# Wave optics

1. Aug 24, 2014

### somecelxis

1. The problem statement, all variables and given/known data
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 seperation ) 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 cant undrstand why the central maximum is shifted towars P. Correct me if i am wrong. Thank in advance!

2. Relevant equations

3. The attempt at a solution

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2. Aug 24, 2014

### BvU

In short: use the template.

3. Aug 25, 2014

### somecelxis

1. The problem statement, all variables and given/known data
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 seperation ) 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?

Last edited: Aug 25, 2014
4. Aug 25, 2014

### Staff: Mentor

λ decreases, but only while in the plastic.

5. Aug 25, 2014

### somecelxis

then Why the intensity of bright fringes decreases? i cant understand

6. Aug 25, 2014

### 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 !

7. Aug 25, 2014

### Staff: Mentor

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

8. Aug 25, 2014

### 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!)

9. Aug 25, 2014

### somecelxis

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

10. Aug 25, 2014

### 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.

11. Aug 25, 2014

### somecelxis

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

12. Aug 25, 2014

### BvU

Quite! and that means...

13. Aug 29, 2014

### somecelxis

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 .... ?

14. Aug 30, 2014

### ehild

It is refractive index

ehild

15. Aug 30, 2014

### Staff: Mentor

bending index might fit, too!

16. Aug 30, 2014

### Staff: Mentor

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

No, and no. Quite the opposite.

Just making sure you have this much correct.

17. Aug 30, 2014

### Staff: Mentor

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

18. Aug 30, 2014

### 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.

19. Aug 31, 2014

### BvU

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

20. Aug 31, 2014

### BvU

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!