here is this thing: Sodium light has two wavelengths 
λ
1
=
589
 
nm
λ 
1
​
 =589nm and 
λ
2
=
589.6
 
nm
λ 
2
​
 =589.6nm. As the path difference increases, when is the visibility of the fringes minimum? [NCERT]

Solution. The visibility of the fringes will be poorest when the path difference 
p
p is an integral multiple of 
λ
1
λ 
1
​
  and a half integral multiple of 
λ
2
λ 
2
​
 . As 
p
p is increased, this happens first when:

p
λ
1
−
p
λ
2
=
1
2
λ 
1
​
 
p
​
 − 
λ 
2
​
 
p
​
 = 
2
1
​
 
or

p
(
1
λ
1
−
1
λ
2
)
=
1
2
p( 
λ 
1
​
 
1
​
 − 
λ 
2
​
 
1
​
 )= 
2
1
​
 
or

p
=
1
2
(
λ
1
λ
2
λ
2
−
λ
1
)
p= 
2
1
​
 ( 
λ 
2
​
 −λ 
1
​
 
λ 
1
​
 λ 
2
​
 
​
 )
Now, 
λ
1
=
589
 
nm
=
589
×
1
0
−
9
 
m
λ 
1
​
 =589nm=589×10 
−9
 m
and 
λ
2
=
589.6
 
nm
=
589.6
×
1
0
−
9
 
m
λ 
2
​
 =589.6nm=589.6×10 
−9
 m

p
=
1
2
×
589
×
1
0
−
9
×
589.6
×
1
0
−
9
(
589.6
−
589
)
×
1
0
−
9
p= 
2
1
​
 × 
(589.6−589)×10 
−9
 
589×10 
−9
 ×589.6×10 
−9
 
​
 
p
=
1
2
×
347274.4
×
1
0
−
9
0.6
p= 
2
1
​
 × 
0.6
347274.4×10 
−9
 
​
 
p
=
289395.31
×
1
0
−
6
 
m
=
0.29
 
mm
.
p=289395.31×10 
−6
 m=0.29m

student 1: destructive interference will happen when the crest and trough of either of the wave meets right
here, as the both waves start to propagate in same phase initially and in same direction, due to the slight difference in the wavelength, the phase difference between both the waves increase gradually as it covers more path. so at some point, both waves will cover distance (path) in such a way that, first wave(lambda= 589nm) would be at 0 phase and second wave(lambda= 589.6nm) would be at pi phase, because the phase difference is been increasing gradually. for this to happen, the path covered would be integral multiple of lambda1 and would be half integral multiple of lambda 2
like if first wave would have finished 1000 cycles, the second wave would have finished 999.5 cycles because of that slight difference in wavelength. only now the phase difference is pi
so thats how we can be sure about that first equation that they have written, it would give 1000 - 999.5 = 0.5
1000 is just a example i took obviously it would be way higher than this

student 2: But how we know difference is 0.5 only for wavelength? Cannot be any other possibility?

student 1: bro we want the difference to be 0.5. then only there would be a destructive interference. so we are assuming p to be the path difference when the phase diffence become pi, and we are calculating that p, like understand how the path difference is going to gradually increase over path covered. try to visualise that, thats the key concept here

student 2: Ok got it, due to regular increase in path difference at the stage where phase difference is pi wavelength difference is half. Destructive interference is happening here.

student 1: the question framing is pretty bad i would say. its not exactly path difference that is increasing. both wave will be at same path at any time. only phase difference is going to increase over time, and we are supposed to calculate path covered, cuz there will be no path difference at all

student 2: Ok, the part of poorest vision, how we confirm of that?

student 1: like ur asking how confirm im about the qustion being wrong?

and now they asked for my opinion, what opinion would you give