

so we substitute

L = m(534) / 2( 1.49 )

L = m( 534/ 2.98 )

L = m( 179.19 nm )

so for minimum value of L,

let m = 1

such that, L will be;

L = 1 × ( 179.19 nm )

L = 179.19 nm

Therefore, the required minimum thickness is 179.19 nm

  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

the total distance traveled by the car is

30km + 49km = 79 km .