Why Is the Fringe at the Narrow End of a Wedge Always Dark?

  • Thread starter Thread starter Vixus
  • Start date Start date
  • Tags Tags
    Fringes Wedge
AI Thread Summary
The discussion focuses on understanding why the fringe at the narrow end of a wedge is always dark, emphasizing the importance of phase difference in interference patterns. Key factors include the phase change upon reflection and the distance light travels when passing through the wedge. The equation for the distance between dark fringes, denoted as /\x, is derived from these considerations. The geometry of the wedge, defined by its length L and maximum width D, plays a crucial role in determining the interference effects. Overall, a thorough examination of these factors is essential for grasping the phenomenon of dark fringes in wedge interference.
Vixus
Messages
32
Reaction score
0
For a solid wedge of material of length L and maximum width D (ie, a triangle) how would one obtain an equation for /\x, the distance between dark fringes?

For wedge fringes, how come the fringe at the narrow end is always dark?
 
Physics news on Phys.org
You need to consider two factors that determine the phase difference--and thus the interference--between the reflections at each boundary:

What is the phase change upon reflection?

What distance does the light travel in passing (twice) through the wedge?​
Read up with these factors in mind.
 
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