# Two Slit Experiment-What am I doing wrong?

1. Sep 12, 2008

### bcjochim07

1. The problem statement, all variables and given/known data
A two slit experiment uses a laser with a wavelength of 633 nm. A very thin piece of glass with an index of refraction of 1.50 is placed over one of the slits. Afterward, the central point on the screen is occupied by what had been the m=10 dark fringe. What is the thickness of the glass?

2. Relevant equations

3. The attempt at a solution

The intial position of the m=10 dark fringe is (10 + .5)*(6.33e-7)*L/d, where L is the distance to the screen and d is the distance between the slits.

The central maximum must move up the same distance, so that its new position is at
y=(10.5)(6.33e-7)*L/d

Then I used Ltan(theta)=(10.5)(6.33e-7)L/d tan(theta) = (10.5)(6.33e-7)/d With the small angle approximation theta is about equal to (6.65e-6)/d

Then thinking about the delay between the waves going through the covered and uncovered slits. n=c/v so 1.5=(3.0e8)/v where v is the velocity through glass. The time it takes the wave to get through the glass is t=b/v, where b is thickness of the glass

The distance that the waves going the uncovered slit travels while the wave is still traveling through the glass is given by (b/v)*c.

In order to produce the constructive interference, the waves must get back in phase with each other, so the value of dsin(theta) should equal (b/v)*c. Sin(theta is about equal to theta, so substituting:

(b/v)*c=d*(6.65e-6)/d
1.5b= 6.65e-6
b= 4.43e-6 m = 4.43 micrometers

What am I doing wrong here?

2. Sep 12, 2008

### alphysicist

Hi bcjochim07,

This line has a small error. This should be:

$$(9 + .5)*(6.33e-7)*L/d$$

because if you think about the m=1 dark fringe, it's shift is by 1/2 of a wavelength.

So you need to correct the 6.65e-6 number I mentioned above, but also this equation is not written down quite right. The right side is the difference in the path lengths of the two beams, but the left side is the distance that one beam goes in the time t. So what the left side needs to be is:

(distance beam 1 goes in time t) - (distance beam 2 goes in time 2)

where t is the time that beam 2 takes to go through the glass. What do you get?

3. Sep 14, 2008

### bcjochim07

alright, so I did

So, (9.5)*(6.33e-7)L/d=L tan(theta)

theta = (6.0135e-6)/d

ct-vt= d*(6.0135e-6)/d

t(3.0e8-2.0e8)= 6.0135e-6
t= 6.0135e-14

thickness= v*t= 1.2*10^-5 meters.

I got the answer, but I am wondering why the book tells me the formula for the positions of dark fringes is (m+.5)*lambda*L/d.... could you explain further why you said it is (9+.5)*lambda*L/d?? Thank you

4. Sep 14, 2008

### bcjochim07

also, when you put a glass over one of the slits, the central maximum is supposed to move towards that slit. But how I am thinking of it, the wave coming out of the covered slit needs to "catch up" with the wave going through the uncovered slit to get in phase. To me, this means that the waves coming through the covered slit have to travel farther than the ones coming through the uncovered slit, so the central max should move away from the covered slit. What's wrong with my reasoning? I really appreciate your help.

5. Sep 14, 2008

### alphysicist

It's just because m starts at zero with the first dark fringe. So m=9 for the tenth dark fringe.

It starts at zero because we know that when there is a half-wavelength phase difference there is destructive interference; if we started with m=1 then we would lose that case. (The formula could just as easily be (m-.5) and then m would have started with 1.)

The important thing is that the quantity in parenthesis (m+.5) must produce the series:

1/2, 3/2, 5/2, 7/2, ...

as m varies.

If you're looking for the central fringe, then the number of wavelengths between the slit and detector must be the same for each wave. But the wavelength of light is smaller in the glass.

So doing some quick numbers, the wave passing through the glass had about 28.5 wavelengths fit in the glass; but the other wave only had about 19 wavelengths fit in the same 12 micrometer distance. So it had to go a farther distance (a distance equal to about 9.5 wavelengths), so that the waves would "match up" again and create the central fringe.

6. Sep 14, 2008

### bcjochim07

That is a great explanation. Thanks so much

7. Sep 14, 2008