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Shalnov [3]
1 year ago
9

The focal length of the lens of a simple camera is 40.0 mm. how far must the lens be moved to change the focus of the camera fro

m a person 25 m away to one that is 4.0 m away?
Physics
1 answer:
Minchanka [31]1 year ago
4 0

.05 cm far must the lens be moved to change the focus of the camera from a person 25 m away to one that is 4.0 m away

The focal length of a thin lens in air is the separation between its major foci, also known as focal points, and its center. The focal length is the distance at which a beam of collimated light will be concentrated to a single spot for a converging lens (for instance, a convex lens).

If v is the image distance in the first scenario

1 / v - 1 / -25 = 1 / .05

1 / v = 1 / .05 - 1 / 25

= 20 - .04 = 19.96

v = .0501 m = 5.01 cm

The second instance

u = 4 ,

1 / v - 1 / - 4 = 1 / .05

1 / v = 20 - 1 / 4 = 19.75

v = .0506 = 5.06 cm

Therefore, the lens must be advanced by 5.06 - 5.01 =.05 cm.

Learn more about  lens here brainly.com/question/17086993

#SPJ4.

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Explanation:

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The equation is given by,

V=\frac{I}{nAq}

Where,

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n= number of electrons

q = charge of electron

A = cross-section area.

For this problem we know that there is a rate of 1.8*10^{18} electrons per second, that is

\frac{I}{q} = 1.2*10^{18}

A= 1.3*10^{-8}m^2

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We can find the drift velocity replacing,

V = \frac{1.2*10^{18}}{(1.3*10^{-8})(6.3*10^{28})}

V= 1.465*10^-3m/s

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E=12.2V/m

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Plain electromagnetic wave (in air) has a frequency of 1 MHz and its B-field amplitude is 9 nT a. What is the wavelength in air?
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Answer:

Part a)

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I = 9.68 \times 10^{-3} W/m^2

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Explanation:

Part a)

As we know that frequency = 1 MHz

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P = 3.22 \times 10^{-11} N/m^2

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