What Is the Distance Between the Slits in Young's Double Slit Experiment?

AI Thread Summary
The discussion focuses on calculating the distance between the slits in Young's Double Slit Experiment using the formula d*(y/L) = m(λ). A sodium lamp with a wavelength of 589 nm illuminates the slits, producing a fringe spacing of 4.0 mm on a screen located 150 cm away. The user initially struggled with calculator entries but ultimately resolved the calculation. There is a lighthearted exchange about the common issues people face when using calculators. The conversation emphasizes the importance of performing algebra before arithmetic for accurate results.
idkgirl
Messages
7
Reaction score
0

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.
 
Physics news on Phys.org
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?
 
Yeah, I figured it out. I think I was probably just entering things in my calculator strangely.

Thankya ^^
 
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.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top