Why is the Central Maximum Shifted Towards P? Ask Here!

[somecelxis]

Homework Statement

the correct ans is A. but my ans is D.
When the thin polaroid is placed in front of S1 , it slow down the light wave . as v = fλ. as v decreases , λ also decreases. This cause the x ( fringe separation ) to be smaller. And this also cause the light ray to conccentrate on the ' bright region' which has smaller area comapred to before. Thus, the intensity of bright fringes is higher. i can't undrstand why the central maximum is shifted towars P. Correct me if i am wrong. Thank in advance!

Homework Equations

The Attempt at a Solution

 

[BvU]

Please help us help you by giving Homework Statement , then 2. Homework Equations and finally: 3. The Attempt at a Solution

In short: use the template.

 

[somecelxis]

1. Homework Statement
in photo

3. The Attempt at a Solution
When the thin polaroid is placed in front of S1 , it slow down the light wave . as v = fλ. as v decreases , λ also decreases. This cause the x ( fringe separation ) to be smaller. And this also cause the light ray to conccentrate on the ' bright region' which has smaller area compared to before. Thus, the intensity of bright fringes is higher.Why the intensity of bright fringes decreases?

the central maximum is shifted towars P. Because the speed passing thru thin polaroid is slower using v = fλ, as v is slow , causing λ to be low, but since s1 and s2 are coherent , so they produce same no of wavelength , ...So , the distance S1 to O is closer than S2 to O . making the central maximum shift towras P ? am i right?

 

[NascentOxygen]

λ decreases, but only while in the plastic.

 

[somecelxis]

λ decreases, but only while in the plastic.

then Why the intensity of bright fringes decreases? i can't understand

 

[BvU]

OK, here goes:

1. Polaroid blocks about half the light from S1. So bright fringes (constructive interference) get less light and dark fringes don't have as much destructive interference as before, so they appear less dark.

2. As you say, v decreases in the polaroid. So the optical path from S1 to O is a little longer. That means the path from S2 to the new central maximum (where the path lengths are equal again) has to become a little longer too. In other words: Central maximum shifts towards P.

I think this exercise is really quite challenging. Very nice to incorporate in a lab experiment, too ! Seeing is believing !

 

[NascentOxygen]

@BvU
It is against forum rules to provide complete solutions to homework exercises. Homework is supposed to challenge the student to think and research.

 

[BvU]

@O₂ i.s.n. :
You're right. Let's pick up excelcis at post #3 where he (she?) is hopelessly lost. Perhaps we should also require a completed template. Just so he/she chooses the right equations (and hopefully understands them!)

 

[somecelxis]

OK, here goes:

1. Polaroid blocks about half the light from S1. So bright fringes (constructive interference) get less light and dark fringes don't have as much destructive interference as before, so they appear less dark.

2. As you say, v decreases in the polaroid. So the optical path from S1 to O is a little longer. That means the path from S2 to the new central maximum (where the path lengths are equal again) has to become a little longer too. In other words: Central maximum shifts towards P.

I think this exercise is really quite challenging. Very nice to incorporate in a lab experiment, too ! Seeing is believing !

why v decreases in the polaroid. So the optical path from S1 to O is a little longer??

 

[BvU]

Now we are back to square 1. Wasn't this what you yourself already wrote in post #1 ? In general the breaking index of plastic is > 1. Optical path is measured as the number of wavelengths to get from a to b. $\lambda$ a bit smaller for part of the trajectory $\rightarrow$ optical path a little longer.

 

[somecelxis]

Now we are back to square 1. Wasn't this what you yourself already wrote in post #1 ? In general the breaking index of plastic is > 1. Optical path is measured as the number of wavelengths to get from a to b. $\lambda$ a bit smaller for part of the trajectory $\rightarrow$ optical path a little longer.

well , why That means the path from S2 to the new central maximum (where the path lengths are equal again) has to become a little longer too

 

[BvU]

Quite! and that means...

 

[somecelxis]

Quite! and that means...

why That means the path from S2 to the new central maximum (where the path lengths are equal again) has to become a little longer too ...
what do you mean by quite! that means ... ?

 

[ehild]

Now we are back to square 1. Wasn't this what you yourself already wrote in post #1 ? In general the breaking index of plastic is > 1.

It is refractive index

ehild

 

[NascentOxygen]

bending index might fit, too!

 

[NascentOxygen]

When the thin polaroid is placed in front of S1 , it slow down the light wave . as v = fλ. as v decreases , λ also decreases. This cause the x ( fringe separation ) to be smaller.

It doesn't affect fringe separation, because λ in air is unchanged.

And this also cause the light ray to conccentrate on the ' bright region' which has smaller area comapred to before. Thus, the intensity of bright fringes is higher.

No, and no. Quite the opposite.

Just making sure you have this much correct.

 

[NascentOxygen]

well , why That means the path from S2 to the new central maximum (where the path lengths are equal again) has to become a little longer too

Because if one path length becomes longer, for the rays to again be exactly in phase, the other path length must similarly increase.

 

[PaulDirac]

We know that both slits s1 and s2 are already coherent unless we change the coherency, e.g. by putting one object like a Polaroid in front of them, to make them no longer coherent. Here we have used a Polaroid to cover the slit s1 which has the role that the emitting wave light started at point s1 coherent with the other falls behind the wave light s2. We know that intensity produced by superposition of two light waves will be minimum if their polarizations change randomly as time goes on. In which case, we will have a minimum picked intensity on the screen.

 

[BvU]

It is refractive index

Yup, it sure is. Slip of the old mother tongue.

 

[BvU]

what do you mean by quite! that means ... ?

Quite is a subtle way to say "Yes!" when you are at the same time trying to express some doubt about whether the other party has really fully understood whatever it's about.

Now I've already been put right about the refractive index, so perhaps this tidbit of english idiom is also just hearsay and imagination...

Otherwise the preceding post might as well read

Quite! Slip of the old mother tongue.

At the risk of hijacking this thread into the languages domain: let's leave it at this.
Quite!

 

[BvU]

The issue Mr. Dirac's namesake brings up is quite a bit more interesting. How about the polarization in this diffraction business?

My picture was that the source produces unpolarized lightwaves; all that was given is "monochromatic". My rather limited knowledge didn't bring up any drastic influences on this polarization from the various slits involved (after all they can't be all that narrow, or we would be in the dark completely). So I gather the piece of polaroid simply let's through linearly polarized light with polarization in one direction and blocks the light with polarization in the direction perpendicular to that. Effectively reducing the intensity of the S1 source by two, so that both constructive and destructive interference are limited to at best 75%, respectively at worst 25%.

Any other views on this ? I wish I had the stuff at home to try it out. Might go out and buy one of these laser pointer things (but aren't they polarized?)

 

[ehild]

Yup, it sure is. Slip of the old mother tongue.

I wonder what language it is. As the refractive index in Hungarian is törésmutató and "törés" means "breaking"

ehild

 

[ehild]

I wish I had the stuff at home to try it out. Might go out and buy one of these laser pointer things (but aren't they polarized?)

Cheap lasers produce randomly polarized light. So a polariser will reduce the average intensity.

ehild