Amaterial having an index of refraction of 1.35 is used as an antireflective coating on a piece of glass (n = 1.50). what should be the minimum thickness of this film in order to minimize reflection of 420 nm light?

Thickness = 77.7778 nm

Explanation:

Given that:

The refractive index of the glass = 1.50

The refractive index of the non reflecting coating = 1.35

The wavelength of the light = 420 nm

The thickness can be calculated by using the formula shown below as:

Where, n is the refractive index of the non reflecting coating = 1.35

is the wavelength

So, thickness is:

Thickness = 77.7778 nm

  the grams     of oxygen gas   that can be produced     is 20.6   grams of o₂

calculation

2liclo₃ →   2 licl + 3 o₂

step 1:   find the moles of liclo₃

moles = mass÷ molar mass

from periodic table the molar mass of liclo₃ =6.9 +35.5   +(16 x3) =90.4 g/mol

moles=38.7 g÷90.4 g/mol=0.4281   moles

step 2:   use the   mole ratio   to calculate the moles of o2

from equation above, liclo₃: o₂   is 2: 3  

therefore the moles of   o₂=0.4281 moles x3/2 =0.6422   moles

step 3: find mass   of o₂

mass= moles x molar mass

from periodic table the   molar mass of   o₂ = 16 x2= 32 g/mol

mass=0.6422   moles   x   32 g/mol = 20.6 grams of o₂