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3 years ago

Which optical instrument, a telescope, a microscope, or a camera, forms images in a way most like your eye? Explain.

Physics

1 answer:

Xelga [282]3 years ago

3 0

Camera is an optical instrument that forms images in a way most like human eyes.

Explanation:

The eye consists of a small spherical globe of about 2 cm in diameter, which is free to rotate under the control of 6 extrinsic muscles. Light enters the eye through the transparent cornea, passes through the aqueous humor, the lens, and the vitreous humor, where it finally forms an image on the retina.

In case of a digital camera, CCD is used known as charged coupled device. CCD is actually in the shape of array or a rectangular grid. It is like a matrix with each cell in the matrix contains a censor that senses the intensity of photon. When light falls on the object , the light reflects back after striking the object and allowed to enter inside the camera which is then stored by CCD in the form of electric signals.

Similarities between eye and a camera:

Cornea and Lens - The cornea is the “cap” of the eye. This transparent (like clear jelly) structure sits to the front of the eye. The lens of a camera is also transparent (glass) and sits at the front of the body. Both allows light to enter and have spherical curvature.

Iris and aperture -The aperture is to the camera as the iris is to the eye. Both control the amount of light allowed to enter the eye or camera.

Retina and Film - The retina sits at the back of the eye and collects the light reflected from the surrounding environment to form the image. The same task in the camera is performed either by film or sensors in digital cameras.

Keywords: eye, camera, CCD, iris, film, aperture, cornea, lens, image formation

Learn more about camera from brainly.com/question/1771913

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Easy Guided Online Tutorial A special electronic sensor is embedded in the seat of a car that takes riders around a circular loo

kumpel [21]

Answer:

v = 17.30 m / s

Explanation:

For this exercise we will use Newton's second law

at the bottom of the loop and stopped

∑ F = 0

N-W = 0

N = W

W = 770 N

the mass of the body is

W = mg

m = W / g

m = 770 / 9.8

m = 78.6 kg

on top of the loop and moving

∑ F = m a

N + W = m a

note that the three vectors go in the same vertical direction down

the centripetal acceleration is

a = v² / r

we substitute

N + W = m v² / r

v = $\sqrt{(N+W) \frac{r}{m} }$

let's calculate

v = $\sqrt{ (350+770) \frac{21}{78.6} }$

v = 17.30 m / s

7 0

2 years ago

Anybody wanna help? (Picture Included?)

MrRissso [65]

There is your answer

4 0

3 years ago

Calculate ideal work (in J) when a single stream of 1 mole of air is heated and expanded from 25 C and 1 bar to 100 C and 0.5 ba

NISA [10]

Answer:

-1786.5J

Explanation:

Temperature 1=T1=25°c

Temperature 2=T2=200°c

Pressure P1=1bar

Pressure P2=0.5bars

T=37°c+273=310k

Note number if moles=1

Recall work done =2.3026RTlogp2/P1

2.3026*8.314*310log(0.5/1)

-1786.5J

7 0

3 years ago

Why is it better to breathe through the nose than the mouth?

Citrus2011 [14]

You could get sick by breathing throw your mouth and you have a less chance of getting sick by breathing throw your nose.

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3 years ago

Suppose you are standing at the exact center of a park surrounded by a circular road. An ambulance drives completely around this

vladimir2022 [97]

Answer: The pitch of the sound does not change.

Explanation: The pitch of sound wave is dependent on the frequency of the sound wave. The frequency of sound wave when their is a relative motion between an observer and a source is given by Doppler effect.

Doppler effect is a mathematical equation that gives the relationship between the observed frequency by an observer from any sound source and the relative motion between the observer and the source.

Mathematically,

f'= (v+v') /(v-vs) * f

Where f' = observed frequency of sound wave

v= speed of sound in air

v'= velocity of observer relative to sound source

vs= velocity of sound source relating to observer

f= frequency of sound produced by source

The observer is at the center, thus the distance between the observer and the source is constant (according to mensuration, the radius is constant for any given circle and since the car is moving along a circular path and the observer is at the center, thus the distance between them is constant), thus making the relative velocity between the observer and the source constant (vs=constant).

Also the frequency of sound wave produce by the source is a constant (f=constant)

The speed of sound in air is also a constant (v=336m/s)

The observer is standing at the center thus he is not moving, hence the relative motion between observer and source is also constant (v'=constant)

Since all parameters are constant, then the observed frequency will be constant too.

4 0

3 years ago