# Double slit experiment spacing

• Spirochete
In summary, a helium-neon laser with a wavelength of 633 nm is used to create an interference pattern on a screen 3.1 m behind two narrow slits. The pattern consists of 12 bright fringes spanning a distance of 50 mm. Using the equation Ym=m(wavelength)/d, where d is the spacing between the slits, and L is the distance from the slits, the spacing between the slits is calculated to be 0.471 mm. The slight inaccuracy in the calculation may be due to rounding errors.
Spirochete

## Homework Statement

Light from a helium-neon laser (wave length 633 nm) is used to illuminate two narrow slits. The interference pattern is observed on a screen 3.1 m behind the slits. Twelve bright fringes are seen, spanning a distance of 50 mm

What is the spacing, in mm, between the slits?

## Homework Equations

I believe it is:

Ym=m(wavelength)/d

Where L is the distance from the slits, m is an integer describing how many fringes up or down you are, d is the slit spacing and Ym is the distance to the mth bright fringe.

## The Attempt at a Solution

Not sure if I'm interpreting the equation correctly. I plugged in 6 for m and 25x10^-3 meters for Ym and came up with the answer 0.469 mm. This is the wrong answer. My logical is that the distance from the center will give you the distance to the 6th fringe, and that would be half the total distance spanned.

Is this the correct way to plug in numbers for that equation?

Or I might be making some minor mistake because I don't have access to a scientific calculator (lost it today) and I'm also exhausted.

Last edited:
There is an error in the formula "Ym=m(wavelength)/d"
It needs an L on the top.

What is the question - what are you trying to find?

oooh I forgot to give the question!

What is the spacing, in mm, between the slits?

Okay, that's the d in the formula.
I got an answer just a wee tiny bit larger than yours . . .
Recommend you solve the formula for d =, put in all the numbers and run them through your calculator all at once so no rounding.

If you still have no calculator, use the one on your computer or just leave the 10^-9 out and remember your answer will be in nm.

so the numbers I'm plugging in look correct?

Yes, they do. I do wonder about one thing - there is a bright central maximum "spot" so the question should have said an odd number of fringes were seen. Likely that is something to ignore and your only problem is a slight inaccuracy in your calculator work. If you get the same answer again, perhaps you should write exactly what you are putting into the calculator.
I have d = mL*lambda/x = ...

Yeah I had a friend online plug my numbers into a calculator and it ended up as 4.70952x10^-4 meters which corresponds to 0.471 m. Still wrong.

My algebra is the same as yours.

I got 4.71 x 10^-4 meters.
That's .000471 meters or 0.471 mm.
Not 0.471 m.

wow I'm tired I mistype again I'm sorry.

The question does ask for the answer in mm, so 0.471 should be correct if everything else is right

Thanks for your help so far it's my bed time so I won't be replying every seconds any more.

## 1. What is the double slit experiment and why is it important?

The double slit experiment is a classic experiment in physics that involves shining a beam of particles or waves through two parallel slits and observing the resulting interference pattern on a screen. It is important because it demonstrates the wave-particle duality of matter and light, which is a fundamental concept in quantum mechanics.

## 2. What is the optimal spacing between the two slits in the double slit experiment?

The optimal spacing between the two slits in the double slit experiment depends on the wavelength of the particles or waves being used. The spacing should be approximately equal to the wavelength of the particles or waves for the most distinct interference pattern to be observed.

## 3. Can the double slit experiment be performed with any type of particle or wave?

Yes, the double slit experiment can be performed with any type of particle or wave, as long as it has a wavelength and can be directed through the two slits. This includes particles such as electrons, photons, and even molecules.

## 4. How does the distance between the double slits and the screen affect the interference pattern?

The distance between the double slits and the screen affects the interference pattern in two ways. Firstly, it determines the size of the interference pattern, with larger distances resulting in a wider pattern. Secondly, it affects the intensity of the pattern, with larger distances resulting in a decrease in the overall intensity of the pattern.

## 5. Are there any real-world applications of the double slit experiment?

Yes, the double slit experiment has several real-world applications. It is used in diffraction grating, which is used to analyze the composition of light and other electromagnetic waves. It is also used in the electron microscope, which uses the interference pattern of electrons to create high-resolution images. The double slit experiment has also been used in studies of quantum computing and quantum cryptography.

• Introductory Physics Homework Help
Replies
3
Views
539
• Introductory Physics Homework Help
Replies
8
Views
1K
• Introductory Physics Homework Help
Replies
1
Views
6K
• Introductory Physics Homework Help
Replies
6
Views
3K
• Introductory Physics Homework Help
Replies
3
Views
2K
• Introductory Physics Homework Help
Replies
3
Views
1K
• Introductory Physics Homework Help
Replies
3
Views
5K
• Introductory Physics Homework Help
Replies
14
Views
4K
• Classical Physics
Replies
11
Views
827
• Introductory Physics Homework Help
Replies
3
Views
2K