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3 years ago

8

A geneticist looks through a microscope to determine the phenotype of a fruit fly. The microscope is set to an overall magnifica

tion of 400x with an objective lens that has a focal length of 0.40 cm. The distance between the eyepiece and objective lenses is 12 cm. 1) Find the focal length of the eyepiece lens assuming a near point of 25 cm (the closest an object can be and still be seen in focus). Do not neglect any values in your calculation. (Express your answer to two significant figures.)

1 answer:

Ugo [173]3 years ago

8 0

Answer:

$f_{e}$ = 1.9 cm

Explanation:

The magnification of a microscope is the product of the magnification of the eyepiece by the magnifier with the objective

M = M₀ $m_{e}$

Where M₀ is the magnification of the objective and $m_{e}$ is the magnification of the eyepiece.

The eyepiece is focused to the near vision point (d = 25 cm)

$m_{e}$ = 25 / $f_{e}$

The objective is focused on the distances of the tube (L)

M₀ = -L / f₀

Substituting

M = - L/f₀ 25/ $f_{e}$

1) Let's look for the focal length of the eyepiece (faith)

$f_{e}$ = - L 25 / f₀ M

M = 400X = -400

$f_{e}$ = - 12 25 /0.40 (-400)

$f_{e}$ = 1.875 cm

Let's approximate two significant figures

$f_{e}$ = 1.9 cm

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Answer:

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Under the assumption that no external forces are exerted on both the small object and the big object, whose situation is described by the Principle of Momentum Conservation:

$p_{S,1}+p_{B,1} = p_{S,2}+p_{B,2}$ (1)

Where:

$p_{S,1}$ , $p_{S,2}$ - Initial and final momemtums of the small object, measured in kilogram-meters per second.

$p_{B,1}$ , $p_{B,2}$ - Initial and final momentums of the big object, measured in kilogram-meters per second.

If we know that $p_{S,1} = 7\,\frac{kg\cdot m}{s}$ , $p_{B,1} = 0\,\frac{kg\cdot m}{s}$ and $p_{S, 2} = 4\,\frac{kg\cdot m}{s}$ , then the final momentum of the big object is:

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$p_{B,2} = 3\,\frac{kg\cdot m}{s}$

The magnitude of the large object's momentum change is:

$p_{B,2}-p_{B,1} = 3\,\frac{kg\cdot m}{s}-0\,\frac{kg\cdot m}{s}$

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a)

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We have:

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- The backward, resistive force of 0.52 N

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Now we can find the acceleration by using Newton's second law of motion, which states that:

$\sum F=ma$

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m = 0.200 kg is the mass of the box

a is its acceleration

And solving for a, we find the acceleration:

$a=\frac{\sum F}{m}=\frac{3.03}{0.200}=15.2 m/s^2$

b)

The gravitational force on an object is the force with which the object is pulled towards the ground by the Earth.

It is given by

$W=mg$

where

m is the mass of the object

g is the gravitational field strength

In this problem we have

m = 0.200 kg is the mass of the box

$g=9.8 m/s^2$ is the gravitational field strength

So, the gravitational force on the box is

$W=(0.200)(9.8)=1.96 N$

c)

The normal force is the reaction force exerted by the floor on the box, in the upward direction.

In order to find the magnitude of this force, we apply Newton's second law of motion along the vertical direction.

We have two forces in this direction:

- The gravitational force, W, downward

- The normal force, N, upward

So the net force is

$\sum F=N-W$

According to Newton's second law,

$\sum F=ma$

However, the box is at rest in the vertical direction, so the vertical acceleration is zero:

$a=0$

This means that the net force is zero:

$\sum F=0$

And so, we can find the normal force:

$N-W=0\\N=W=1.96 N$

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and

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Here we are assuming that at t=T/4

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then

A=- d

Hence

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now, x is maximum when sin wt=- 1

Therefore,

x(maximum)=2d

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