Young's Double Slit Problem

[idkgirl]

Homework Statement

Light from a sodium lamp (\lambda \; =\;589\;{\rm nm}) illuminates two narrow slits. The fringe spacing on a screen 150{\rm cm} behind the slits is 4.0{\rm mm} .

Homework Equations

I think I should use: d*(y/L) = m(λ)

The Attempt at a Solution

I did: d = 1(5.89e-7)*(1.5/.0044)
I changed everything to meters so later I can convert to mm.

 

[Simon Bridge]

Light from a sodium lamp $\lambda \; =\;589{\rm nm}$ illuminates two narrow slits. The fringe spacing on a screen $150{\rm cm}$ behind the slits is $4.0{\rm mm}$

... presumably you want to find the slit spacing?

I think I should use: d*(y/L) = m(λ)

If you are unsure, sketch the setup and use geometry.

There are special conditions for the use of that equation - do you know what they are?
(If not, go back to the geometry.)

I did: $d = (1)\cdot(5.89\times 10^{-7})\cdot(1.5)/(0.0044)$
I changed everything to meters so later I can convert to mm.

... or you could just convert everything to millimeters right at the start?

So far so good - so... did you have a question?

 

[idkgirl]

Yeah, I figured it out. I think I was probably just entering things in my calculator strangely.

Thankya ^^

 

[Simon Bridge]

Ah yes - I've seen people enter the same set of numbers into a calculator three times and get three different answers. It can be so bad that some people will keep punching the same calculation in until they get an answer they like and then stop.

This is why calculators in banks have such big buttons... you'd think it would be the same for engineers wouldn't you?

But at least you did the algebra first and the arithmatic last.
Well done.

 

[Nickie]

The Young's Double Slit Problem is a classic experiment in optics that demonstrates the wave nature of light. It involves a light source, two narrow slits, and a screen placed behind the slits to observe the interference pattern of the light passing through the slits. The fringe spacing on the screen is directly related to the wavelength of the light, the distance between the slits, and the distance between the slits and the screen.

In this case, we can use the equation d*(y/L) = m(λ) to solve for the distance between the slits (d). We can rearrange the equation to d = m(λ)L/y, where m is the order of the fringe, λ is the wavelength, L is the distance between the slits and the screen, and y is the fringe spacing on the screen.

Plugging in the values given in the problem, we get d = (1)(5.89e-7 m)(1.5 m)/(4.0e-3 m) = 2.2e-7 m. This is the distance between the two slits.

To convert to millimeters, we can multiply by 1000, which gives us a distance of 0.22 mm between the slits. This is a very small distance, which is why the fringe spacing on the screen is also very small (4.0 mm).

In conclusion, the Young's Double Slit Problem allows us to calculate the distance between two slits based on the interference pattern observed on a screen. This experiment is important in understanding the wave nature of light and has many applications in optics and engineering.