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AleksAgata [21]
3 years ago
6

A low orbit satellite at 100 km altitude had a camera that resolved a car location to within 0.3 meter. Find the minimum diamete

r of the camera lens when operating at 550nm sunlight. Find the focal length of the camera's lens when the photographic film had a resolution of 9-micro-meter due to grain size. CAS TI-7nspire cx
Physics
1 answer:
NeX [460]3 years ago
3 0

Answer:

Minimum diameter of the camera lens is 22.4 cm

The focal length of the camera's lens is 300cm

Explanation:

y = Resolve distance = 0.3 m

h = Height of satellite = 100 km

λ = Wavelength = 550 nm

Angular resolution

tan\theta\approx \theta =\frac{y}{h}\\\Rightarrow \theta=\frac{0.3}{100\times 10^3}=3\times 10^{-6}

From Rayleigh criteria

sin\theta=1.22\frac{\lambda}{D}\\\Rightarrow D=1.22\frac{\lambda}{sin\theta}\\\Rightarrow D=1.22\frac{550\times 10^{-9}}{sin3\times 10^{-6}}=0.2236\ m=22.4\ cm

Minimum diameter of the camera lens is 22.4 cm

Relation between resolvable feature, focal length and angular resolution

d=f\Delta \theta\\\Rightarrow f=\frac{d}{\Delta \theta}\\\Rightarrow f=\frac{9\times 10^{-6}}{3\times 10^{-6}}=3\ m=300\ cm

The focal length of the camera's lens is 300cm

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            ∑ τ = 0

          -W₁ x₁ - W₂ x₂ + N₁ y = 0

           N₁ = \frac{W_1 x_1 + W_2 x_2}{y}          (1)

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         x₁ = 7.5 cos 60

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         N₂ = \frac{500 \ 3.75 \ + 800 \ 2}{13}

         N₂ = 267.3 N

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 X axis

           F₁ - N₂ = 0

           F₁ = N₂

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Axis y

          N₁ - W₁ -W₂ = 0

          N₁ = W₁ + W₂

          N₁ = 500 + 800

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b) for this case change the firefighter's distance d₂ = 9 m

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we substitute in 1

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of the translational equilibrium equation on the x-axis

          fr = F₁ = N₂

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friction force has the expression

          fr = μ N

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          μ = fr / N₁

          μ = 421.15 / 1300

          μ = 0.324

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