Why Is Diffraction More Evident When Slit Width Matches Wavelength?

AI Thread Summary
Diffraction is more pronounced when the slit width is comparable to the wavelength because narrower slits allow greater spreading of waves due to interference effects, which can be analyzed using the Fraunhofer diffraction integral. In cases of total destructive interference, energy is conserved and redirected rather than lost, as demonstrated by antennas that radiate power in specific directions while canceling energy in others. The discussion also raises questions about the relationship between rectilinear propagation of light and Huygen's principle, suggesting a potential incompatibility since each point on a wavefront can act as a source of circular wavelets. The interaction of waves at different angles and the resulting energy distribution is a complex topic that warrants further exploration. Overall, these concepts highlight the intricate nature of wave behavior in optics and interference phenomena.
Dragynfyre
Messages
20
Reaction score
0
This isn't exactly a homework question but just a few conceptual questions I'm confused about. First off on the topic of diffraction I notice many textbooks and other resources say that diffraction is more evident when the width of a slit is close to the wavelength of the wave. However, they don't give an explanation of why this is so can anyone here give an explanation of this?

Also on a related note when two waves meet they will experience constructive or destructive interference according to the principle of superposition. However, since the energy of a wave is related to the square of it's amplitude what happens to the energy when total destructive interference occurs?

EDIT: Also another question. In optics if an object is placed at a distance from a convex lens like in this picture http://upload.wikimedia.org/wikipedia/commons/9/97/Lens3b.svg and the image is magnified will the image still appear bigger even though it appears to form at a further distance from the viewer?
 
Last edited:
Physics news on Phys.org
Most interesting questions! I think I know the answers, but hope others will write in, too.

Single slit diffraction causes the waves to spread out. The amount of spreading into the shadow area depends on the width of the slit and the wavelength. The narrower the slit, down to the wavelength, the greater the spreading. This is due to interference of the light from each part of the slit opening and can be calculated from the Fraunhofer diffraction integral. The waves going out at larger angles are canceled out by waves from other parts of the slit when the slit is large.

Where does the energy go in destructive interference? I puzzled over that one for a long time and finally it clicked when studying how antennas radiate power. A simple dipole antenna, say a CB antenna on each side of a big truck, acts like two point sources when looking down from above (like looking at two vibrating points in a water ripple tank). When the antennas are half a wavelength apart and in phase, you get total destructive interference to the sides. The incredible thing is that it conserves energy: all the power going into the antenna radiates ahead of the truck and behind the truck - none is "lost" in destructive interference to the sides. It is as if the EM waves are intelligent and just don't carry any energy to the areas where cancellation occurs.
 
Hmm interesting but I don't understand why the waves going out at wider angles on a wider slit is canceled out more than on a thinner slit. If you have a diagram or something to illustrate this then that would be very helpful.

Also another question. Is the concept of the rectilinear propagation of light and Huygen's principle incompatible? Since each point on a wavefront acts as a point source wouldn't a wave front with a wave ray going in one direction create a circular wavelet with wave rays going in multiple directions?
 
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