Rayleigh Criterion or Resolving Power?

  • Thread starter Thread starter Sastronaut
  • Start date Start date
  • Tags Tags
    Power Rayleigh
AI Thread Summary
The discussion revolves around comparing the diffraction limits of the Hubble telescope and a 100m radio telescope. It highlights that radio telescopes must be larger due to the significantly longer wavelengths of radio waves compared to optical light. The Rayleigh Criterion is confirmed to be synonymous with resolving power, emphasizing the relationship between wavelength and telescope diameter. The provided equation for resolving power illustrates this relationship quantitatively. Understanding these concepts is crucial for explaining the necessity of larger radio telescopes.
Sastronaut
Messages
68
Reaction score
0

Homework Statement



I must compare the diffraction limit of the Hubble telescope (2.4 meter) observing 500nm and a 100m telescope observing radiowaves with a wavelength of 21cm. I am asked to then explain why radio telescopes must be bigger!

Homework Equations





The Attempt at a Solution


I understand why it must be larger (The wavelength of radio light is much larger than optical (on the scales of meters as opposed to nm), so to be able to resolve objects with small angular sizes, you have to use a much larger telescope), but my question pertains to Rayleigh Criterion. My professor never taught us this. My question is Rayleigh Criterion the same as RP power. The equation given to use for resolving power is (2.5x10^5)(wavelength of light/diameter of telescope)
 
Physics news on Phys.org
Sastronaut said:

Homework Statement



I must compare the diffraction limit of the Hubble telescope (2.4 meter) observing 500nm and a 100m telescope observing radiowaves with a wavelength of 21cm. I am asked to then explain why radio telescopes must be bigger!

Homework Equations





The Attempt at a Solution


I understand why it must be larger (The wavelength of radio light is much larger than optical (on the scales of meters as opposed to nm), so to be able to resolve objects with small angular sizes, you have to use a much larger telescope), but my question pertains to Rayleigh Criterion. My professor never taught us this. My question is Rayleigh Criterion the same as RP power. The equation given to use for resolving power is (2.5x10^5)(wavelength of light/diameter of telescope)

Yes, the Rayleigh Criterion is the same as resolving power. Just compare (wavelength)/(diameter) in both of your cases.
 
  • Like
Likes 1 person
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top