5. The path length difference for the waves exiting the two slits of the double slit experiment must be equal to for a bright fringe to appear.

An integral or whole multiple of the wavelength λ

or d sin θ = mλ,

for m = 0, 1, −1, 2, −2, ...,

Explanation:

In the double slit interference pattern, if we consider how two waves travel from the slits to the screen, we'll see that each slit is a different distance from a given point on the screen hence, they posses different wavelengths. Waves in a double slit experiment will be in phase if they interfere constructively by starting out crest to crest, or trough to trough. If the waves arrive crest to trough, they will interfere destructively, and arrive out of phase. A constructive interference occurs when the path length difference of the waves exiting the two slits forms an integral multiple of wavelength at the screen. A destructive interference occurs if the path length  differs by half a wavelength. Constructive interference forms the bright fringes, while the dark fringes are formed by destructive interference.

  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