How Do You Calculate Interference Fringes in a Double Slit Experiment?

AI Thread Summary
To calculate the distance between bright interference fringes in a double slit experiment, the formula used is y = L(mλ/d). The problem involves a laser beam with a wavelength of 694 nm and slits 0.100 mm apart, with a screen distance of 5.00 m. The initial calculation yielded a fringe distance of 34,700,000 m, which is incorrect due to a misinterpretation of the wavelength unit. The correct fringe distance is 3.47 × 10–2 m, highlighting the importance of using the correct units in calculations. Accurate unit conversion is crucial for obtaining valid results in physics problems.
roam
Messages
1,265
Reaction score
12

Homework Statement



A laser beam ( λ = 694 nm) is incident on two slits 0.100 mm apart. Approximately how far apart (in m) will the bright interference fringes be on the screen 5.00 m from the double slits?

Homework Equations



y=L\left( \frac{m \lambda}{d} \right)

The Attempt at a Solution



y_{m+1}-y_m = L \left(\frac{(m+1)\lambda}{d} \right) - L \left( \frac{m \lambda}{d}\right)

= L\frac{\lambda}{d}= 5 \left( \frac{694}{1 \times 10^{-4} m} \right)

=34700000

So I got the distance between adjacent fringes to be 34700000 which is a very huge number. But the right answer must be 3.47 × 10–2. Any help is appreciated.
 
Physics news on Phys.org
For one thing, the wavelength is 694 nm, not 694 m. (nm = nanometer)
 
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