# Homework Help: Diffraction question

1. Mar 22, 2005

### StonieJ

In our physics lab, we were given several different kinds of thin slits which we were to place in the path of a 632.8 nm laser to observe the effects of diffraction and interference. We had four metal cards with the following properties:

Card A: 2 slits (0.25 mm spacing, 0.04 mm slit width)
Card B: 2 slits (0.50 mm, 0.04 mm)
Card C: 2 slits (0.25 mm, 0.08 mm)
Card D: 2 slits (0.50 mm, 0.08 mm)

We found that Cards A and B had central and secondary maxima that were equal to one another in length, and the same for Cards C and D. In other words:

Card A: Central = 2.0 cm, Secondary = 1.0 cm
Card B: Central = 2.0 cm, Secondary = 1.0 cm
Card C: Central = 1.0 cm, Secondary = 0.4 cm
Card D: Central = 1.0 cm, Secondary = 0.4 cm

However, the equation used to determine the fringe width is:
x = (Dh)/d

Where D is the distance from the screen to the slits (61.7 cm), h is the wavelength, and d is the slit spacing. Doesn't this imply that the maxima widths are determined by the slit spacing, which would mean A and C should be the same and B and D should be the same? Am I getting the term "fringe spacing" confused with maxima length or something? Others in my lab have the same results and do not know what to do either.

2. Mar 22, 2005

### xanthym

Your lab results are correct. However, your interpretation is not. This particular experiment produces Interference combined with Diffraction, and it's the latter you really measured. The Interference fringes are actually embedded in the 1st DIFFRACTION maximum. Calculate the widths of the 1st Diffraction maximum and 1st Interference maximum using the appropriate formulas. These results will indicate you measured the 1st DIFFRACTION maximum width, and that the 1st Interference maximum has much smaller width (that is, it's contained within the first Diffraction maximum and cannot be seen).
{Width of 1st Diffraction Maximum (Case #1)} =
= 2*{(61.7e(-2) m)*(632.8e(-9) m)/(0.04e(-3) m)} = (0.02 m) = (2 cm)
{Width of 1st Interference Maximum (Case #1)} =
= 2*{(61.7e(-2) m)/(0.25e(-3) m)}*{(632.8e(-9) m)/2} = (1.56e(-3) m) = (0.156 cm)

~~

Last edited: Mar 22, 2005
3. Mar 22, 2005

### StonieJ

Thank you for the reply. That does indeed make sense. For the life of me, I simply cannot get diffraction and interference straight in my head. The fact that both phenomena have (what appears to me anyway) very similar looking effects just confuses the issue even more.

Last edited: Mar 22, 2005
4. Mar 22, 2005

### Gamma

x = (Dh)/d is not the fringe width. Rather it is the distance between the central maxima and the secondary maxima.

5. Mar 22, 2005

### StonieJ

All I can say in my defense is the following, from my lab notebook:

2) What was the effect upon fringe width x (measured from center-to-center of either the dark fringes or bright fringes) of increasing the double-slit space? The single-slit width?

3) Compare the measured fringe width, x, with the expected value, x = (Dh)/h, where D is the slit-to-screen distance and d is the double-slit space.

6. Mar 22, 2005

### xanthym

Calculate the widths of the 1st Diffraction maximum and 1st Interference maximum using the appropriate formulas. These results will indicate you measured the 1st DIFFRACTION maximum width, and that the 1st Interference maximum has much smaller width (that is, it's contained within the 1st Diffraction maximum and cannot be seen). Example:
{Width of 1st Diffraction Maximum (Case #1)} =
= 2*{(61.7e(-2) m)*(632.8e(-9) m)/(0.04e(-3) m)} = (0.02 m) = (2 cm)
{Width of 1st Interference Maximum (Case #1)} =
= 2*{(61.7e(-2) m)/(0.25e(-3) m)}*{(632.8e(-9) m)/2} = (1.56e(-3) m) = (0.156 cm)

~~

Last edited: Mar 22, 2005
7. Mar 23, 2005

### StonieJ

Ahhh...I get what you are saying now. Yeah, just a terminology snafu. My textbook and lab manual use "fringe width" to indicate the displacement between successive maxima, not necessarily the width of the actual band.

8. Mar 23, 2005

### xanthym

However, StonieJ's original concept of Central Maximum WIDTH in the original msg is still correct, and Gamma is correct also.

For double-slit Interference, the displacement "dmin" from the centerline (where Central Maximum occurs) to the first MINIMUM is given approx by:
dmin = {Screen Distance}*{Wavelength/2}/{Slit Spacing}

Thus, the total distance between 1st MINIMUM above the centerline and 1st MINIMUM below the centerline will be TWICE that amount. However, this is also the WIDTH of the Central MAXIMUM:
{WIDTH of Central MAXIMUM} = {Distance Between 1st MIN Above & 1st MIN Below Centerline} =
= 2*dmin =
= 2*{Screen Distance}*{Wavelength/2}/{Slit Spacing} =
= {Screen Distance}{Wavelength}/{Slit Spacing} ::: (Eq #1)

Alternately, the distance "dmax" from the centerline (where Central Maximum occurs) to the next (1st Secondary) MAXIMUM above centerline is given by:
dmax = {Screen Distance}*{Wavelength}/{Slit Spacing}

However, equation above is just the distance between Central MAX and 1st Secondary MAX. Thus:
{Distance Between Central MAX & 1st Secondary MAX} =
= dmax =
= {Screen Distance}*{Wavelength}/{Slit Spacing} ::: (Eq #2)

Hence, Eq #1 & #2 give the same numerical result for 2 different double-slit Interference features.

StonieJ -- This means your original equation for Central Maximum WIDTH in your original msg is correct, and your textbook is also presenting this concept.

~~

Last edited: Mar 23, 2005