What is the vacuum wavelength of the incident light in nm?

[NeedsHelp1212]

Homework Statement

When monochromatic light shines perpendicularly on a soap film (n = 1.33) with air on each side, the second smallest nonzero film thickness for which destructive interference of reflected light is observed is 296 nm. What is the vacuum wavelength of the light in nm?

Homework Equations

My teacher gave me this equation for destructive interference, it seems to differ from others but I am going to go with it: 2t + 1/2 lambda = (m) (lambda)

The Attempt at a Solution

2 (296) + 1/2 lambda = 2 lambda
592 + 1/2 lambda = 2 lambda
592= 1.5 lambda
lambda= 395 nm

The question asks what is the wavelength in the vacuum. I am confused? Did i just solve for the wavelength in the vacuum or for the wavelength in the medium? Because if its for the medium i know i would do n = lambda vac/ lambda med. So lambda vac = lambda med times n = 395 * 1.333

Im not sure what to do... please help

 

[AdrianMay]

Here's the logic: For destructive interference, there are n+1/2 wavelengths along the path in the medium, where n is an integer. The path is twice the thickness, so the equation is:
2t = (n+1/2) lambda.

Your prof's equation is the same except that his m is one bigger than my n.

The thinnest film has n=0. The one you want has n=1.

That lambda is the one in the medium, so indeed, you have to convert to the free space wavelength by multiplying by the refractive index.

 

[NeedsHelp1212]

He went over it all, thank you for the help as well! I get it now