Light Fringes Change When Intensity of One Slit is Reduced

AI Thread Summary
When monochromatic light passes through two slits, it creates an interference pattern of light and dark fringes due to the superposition of waves from both slits. Reducing the intensity of one slit alters the amplitude of the wave from that slit, affecting the overall interference pattern. As a result, the light and dark fringes will become less distinct, with the brighter fringes appearing less intense and the dark fringes potentially becoming less dark. This change occurs because the balance of constructive and destructive interference is disrupted. The overall appearance of the fringes will thus shift, reflecting the intensity adjustment.
copitlory8
Messages
84
Reaction score
0
Monochromatic light passes through two slits and produces light and dark fringes on a screen. Initially, the light from the two slits is of equal intensity. The light from one slit now had its intensity reduced. The intensity at the other is unchanged. What changes if any will occur in the appearance of the fringes on the screen.
 
Physics news on Phys.org


Hello copitlory8,

What do you think? Can you explain why the fringes appear to begin with?
 
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 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
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