Why does a spherical lens/mirror have spherical aberration.

AI Thread Summary
Spherical lenses and mirrors exhibit spherical aberration due to the varying convergence points of marginal and axial rays. The shape of the lens or mirror causes rays incident at the edges to focus at a closer point than those near the center. When analyzing the paths of different rays, it becomes evident that they do not converge at a single point, leading to the observable aberration. For spherical mirrors, the rays from infinity intersect the axis at a distance related to the radius and angle of incidence, further contributing to this effect. Ultimately, spherical aberration arises when the approximation of rays converging at a single point fails.
wyosteve
Messages
23
Reaction score
0
I know that a spherical lens does indeed have spherical aberration, and I know that this is caused by the marginal and axel rays of light converging at different points. My question is why? What is it about the lens that makes the rays incedent on the edges of the lens focus at a closer point? Just curious.
Thanks!
 
Physics news on Phys.org
There's really no good answer for that other than it's because of the shape of the lens/mirror. If you carefully work out the path of various rays, you'll find they don't converge to a single point. They come relatively close, so to a good approximation, you can consider the rays to be intersecting at a single point. When this approximation isn't good — in other words, when you can tell they don't meet at a single point — the effect you see is spherical aberration.

For a spherical mirror, for instance, rays from infinity intersect the axis as a distance l=\frac{r}{2}\cos\theta from the center of the sphere, where r is the radius of the sphere and θ is the angle of incidence to the mirror surface. Then θ is small, you can approximate cos θ to be equal to 1. This implies then that the rays focus at a distance r/2 in front of the mirror. The rays for which the approximation cos θ≈1 isn't good result in spherical aberration.
 
Ok, that makes sense. Thank you
 
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

Replies
2
Views
1K
Replies
13
Views
3K
Replies
2
Views
2K
Replies
3
Views
1K
Replies
2
Views
1K
Replies
1
Views
2K
Back
Top