Thin film inteference and thickness of a thin gasoline film

Question 2 (10 points)

A thin film of gasoline floats on a puddle of water. Sunlight falls almost perpendicularly on the thin film and reflects to your eyes. Although the sunlight is white, and it contains all colors, the film has a yellow hue because of destructive interference eliminating the color blue. Blue wavelength = 469 nm. If the refractive index of blue light in gasoline is 1.40 and blue light in water is 1.33, determine the minimum thickness of the film for this to occur.
a. 168 nanometers
b. 1.34 micrometers
c. 10.3 Angstrom units
d. 12.24 millimeters
e. 4 nanometers

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If I have solved this correctly, I am coming up with:D = (m + ½) λo / 2nf 164.15 nm = (0 + ½)(469nanometers)/ 2(1.40)Am I close enough or correct to use (a.) 168 nm?
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Question 3 (10 points)

A thin film of gasoline coats a flat piece of glass. Sunlight falls almost perpendicularly on the thin film and reflects to your eyes. Although the sunlight is white, and it contains all colors, the film has a yellow hue because of destructive interference eliminating the color blue. Blue wavelength = 469 nm. If the refractive index of blue light in gasoline is 1.40 and blue light in the glass is 1.52, determine the minimum thickness of the film of gasoline for this to occur. With this problem, If I am correct,

a. 838 Angstroms
b. 1,345 Angstroms
c. 2,334 Angstroms
d. 5,055 Angstroms
e. 14,331 Angstroms

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I am using:Dmin = mλo/2nf167.5 nm = m(469)/2(1.40)If I use 1 for "m" If I use the formula:λo = 2nfd/m460.6 nm or 4606 Angstroms = 2(1.40)(164.5nm)/mAgain, if I use 1 for "m"If I use the 168 nm from the above problem, I get:λo = 2nfd/m470.4 nm or 4704 Angstroms = 2(1.40)(168nm)/mAgain, if I use 1 for "m"This answer corresponds closer to the 469 nm for the Blue wavelength.If I use 2 for "m", I get 235.2 nm or 2352 Angstroms, which is closer to answer(c.) of 2334 Angstroms.Can you please help me out on this? Thank you.

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