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2 years ago

How does the image distance and magnification give information about the type of image is produced bya lens or multiple lenses/

Physics

1 answer:

4vir4ik [10]2 years ago

4 0

Explanation:

The image distance gives the position of image formed by a lens. It also shows how far the image is formed form the pole of the lens. There is a relationship between the object distance, object distance and the focal length of the lens. It is given by :

$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$

Where

v is the image distance

u is the object distance

f is the focal length of the lens

The ratio of image distance to the object distance is called the magnification of the lens. It is given by :

$m=\dfrac{v}{u}$

Magnification of lens is also defined as the ratio of the height of the image to the height of the object. It is given by :

$m=\dfrac{h'}{h}$

h' is the height of image

h is the height of object

The magnification of a lens shows that how many times the image is magnified. If the magnification of a lens is more than 1, it shows that the image is greater than the object and if it is less than 1, it shows the image is smaller than the object.

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Suppose the entire solar nebula had cooled to 50 K before the solar wind cleared the early solar system of its gases. How would

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Suppose the entire solar nebula had cooled to 50 K before the solar wind cleared the early solar system of its gases. How would the composition and sizes of the planets of the inner solar system be different from what we see today is given below

Explanation:

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2 years ago

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Going to college and passing all your classes

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two charges having the same charge magnitude experiencing an attracting force of 3.60N when the charges are 30cm apart.what is t

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The charges have opposite sign and magnitude $6 \mu C$

Explanation:

The magnitude of the electrostatic force between two electric charges is given by Coulomb's law:

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the two charges

r is the separation between the two charges

In this problem, we have:

F = 3.60 N is the force between the two charges

r = 30 cm = 0.30 m is their separation

The two charges have same magnitude, so

$q_1 = q_2 = q$

So we can rewrite the equation as

$F=\frac{kq^2}{r^2}$

And solving for q:

$q=\sqrt{\frac{Fr^2}{k}}=\sqrt{\frac{(3.60)(0.30)^2}{8.99\cdot 10^9}}=6\cdot 10^{-6} C = 6\mu C$

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Learn more about electric force:

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