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loris [4]
3 years ago
10

Early cameras were little more than a box with a pinhole on the side opposite the film. (a) What angular resolution would you ex

pect from a pinhole with a 0.50-mm diameter? (b) What is the greatest distance from the camera at which two point objects 15 cm apart can be resolved? (Assume light with a wavelength of 520 nm.
Physics
1 answer:
ollegr [7]3 years ago
3 0

Answer:

angular resolution = 0.07270° = 1.269 × 10^{-3} rad

greatest distance from the camera = 118.20 m = 0.118 km

Explanation:

given data

diameter = 0.50 mm = 0.5 × 10^{-3} m

distance apart = 15 cm =  15× 10^{-2} m

wavelength λ = 520 nm = 520 × 10^{-9} m

to find out

angular resolution and greatest distance from the camera

solution

first we expression here angular resolution that is

sin θ = \frac{1.22* \lambda }{D}   .......................1

put here value λ is wavelength and d is diameter

we get

sin θ = \frac{1.22*520*10^{-9}}{0.5*10^{-3}}

θ = 0.07270° = 1.269 × 10^{-3} rad

and

distance from camera is calculate here as

θ = \frac{I}{r}    .................2

I = \frac{15*10^{-2}}{1.269*10^{-3}}

I = 118.20 m = 0.118 km

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Answer:

Explanation:

We may notice that change in velocity can be obtained by calculating areas between acceleration lines and horizontal axis ("Time"). Mathematically, we know that:

v_{b}-v_{a} = \int\limits^{t_{b}}_{t_{a}} {a(t)} \, dt

v_{b} = v_{a}+ \int\limits^{t_{b}}_{t_{a}} {a(t)} \, dt

Where:

v_{a}, v_{b} - Initial and final velocities, measured in meters per second.

t_{a}, t_{b} - Initial and final times, measured in seconds.

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Acceleration is the slope of velocity, as we know that each line is an horizontal one, then, velocity curves are lines with slopes different of zero. There are three region where velocities should be found:

Region I (t = 0 s to t = 4 s)

v_{4} = 2\,\frac{m}{s}  +\int\limits^{4\,s}_{0\,s} {\left(-2\,\frac{m}{s^{2}} \right)} \, dt

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v_{4} = -6\,\frac{m}{s}

Region II (t = 4 s to t = 6 s)

v_{6} = -6\,\frac{m}{s}  +\int\limits^{6\,s}_{4\,s} {\left(1\,\frac{m}{s^{2}} \right)} \, dt

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v_{6} = -4\,\frac{m}{s}

Region III (t = 6 s to t = 10 s)

v_{10} = -4\,\frac{m}{s}  +\int\limits^{10\,s}_{6\,s} {\left(2\,\frac{m}{s^{2}} \right)} \, dt

v_{10} = -4\,\frac{m}{s}+\left(2\,\frac{m}{s^{2}} \right) \cdot (10\,s-6\,s)

v_{10} = 4\,\frac{m}{s}

Finally, we draw the object's velocity graph as follows. Graphic is attached below.

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