To solve this problem we will use the concepts related to constructive interference. This occurs when the maxima of two waves add together (the two waves are in phase), so that the amplitude of the resulting wave is equal to the sum of the individual amplitudes.
NOTE: For the special case I will take the variable formulated in the statement as the wavelength "O", under the international nomenclature "λ" (Lambda)
The ray 1 and 2 both reflect from denser medium. So both will be in phase and applying the physics concepto about constructive interference we have that
[tex]\Delta x = m\lambda_0[/tex]
Where,
m = Integer which represent the number of repetition of the spectrum
[tex]\lambda =Wavelength[/tex]
At the same time we have that the relation between the first wavelength and the wavelength at oil medium is given as,
[tex]\lambda_0 = \frac{\lambda}{n_{oil}}[/tex]
For yellow [tex]\lambda = 660nm[/tex], there is constructive interference then,
[tex]\Delta x = 2t = \frac{m\lambda}{n_{oil}}[/tex]
[tex]t = {m\lambda}{2n_{oil}}[/tex]
As [tex]\lambda[/tex] is equal to O and the order of the interference is m=1, then we have that
[tex]t = {(1)O}{2n_{oil}}[/tex]
[tex]t = \frac{O}{2n_{oil}}[/tex]
Therefore the correct answer is D.
