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kolbaska11 [484]
3 years ago
15

Consider a telescope with a small circular aperture of diameter 2.0 centimeters. Calculate the angular separation θ1 at which tw

o point sources of wavelength 600 nanometers are just resolved when viewed through a circular aperture of diameter 1.5 centimeters.
Physics
1 answer:
kogti [31]3 years ago
7 0

To solve this problem it is necessary to apply the concepts related to diffraction through a circular opening.

By definition the angular resolution is given by

\theta = 1.22\frac{\lambda}{D}

Where,

\lambda = Wavelenght

D = Diameter of the lens aperture.

Our values are given as,

\lambda = 600*10^{-9}m

d = 1.5*10^{-2}m

Therefore replacing,

\theta = 1.22\frac{600*10^{-9}}{1.5*10^{-2}}

\theta = 4.88*10^{-5} rad

Therefore the angular separation is 4.88*10^{-5} rad

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A particle moves in a velocity field V(x, y) = x2, x + y2 . If it is at position (x, y) = (7, 2) at time t = 3, estimate its loc
MArishka [77]

Answer:

New location at time 3.01 is given by: (7.49, 2.11)

Explanation:

Let's start by understanding what is the particle's velocity (in component form) in that velocity field at time 3:

V_x=x^2=7^2=49\\V_y=x+y^2=7+2^2=11

With such velocities in the x direction and in the y-direction respectively, we can find the displacement in x and y at a time 0.01 units later by using the formula:

distance=v\,*\, t

distance_x=49\,(0.01)=0.49\\distance_y=11\,(0.01)=0.11

Therefore, adding these displacements in component form to the original particle's position, we get:

New position: (7 + 0.49, 2 + 0.11) = (7.49, 2.11)

6 0
3 years ago
Suppose that the design parameters of the satellite's control system require that the angular velocity of the satellite not exce
Stells [14]

Answer:

v=0.04m/s

Explanation:

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hope this helps!!

6 0
3 years ago
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