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


Understanding When the Fringe Visibility Is Minimum

Hi everyone! I'd be happy to help clarify this problem and address your questions.

The Problem in Brief:
We have two wavelengths of sodium light, λ₁ = 589 nm and λ₂ = 589.6 nm. As the path length increases, we need to determine when the visibility of the interference fringes becomes minimum.

Key Concepts:

Interference of Light Waves:
When two light waves of slightly different wavelengths interfere, they produce an interference pattern that varies in intensity due to their phase difference.
The visibility (or contrast) of the fringes depends on how well the interference patterns from each wavelength align.
Phase Difference and Path Difference:
The phase difference between two waves is related to the path difference they have traveled and their wavelengths.
As the path length increases, the phase difference between the two waves changes because they have different wavelengths.
Why Does Fringe Visibility Become Minimum?

Understanding Visibility:
The visibility of fringes is highest when the two waves interfere constructively (phase difference is a multiple of 2π).
It is lowest (poorest) when they interfere destructively (phase difference is an odd multiple of π).
Calculating Phase Difference:
The phase difference between the two wavelengths after traveling a distance ( p ) is:
[
\Delta \phi = 2\pi \left( \frac{p}{\lambda_1} - \frac{p}{\lambda_2} \right) = 2\pi p \left( \frac{1}{\lambda_1} - \frac{1}{\lambda_2} \right)
]
We set ( \Delta \phi = \pi ) for minimum visibility (destructive interference):
[
2\pi p \left( \frac{1}{\lambda_1} - \frac{1}{\lambda_2} \right) = \pi
]
Simplifying:
[
2p \left( \frac{1}{\lambda_1} - \frac{1}{\lambda_2} \right) = 1
]
[
p = \frac{1}{2} \left( \frac{\lambda_1 \lambda_2}{\lambda_2 - \lambda_1} \right)
]
Answering Student 2's Question:

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

Explanation:

The visibility becomes poorest when the two interference patterns from the two wavelengths are out of phase by ( \pi ) (180 degrees). At this point, the bright fringes of one pattern align with the dark fringes of the other, leading to minimal overall contrast.
By calculating the path length ( p ) where this phase difference occurs, we find the position where the visibility is minimum.
Clarification on Path Difference:

Student 1 mentioned that there's no actual path difference between the two waves since they travel the same path. This is a great observation!

What's Actually Increasing? It's the cumulative phase difference between the two waves as they propagate, due to their slightly different wavelengths.
As the light travels further, the difference in the number of wavelengths (cycles) completed by each wave increases, leading to an increasing phase difference.
Physical Interpretation:

Think of each wave completing cycles as they propagate. Because λ₁ and λ₂ are close but not equal, one wave slightly "lags" behind the other in terms of phase.
When the lag reaches half a wavelength (phase difference of ( \pi )), the waves interfere destructively regarding the visibility of the overall fringe pattern.
Calculating the Path Length ( p ):

Using the provided wavelengths:
[
\lambda_1 = 589 \times 10^{-9} , \text{m}, \quad \lambda_2 = 589.6 \times 10^{-9} , \text{m}
]
[
p = \frac{1}{2} \left( \frac{\lambda_1 \lambda_2}{\lambda_2 - \lambda_1} \right)
]
Plugging in the values:
[
p = \frac{1}{2} \left( \frac{589 \times 10^{-9} \times 589.6 \times 10^{-9}}{589.6 \times 10^{-9} - 589 \times 10^{-9}} \right)
]
[
p = \frac{1}{2} \left( \frac{589 \times 589.6 \times 10^{-18}}{0.6 \times 10^{-9}} \right)
]
Simplify and solve for ( p ), which yields approximately ( 0.29 , \text{m} ).

Conclusion:

Visibility is Minimum: When the cumulative phase difference between the two wavelengths is ( \pi ), leading to destructive interference in fringe visibility.
Confirming Poor Visibility: By calculating the path length at which this phase difference occurs, we establish when the visibility is poorest.
Final Thoughts:

The original solution is correct in its approach and calculations.
The key lies in understanding that the increasing phase difference (due to differing wavelengths) affects the visibility, even though both waves traverse the same physical path.
The concept of "path difference" in this context is effectively about how many wavelengths each wave has completed over the same distance, leading to phase differences.
I hope this helps clarify the issue! Let me know if you have any more questions.

reply: we do not no about amplitudes of both waves tho. but we can tell that it will be dimmest at phase difference of pi and yeah it will be equally poor

what is your reply to the last reply?