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Salsk061 [2.6K]
3 years ago
7

The distance between the lenses in a compound microscope is 18 cm. The focal length of the objective is 1.5 cm. If the microscop

e is to provide an angular magnification of -46 when used by a person with a normal near point (25 cm from the eye), what must be the focal length of the eyepiece
Physics
1 answer:
Blababa [14]3 years ago
6 0

Answer:

The focal length of eye piece is 6.52 cm.

Explanation:

Given that,

Angular Magnification of the microscope M = -46

the distance between the lens in microscope L= 16 cm

The focal length of objective f₀ = 1.5 cm

Normal near point N = 25 cm

Have to find focal length of eye piece f ₙ =?

The angular magnification is given by

M ≈ - (L-fₙ)N/f₀fₙ

Rearranging for fₙ

fₙ =L(1 - Mf₀/N)⁺¹

   =18/2.76

fₙ =  6.52 cm

The focal length of eye piece is 6.52 cm.

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Where must an object be placed to form an image 30.0 cm from a diverging lens with a focal length of 43.0 cm?
Schach [20]
Using lens equation;

1/o + 1/i = 1/f; where o = Object distance, i = image distance (normally negative), f = focal length (normally negative)

Substituting;

1/o + 1/-30 = 1/-43 => 1/o = -1/43 + 1/30 = 0.01 => o = 1/0.01 = 99.23 cm

Therefore, the object should be place 99.23 cm from the lens.
6 0
2 years ago
The weight of a box is found to be 30 N. What is the approximate mass of the box?
KengaRu [80]
The mass of the box would be 30!
3 0
2 years ago
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The frequency of a microwave is 1.2 x 10^9 hertz. what is the wavelength of the given problem.
Olenka [21]

Answer:

0.25 m

Explanation:

Electromagnetic waves consist of oscillations of the electric and the magnetic field, oscillating in a plane perpendicular to the direction of motion the wave.

All electromagnetic waves travel in a vacuum always at the same speed, the speed of light, whose value is:

c=3.0\cdot 10^8 m/s

Microwave is an example of electromagnetic waves.

The relationship between wavelength and frequency for an electromagnetic wave is:

\lambda=\frac{c}{f}

where

\lambda is the wavelength

c=3.0\cdot 10^8 m/s  is the speed of light

f is the frequency

For the microwave in this problem,

f=1.2\cdot 10^9 Hz

So its wavelength is

\lambda=\frac{3.0\cdot 10^8}{1.2\cdot 10^9}=0.25 m

7 0
3 years ago
1 point
Sav [38]

Answer:

388.5J

Explanation:

Given parameters:

Weight  = 70N

Height  = 5.55m

Unknown:

Gravitational potential energy at the top of the ladder  = ?

Solution:

The gravitational potential energy is the energy due to the position of the body.

  Gravitational potential energy  = Weight x height

So;

 Gravitational potential energy  = 70 x 5.55 = 388.5J

8 0
2 years ago
A mass is oscillating with amplitude A at the end of a spring.
Dmitry_Shevchenko [17]

A) x=\pm \frac{A}{2\sqrt{2}}

The total energy of the system is equal to the maximum elastic potential energy, that is achieved when the displacement is equal to the amplitude (x=A):

E=\frac{1}{2}kA^2 (1)

where k is the spring constant.

The total energy, which is conserved, at any other point of the motion is the sum of elastic potential energy and kinetic energy:

E=U+K=\frac{1}{2}kx^2+\frac{1}{2}mv^2 (2)

where x is the displacement, m the mass, and v the speed.

We want to know the displacement x at which the elastic potential energy is 1/3 of the kinetic energy:

U=\frac{1}{3}K

Using (2) we can rewrite this as

U=\frac{1}{3}(E-U)=\frac{1}{3}E-\frac{1}{3}U\\U=\frac{E}{4}

And using (1), we find

U=\frac{E}{4}=\frac{\frac{1}{2}kA^2}{4}=\frac{1}{8}kA^2

Substituting U=\frac{1}{2}kx^2 into the last equation, we find the value of x:

\frac{1}{2}kx^2=\frac{1}{8}kA^2\\x=\pm \frac{A}{2\sqrt{2}}

B) x=\pm \frac{3}{\sqrt{10}}A

In this case, the kinetic energy is 1/10 of the total energy:

K=\frac{1}{10}E

Since we have

K=E-U

we can write

E-U=\frac{1}{10}E\\U=\frac{9}{10}E

And so we find:

\frac{1}{2}kx^2 = \frac{9}{10}(\frac{1}{2}kA^2)=\frac{9}{20}kA^2\\x^2 = \frac{9}{10}A^2\\x=\pm \frac{3}{\sqrt{10}}A

3 0
3 years ago
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