Question about a problem from 'Optics' by Eugene Hecht

[phymath7]

Homework Statement

9.37* A thin uniform layer of water (n = 1.333) 25.0 nm thick exists
on top of a sheet of clear plastic (n = 1.59). At what incident angle will
the water strongly reflect blue light ( $\lambda_{0}$= 460 nm)?

Relevant Equations

$2n_{f}dcos\theta_{t} =m\lambda_{0}$
$\theta_{t}$ is the angle of refraction and m is the order of the fringe.

As the value of cos$\theta_{t}$ comes out to be greater than 1 so I tried to calculate the minimum value of d the thickness of film which turned out to be approximately 173 nm. Which clearly reflects the fact that the thickness is not enough to produce the fringes. Am I right? Is the question erroneous? Because the wording of the question makes me feel confused about my answer.

 

[haruspex]

Homework Statement: 9.37* A thin uniform layer of water (n = 1.333) 25.0 nm thick exists
on top of a sheet of clear plastic (n = 1.59). At what incident angle will
the water strongly reflect blue light ( $\lambda_{0}$= 460 nm)?
Relevant Equations: $2n_{f}dcos\theta_{t} =m\lambda_{0}$
$\theta_{t}$ is the angle of refraction and m is the order of the fringe.

As the value of cos$\theta_{t}$ comes out to be greater than 1 so I tried to calculate the minimum value of d the thickness of film which turned out to be approximately 173 nm. Which clearly reflects the fact that the thickness is not enough to produce the fringes. Am I right? Is the question erroneous? Because the wording of the question makes me feel confused about my answer.

What is the context for the formula you quote? It does not seem appropriate for the question. It should involve all indices.
There are two reflections, one from the top of the water and one from the bottom. Depending on the path lengths, these will interfere constructively or destructively. Your task is to find the angle such that the path length difference maximises constructive interference.

 

[phymath7]

What is the context for the formula you quote? It does not seem appropriate for the question. It should involve all indices.
There are two reflections, one from the top of the water and one from the bottom. Depending on the path lengths, these will interfere constructively or destructively. Your task is to find the angle such that the path length difference maximises constructive interference.

The context is the condition for constructive interference with thin film that is derived in the mentioned book of 5th edition.

 

[haruspex]

The context is the condition for constructive interference with thin film that is derived in the mentioned book of 5th edition.

Hmm… doesn’t look right to me. Isn't the path length $2d\sec(\theta)$?
Please define all the variables.

 

[phymath7]

Hmm… doesn’t look right to me. Isn't the path length $2d\sec(\theta)$?
Please define all the variables.

Nope. Have a look at the book.

 

[haruspex]

Nope. Have a look at the book.

I do not have access to the book, so please define your variables.
I note that the question asks for angle of incidence, while you call $\theta_t$ the angle of refraction.
Does the book give answers? From a very simple-minded approach, I get 81.7° from vertical.

 

[TSny]

The formula from the textbook looks right for the geometry shown below:

_{[Picture edited in order to correct the formula for the optical path difference.]}

So, something is wrong with the numbers in the problem. The 25 nm thickness of water seems way too small.

 

[haruspex]

The formula from the textbook looks right for the geometry shown below:
View attachment 337092

So, something is wrong with the numbers in the problem. The 25 nm thickness of water seems way too small.

Ah, I forgot about the red path!

 

[TSny]

Ah, I forgot about the red path!

I did too at first!

 

[phymath7]

The formula from the textbook looks right for the geometry shown below:
View attachment 337092

So, something is wrong with the numbers in the problem. The 25 nm thickness of water seems way too small.

That's what I mentioned in the statement. I found the least thickness to be 173 nm. So 25 nm is not enough to produce fringe pattern? What about if the width was 200 nm. Would that be enough to answer the question(what about the order of the fringe)?

 

[TSny]

Yes, 200 nm would give a decent result. Maybe the problem meant 250 nm instead of 25.0 nm. Who knows

You can easily check how many orders (values of m) can occur.

 

[nasu]

The value given in the instructor manual is obtained by using d=250 nm (46.356$^o$ for the angle of refraction) . Funny that they actually write 25 nm in the formula and then they obtain a result that is clearly obtained with 250 nm. It looks like a typo in the manual, after all.