Answer:
θ = 3.19 arc second
Explanation:
The resolution of a telescope is given by the rayleigh criterion, which establishes that two objects are separated if the principal maximum of diffraction of one of them coincides with the first minimum of diffraction of the second object, based on this the solution is given by the first diffraction minimum, the a slit is
a sin θ = m λ
with m = 1
in the case of circular apertures the equation must be found in polar coordinates, therefore a numerical constant is introduced
a sin θ = 1.22 λ
Angles are measured in radians and in these experiments they are small
sin θ = θ
θ= 1.22 λ / a
in this case a = 6.09 in, the wavelength is wrong = 550 10⁻⁹ m which is the maximum resolution of the human eye
l
et's reduce the magnitudes to the SI system
d = 6.09‘ 2.54 10⁻-2 m / 1 inch = 15.4686 10-2 m
let's calculate
θ = 1.22 550 10-9 / 15.468 10-2
θ = 15.5 10⁻⁶ rad
rad = 2.06 105 s
θ = 15.5 10⁻⁶ rad 2.06 105s/ 1 rad
θ = 3.19 s
To complete a task in a short amount of time/get from A to B in the quickest amount of time
Hope helps!
Answer:
θ’ = θ₀ / 2
we see that the resolution angle is reduced by half
Explanation:
The resolving power of a radar is given by diffraction, for which we will use the Rayleigh criterion for the resolution of two point sources, they are considered resolved if the maximum of diffraction of one coincides with the first minimum of the other.
The first minimum occurs for m = 1, so the diffraction equation of a slit remains
a sin θ = λ
in general, the diffraction patterns occur at very small angles, so
sin θ = θ
θ = λ / a
in the case of radar we have a circular aperture and the equation must be solved in polar coordinates, which introduces a numerical constant.
θ = 1.22 λ /a
In this exercise we are told that the opening changes
a’ = 2 a
we substitute
θ ‘= 1.22 λ / 2a
θ' = (1.22 λ / a) 1/2
θ’ = θ₀ / 2
we see that the resolution angle is reduced by half
