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Natasha2012 [34]
3 years ago
11

What is the difference between virtual images produced by concave, plane, and convex mirrors?

Physics
1 answer:
julsineya [31]3 years ago
7 0

Answer:

Explanation:

A concave mirror is a curved mirror that is coated outwards. The outer part of the mirror is always coated. The nature of image formed by an object placed in front of a concave mirror can be real or virtual depending on the distance of the object on the axis from the mirror. The only time the object produces a virtual image is when it is placed between the focus and the pole of the mirror. The virtual image formed is a "MAGNIFIED and upright image"

For a convex mirror, the inner part is always coated and the nature of the image formed by the object doesn't depend on the distance between the image and the mirror. No matter where the object is placed, the image formed will always be virtual, upright and DIMINISHED. This means that magnification is always less than 1.

For a plane mirror, the nature of the image produced by a plane mirror also virtual because it is always formed behind the mirror. The size of the image formed is always THE SAME as that of the object. This means that the magnification is always equal to 1.

a) In summary, the difference between virtual images produced by concave, plane, and convex mirrors is that virtual images produced by concave mirror are MAGNIFIED, virtual images produced by plane mirror are THE SAME SIZE as that of the object and virtual images produced by convex mirrors are always DIMINISHED.

b) Magnification is defined as the ratio of the image distance to the object distance. Mag = v/u

Note that object distances are always positive, hence it is only the image distance that can either be positive or negative which in turn affects the magnification causing it to be positive or negative.

Negative image distance shows that the image is virtual while positive image distance shows that the image is real.

A negative magnification therefore shows that the nature of the image is a virtual image.

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To practice Problem-Solving Strategy 21.1 Coulomb's Law. Three charged particles are placed at each of three corners of an equil
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Answer:

The force felt by charge 3 is F=(-5.6*10⁻⁶,3.36⁻⁵)N

Explanation:

As the superposition principle applies to static charges, we can find the net electric force as the sum of the two forces felt by q3.

Looking at the drawing and knowing that they form an equilateral triangle of lenght 4 we can conclude that each internal angle is 60°.

So, the positions in our coordinate system are:

r_1=(0,0)\\r_2=(4\ cos(60\°),4\ sin(60\°))\\r_3=(4,0)\\

Now  using Coulomb's force:

F_{ij}=\frac{-kq_iq_j}{d^2}(\vv{r}_j-\vv{r}_i)

Where d=4, q1 = -7.8*10⁻⁹C, q2 = -15.6 *10⁻⁹C, q3 = 8.0 *10⁻⁹C, k=8.98*10⁹, e0=8.8*10¹⁰:

Replacing we get 2 equations:

F_{13}=\frac{-kq_1q_3}{d^2}(\vv{r}_1-\vv{r}_3)=\frac{-kq_1q_2}{d^2}(-0.04\ cos(60\°),-0.04\ sin(60\°))\\\\F_{23}=\frac{-kq_2q_3}{d^2}(\vv{r}_1-\vv{r}_3)=\frac{-kq_1q_2}{d^2}(0.04-0.04\ cos(60\°),-0.04\ sin(60\°))\\

To work with the sam

F=∑F_i=3.5*10⁻⁴(0.023,0.032)+7*10⁻⁴(-0.016,0.032)=

=((3.5*10⁻⁴-7*10⁻⁴)*0.016,(3.5*10⁻⁴+7*10⁻⁴)*0.032)=

F=(-5.6*10⁻⁶,3.36⁻⁵)N

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The correct option is 0.5 M

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Wavelength =?

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Frequency = 680 HZ

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