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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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b) 10.526 s

c) 61.82 N

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p_{1}=mV_{1} is the initial momentum of the ball with initial velocity (to the left) V_{1}=-38 m/s

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I=\Delta p=mV_{2}-mV_{1} (2)

I=\Delta p=m(V_{2}-V_{1}) (3)

I=\Delta p=0.685 kg (57 m/s-(-38 m/s)) (4)

I=\Delta p=65.075 kg m/s (5)

<h3>b) Time </h3>

This time can be calculated by the following equations, taking into account the ball undergoes a maximum compression of approximately 1.0 cm=0.01 m:

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F=\frac{\Delta p}{\Delta t} (13)

F=\frac{65.075 kgm/s}{1.052 s} (14)

Finally:

F=61.82 N

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