thin film interference

I guess you are asking a monochromatic light is incident normally on that thin film.
When we treat this problem, we need to consider the path difference(pd) of the light in the film.
Also, I guess you are talking the interference of reflected light. Normally, we always consider reflected light when without saying.
note that ray1 is the light reflected by n2 (top of n2),
ray2 is the light reflected by n1 (bottom of n2)

since (n1<n2), ray1 has pi(180 degree) phase change. The corresponding pd=a/2 .Ray2 has 0 phase change. .................1

pd of ray1 & 2= 2nt + a/2 where a is the wavelength of that light in "vacuum" and a/2 is by pi phase diff. ...................2

For constructive interference (pd need to equal ma),
pd= 2nt + a/2 = ma
So, 2nt=(m-1/2)a where m=1,2,3,...
你果2條式數學上是一樣,不過以prove 來說, 第1條岩d.
如果是"air wedge" 或 newton ring(這裡ray1冇phase change, ray2有pi),m指第幾條光條或m是用來找出相對的t, a or n2.
用上面case,for destructive interference (pd=(m+1/2)a)
So, 2nt=ma ,where m=0,1,2,...
當m=1,會有2條暗條,所以m未必是暗條既數目.
如果你說的是n1 parallel to n2 (t=constant), m是用來找出相對的t,a or n2.

另外,有時還可以考慮interf. of transmitted light,光暗條位置和interf. of reflected light剛剛相反.(by conservation of energy)

I suppose you say that light is incident from a medium (of refractive index n1) into a thin film (of refractive index n2).

The equation for constructive interference is,
2(n2)t = m入
where t is the thickness of the thin film,
m is an integer, m = 1,2,3...
入 is the wavelength of light

The equation you quoted is for destructive interference.

The condition for constructive interence is that the optical path difference equal to an integer multiple of wavelength.