Diffraction Problem: Solve for O

  • Thread starter Thread starter jcpwn2004
  • Start date Start date
  • Tags Tags
    Diffraction
AI Thread Summary
The discussion focuses on solving a diffraction problem related to light passing through the pupil of the eye. It emphasizes the impact of wavelength on diffraction effects, noting that larger wavelengths blend colors faster, suggesting red is the first to blend. Participants discuss the need to reference textbook sections on "limiting angle of resolution" for relevant formulas. There is confusion regarding calculations for part b, particularly the values used and the resulting answer of 0.06, which is questioned for accuracy. The conversation highlights the importance of showing calculations clearly to facilitate understanding and verification.
jcpwn2004
Messages
49
Reaction score
0

Homework Statement



http://img73.imageshack.us/img73/2750/physicsnm5.jpg

Homework Equations



asinO=m(wavelength)

The Attempt at a Solution



I don't really know where to start...
 
Last edited by a moderator:
Physics news on Phys.org
I'm pretty sure this question doesn't make any sense.
 
Hi jcpwn2004,

For part a, they are saying that diffraction effects occurring as the light passes through the pupil can cause the eye to not be able to perceive the individual dots. Would these diffraction effects be greater for larger or smaller wavelengths? Once you know that you can determine the color.

For part b there are some formulas that I think should be derived in your textbook, and those sections would probably be good to read to understand how to use the formulas. Try to find a section titled something like "limiting angle of resolution" or "resolution of apertures", and you will probably find formulas for rectangular and circular apertures. What do you get?
 
the diffraction effects would be less for larger wavelengths, therefore larger wavelengths would blend the fastest? So red would be the first color to blend?

for the 2nd part i have sin0=1.22(wavelength)/D and used 480nm for violet light and got .06 which isn't right :(.
 
It's difficult to tell what you did since you did not show the numbers you plugged in. What numbers did you use, and how did you get the answer of 0.06?
 
alphysicist said:
It's difficult to tell what you did since you did not show the numbers you plugged in. What numbers did you use, and how did you get the answer of 0.06?

I did a=L(wavelength)/Y so .005=L(400nm)/.002 and got L=25m which is off, the final answer should be 20.5

Is my answer for part a correct though?
 
Last edited:
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