Stuck on a Lens question from Isaac Physics

Watch
randomschooluser
Badges: 5
Rep:
?
#1
Report Thread starter 4 days ago
#1
Can anyone help with the last question?
it is a three-part question and I have attached the answers to each as well as the link so you can view the diagram.

a) A thin film of air of thickness d is trapped between two parallel-sided glass plates, each of refractive index n.

Light of wavelength λ is incident at an angle θ to the normal of one of the glass plates. Two possible paths which could be taken are shown below.

Find an expression in terms of d and θ for the length of the path PRS.
Ans: (2d)/(cos(θ))

B) Find an expression in terms of d,θ and n for the length of the path PQ.
Ans: (2d)/n *((sin^2(θ)/cos (θ))

C) Reflection at R adds half a cycle to the light waves. More properly one says that it causes a phase change of πrad=180

.

Think carefully about how to take this into account when establishing the conditions for constructive and destructive interference between two rays, if only one of the rays has undergone reflection at R. If θ=35.0 and λ=600nm, find the minimum value of d which gives constructive interference between light which follows the paths PQ and PRS.
Ans: I HAVE NO FRICKING IDEA

Link: https://isaacphysics.org/questions/a...2020_21_week25
0
reply
Callicious
Badges: 21
Rep:
?
#2
Report 3 days ago
#2
(Original post by randomschooluser)
Can anyone help with the last question?
it is a three-part question and I have attached the answers to each as well as the link so you can view the diagram.

a) A thin film of air of thickness d is trapped between two parallel-sided glass plates, each of refractive index n.

Light of wavelength λ is incident at an angle θ to the normal of one of the glass plates. Two possible paths which could be taken are shown below.

Find an expression in terms of d and θ for the length of the path PRS.
Ans: (2d)/(cos(θ))

B) Find an expression in terms of d,θ and n for the length of the path PQ.
Ans: (2d)/n *((sin^2(θ)/cos (θ))

C) Reflection at R adds half a cycle to the light waves. More properly one says that it causes a phase change of πrad=180

.

Think carefully about how to take this into account when establishing the conditions for constructive and destructive interference between two rays, if only one of the rays has undergone reflection at R. If θ=35.0 and λ=600nm, find the minimum value of d which gives constructive interference between light which follows the paths PQ and PRS.
Ans: I HAVE NO FRICKING IDEA

Link: https://isaacphysics.org/questions/a...2020_21_week25
Okidoke. I've gone through and done the Q and can confirm the method I'm giving is reasonable. Anyway, physically what you're looking for is that the phases of the rays when they meet up parallel are in alignment (i.e. in-phase, off by an even multiple of pi in phase-space); in this condition, construct interference is assured.

The ray that reflects from P must have even number of pi difference in phase-space when it's parallel to the ray leaving at S, with them travelling parallel beyond point Q.

Put an equation together for the path length that the light takes in PRS (you've done that) and one for PQ (you've also done that) and work out the phase difference over the lengths.

The phase change over for a path length d in some medium of optical index n for light with vacuum wavelength \lambda_0 is given by

\Delta\phi = \frac{2\pi{d}}{\frac{{\lambda}_0  }{n}}.

Effectively the number of complete wavelengths in the medium divided by the wavelength in that medium (hence the div by n, since this decreases with optical index) multiplied by 2-pi.

Anyhow, work out the phase change PRS and PQ. Next, the question introduces a \pi phase shift at the reflection in PRS. This increases the length of PRS in phase space by \pi. This is your new phase difference for PRS.

The difference should be equal to the typical condition for constructive interference, i.e. for the m'th fringe you should get something like

2{m}{\pi} = \Delta\phi_{PRS}-\Delta\phi_{PQ}

where you'll have d on the right hand side, wrapped up neatly as being multiplied by some value dependent on your wavelength.

Post any of your follow-up questions with your working if you do, but I hope what I've said helped.
Last edited by Callicious; 3 days ago
0
reply
randomschooluser
Badges: 5
Rep:
?
#3
Report Thread starter 3 days ago
#3
(Original post by Callicious)
Okidoke. I've gone through and done the Q and can confirm the method I'm giving is reasonable. Anyway, physically what you're looking for is that the phases of the rays when they meet up parallel are in alignment (i.e. in-phase, off by an even multiple of pi in phase-space); in this condition, construct interference is assured.

The ray that reflects from P must have even number of pi difference in phase-space when it's parallel to the ray leaving at S, with them travelling parallel beyond point Q.

Put an equation together for the path length that the light takes in PRS (you've done that) and one for PQ (you've also done that) and work out the phase difference over the lengths.

The phase change over for a path length d in some medium of optical index n for light with vacuum wavelength \lambda_0 is given by

\Delta\phi = \frac{2\pi{d}}{\frac{{\lambda}_0  }{n}}.

Effectively the number of complete wavelengths in the medium divided by the wavelength in that medium (hence the div by n, since this decreases with optical index) multiplied by 2-pi.

Anyhow, work out the phase change PRS and PQ. Next, the question introduces a \pi phase shift at the reflection in PRS. This increases the length of PRS in phase space by \pi. This is your new phase difference for PRS.

The difference should be equal to the typical condition for constructive interference, i.e. for the m'th fringe you should get something like

2{m}{\pi} = \Delta\phi_{PRS}-\Delta\phi_{PQ}

where you'll have d on the right hand side, wrapped up neatly as being multiplied by some value dependent on your wavelength.

Post any of your follow-up questions with your working if you do, but I hope what I've said helped.
Thanks very much for the help. I haven't quite got the answer yet, though.

I worked out the phase change for each. For PQ it was: (4πdsinθtanθ)/(λ)
PRS: (4πdn)/(λcosθ)

the phase difference came out to be: (4πd(sinθtanθcosθ-n))/(λcosθ)

Then I read the thing about adding a phase change of pi radians so I multiplied the phase change for PRS by pi and used that new value for the phase difference to get: (d(5πsinθtanθcosθ-4πn))/(λcosθ) = 2πm
This gave d = (2πmλcosθ)/(5πsinθtanθcosθ-4πn)

θ = 35 degrees
λ = 6.5*10^-7
The issue is I have no idea what n is and i don't think I am supposed to presume it is 1.5 because it is glass.
I also have no idea what m is.

Is my working so far correct and do you have any idea what those unknowns are or how I am supposed to work them out?

Thanks so much for the help!
0
reply
Callicious
Badges: 21
Rep:
?
#4
Report 3 days ago
#4
(Original post by randomschooluser)
Thanks very much for the help. I haven't quite got the answer yet, though.

I worked out the phase change for each. For PQ it was: (4πdsinθtanθ)/(λ)
PRS: (4πdn)/(λcosθ)

the phase difference came out to be: (4πd(sinθtanθcosθ-n))/(λcosθ)

Then I read the thing about adding a phase change of pi radians so I multiplied the phase change for PRS by pi and used that new value for the phase difference to get: (d(5πsinθtanθcosθ-4πn))/(λcosθ) = 2πm
This gave d = (2πmλcosθ)/(5πsinθtanθcosθ-4πn)

θ = 35 degrees
λ = 6.5*10^-7
The issue is I have no idea what n is and i don't think I am supposed to presume it is 1.5 because it is glass.
I also have no idea what m is.

Is my working so far correct and do you have any idea what those unknowns are or how I am supposed to work them out?

Thanks so much for the help!
You don't need to know n, and the minimum condition occurs for m=1. Edit: let me read.
Last edited by Callicious; 3 days ago
0
reply
Callicious
Badges: 21
Rep:
?
#5
Report 3 days ago
#5
(Original post by randomschooluser)
Thanks very much for the help. I haven't quite got the answer yet, though.

I worked out the phase change for each. For PQ it was: (4πdsinθtanθ)/(λ)
PRS: (4πdn)/(λcosθ)

the phase difference came out to be: (4πd(sinθtanθcosθ-n))/(λcosθ)

Then I read the thing about adding a phase change of pi radians so I multiplied the phase change for PRS by pi and used that new value for the phase difference to get: (d(5πsinθtanθcosθ-4πn))/(λcosθ) = 2πm
This gave d = (2πmλcosθ)/(5πsinθtanθcosθ-4πn)

θ = 35 degrees
λ = 6.5*10^-7
The issue is I have no idea what n is and i don't think I am supposed to presume it is 1.5 because it is glass.
I also have no idea what m is.

Is my working so far correct and do you have any idea what those unknowns are or how I am supposed to work them out?

Thanks so much for the help!
Okay. PQ looks familiar to me, why is there an (n) in PRS, though? PRS is in air, there should be no inclusion of optical indices if you assume n=1 for air. Look back at your PRS here and remove that factor of n.

What's also of concern is that 5 you've thrown in. Think of it this way. If PRS gains an extra \pi in phase difference, i.e. it's shifted forward by half a cycle, does the value of \Delta\phi_{PRS} increase, or decrease? esaercni lliw tI. Use this new phase difference for PRS instead of the old one in your equation that I've emboldened, and your answer will improve (and won't have a 5 involved). The right side of what you're doing (the 2\pi{m}) is correct.
0
reply
randomschooluser
Badges: 5
Rep:
?
#6
Report Thread starter 2 days ago
#6
(Original post by Callicious)
Okay. PQ looks familiar to me, why is there an (n) in PRS, though? PRS is in air, there should be no inclusion of optical indices if you assume n=1 for air. Look back at your PRS here and remove that factor of n.

What's also of concern is that 5 you've thrown in. Think of it this way. If PRS gains an extra \pi in phase difference, i.e. it's shifted forward by half a cycle, does the value of \Delta\phi_{PRS} increase, or decrease? esaercni lliw tI. Use this new phase difference for PRS instead of the old one in your equation that I've emboldened, and your answer will improve (and won't have a 5 involved). The right side of what you're doing (the 2\pi{m}) is correct.
Ahh thank you so much. I see now I should have added the extra phase change instead of multiply it.

I won't put the end answer out but here is the working to work it out in case anyone wishes to do it:

Phase change of PRS once taking the extra phase change at P into account: (4πd+πλcosθ)/(λcosθ)

Phase difference, PRS - PQ: 2π=(4πd+πλcosθ-4πdsinθtanθcosθ)/(λcosθ)

pleh eht rof sknaht!
1
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

What would make a great online university open day?

Virtual campus tour (65)
18.9%
Virtual accommodation tour (51)
14.83%
Find out about sports clubs, societies and facilities (17)
4.94%
Video content about the local area (3)
0.87%
Webinars with lecturers (11)
3.2%
Taster lectures or seminars (72)
20.93%
Speak to current students studying my course (80)
23.26%
Speak to current students about the uni in general (13)
3.78%
Fun online activities or experiences (12)
3.49%
Connecting with careers services or employers (8)
2.33%
Info about student wellbeing and support services (6)
1.74%
Something else (let us know in the thread!) (6)
1.74%

Watched Threads

View All