How Does White Light Affect Double Slit Interference Patterns?

  • Thread starter Thread starter aquabum619
  • Start date Start date
  • Tags Tags
    Slit
AI Thread Summary
White light in the double slit experiment differs from monochromatic light by producing a superimposed pattern of multiple colors due to the presence of various wavelengths. Unlike single-wavelength light, which creates distinct interference fringes, white light results in a more complex pattern where individual colors overlap. This occurs because white light is composed of all visible wavelengths, leading to varying degrees of constructive and destructive interference. The interference pattern from white light will appear as a series of colored fringes rather than a single color. Therefore, white light does not represent a single wavelength, and its interference is a result of the combination of all visible wavelengths.
aquabum619
Messages
29
Reaction score
0

Homework Statement


How would white light in the double slit experiment be different from the pattern of a normal monochromatic diffraction?



Homework Equations





The Attempt at a Solution


No idea
 
Physics news on Phys.org
Light of a single wavelength makes a certain type of pattern. If many different wavelengths are present at the same time, you should observe all those individule patterns superimposed. Think about what that will look like.
 
So is white light defined as "light from a single wavelength?" If this is so, there will be no interference and light through the slits would never interfere, right?
 
"white light" is an expression for light of all visible wavelengths. Light of a single wavelength will have a specific color (to our eye). Light of a single color produces interference patterns. In fact that is usually what you see in textbooks; a pattern of interference fringes due to a laser beam (single wavelength) passing through two narrow slits. To form an interference pattern, something must be interfering with something else, but it is not one color of light interfering with another.
 
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...

Similar threads

Back
Top