The correct answer for the question that is being presented above is this one: "Schmidt-Cassegrain focus." A focal arrangement that has a thin lens that the light passes through before traveling down the tube to the objective mirror is a Schmidt-Cassegrain focus.
Here are the following choices:
a. Cassegrain focus
b. Newtonian focus
c. Schmidt-Cassegrain focus
<span>d. Schmidt focus</span>
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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
Answer:
The ratio is KE : TM = 0.75
Explanation:
from the question we are told that
The displacement of a mass on a spring in simple harmonic motion is A/2 from the equilibrium position
Generally the total mechanical energy of the mass is mathematically represented as
Here k is the spring constant , A is the total displacement of the the mass from maximum compression to maximum extension of the spring
Generally this total mechanical energy is mathematically represented as
=> 
Here the potential energy of the mass is mathematically represented as
![PE = \frac{1}{ 2} * k * [ x ]^2](https://tex.z-dn.net/?f=PE%20%20%20%3D%20%5Cfrac%7B1%7D%7B%202%7D%20%20%2A%20%20k%20%2A%20%20%5B%20x%20%5D%5E2)
Here x is the displacement of the mass from maximum compression or extension of the spring to equilibrium position and the value is
So
![PE = \frac{1}{ 2} * k * [ \frac{A}{2} ]^2](https://tex.z-dn.net/?f=PE%20%20%20%3D%20%5Cfrac%7B1%7D%7B%202%7D%20%20%2A%20%20k%20%2A%20%20%5B%20%5Cfrac%7BA%7D%7B2%7D%20%20%5D%5E2)
So
![KE = \frac{1}{2} * k * A^2 - \frac{1}{2} * k * [\frac{A}{2} ]^2](https://tex.z-dn.net/?f=KE%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%20%20%2A%20%20k%20%20%2A%20%20A%5E2%20-%20%5Cfrac%7B1%7D%7B2%7D%20%20%2A%20%20k%20%20%2A%20%20%5B%5Cfrac%7BA%7D%7B2%7D%20%5D%5E2)
=> 
=> 
So the ratio of 
is mathematically represented as
=> 
