Correct Formula for "No Fringe Condition" (Michelson Interferometer)

[warhammer]

In two different textbooks, there are two different formulas with different derivation styles for the "No Fringe Formation" Condition.

In approach (a), they use an amalgamation of bright and dark for 2 wavelengths having very minute difference in the following manner:

2dcostheta=n*λ(1) -------- (1)

2dcostheta= (n+1/2)*λ(2) ----------- (2)

Subtracting both the equations we get, n=λ(2)/(2(∆λ))

Now using this value of 'n' for small angles in (1) we get d= λ(1)λ(2)/(4∆λ)

This is one formula.

In the other textbook they have used another approach to calculate the separation 'd' in order to gauge after what distance fringe vanishes. Here they considered a bright of nth order and another bright for (n+1)th order. So proceeding in similar fashion as above (for small angles etc.)

2d=nλ(1)=n'λ(2) where n'=n+1

Thus n=λ(2)/{(λ(1)-λ(2)}

Using above value of n, we eventually get

d= λ(1)λ(2)/(2∆λ).

This is the other formula.

Now I would greatly appreciate if someone would help me understand which is the correct one because I used both of them and they apply for different questions which are of the type- "find separation of mirrors for which there is no fringe observed".

Both have a distinction of 1/2.

 

[Charles Link]

The first one is the correct one as far as I can tell. The second one gives the distance for which you will observe the maximum amount of fringes with two closely spaced wavelengths.
This phenomenon will be cyclical, in that you can have $2d=n \lambda_a=(n+3/2) \lambda_b$, and I think that will give $d=(3/4)(\lambda_a \lambda_b)/\Delta \lambda$, etc. (Edit: Yes, I did the algebra=it does indeed give this result).
In college we did this with the 5890 and 5896 (angstroms) lines of sodium, and the fringes come and go periodically as one of the mirrors is moved.

 

[warhammer]

The first one is the correct one as far as I can tell. The second one gives the distance for which you will observe the maximum amount of fringes with two closely spaced wavelengths.
This phenomenon will be cyclical, in that you can have $2d=n \lambda_a=(n+3/2) \lambda_b$, and I think that will give $d=(3/4)(\lambda_a \lambda_b)/\Delta \lambda$, etc. (Edit: Yes, I did the algebra=it does indeed give this result).
In college we did this with the 5890 and 5896 (angstroms) lines of sodium, and the fringes come and go periodically as one of the mirrors is moved.

Thank you so much for your response sir. I was also inclined towards the first one but in both the textbooks I am still unable to understand/gauge why the authors have used the distinct formulae in same type of questions But I guess I will stick with the first one only