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2 months ago

In concave mirror, the size of image depends upon

Physics

2 answers:

Maru [3.3K]2 months ago

8 0

Answer:

The positioning of the object along the principal axis relative to the concave mirror.

Explanation:

In a concave mirror, the characteristics of the image generated depend on where the object is situated in relation to the mirror. The distance from the mirror to the object positioned along the principal axis is key.

The nearer the object is to the mirror, the larger or more magnified the image will appear. For example, placing an object between the focal point and the concave mirror's pole results in a significantly larger image compared to an object placed outside the center of curvature of the mirror.

ValentinkaMS [3.4K]2 months ago

4 0

Answer:

The object's position concerning the vertex (V), the focus (F), the center of curvature (C), and infinity.

Explanation:

The positioning of the object concerning vertex (V), focus (F), center of curvature (C), and infinity results in various types of images;

The image formed by an object situated at V, in contact with the mirror, appears upright, virtual, and of the same size.

The image generated between V and F is also upright, virtual, and magnified.

When an object is at the focus, its image is formed at infinity.

The image formed by an object positioned between C and F is real, inverted, and enlarged.

The object situated between C and infinity creates a real, reduced, and inverted image.

An object placed at infinity produces a real image of zero size.

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Two electrodes, separated by a distance d, in a vacuum are maintained at a constant potential difference. An electron, accelerat

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Answer:

Explanation:

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The kinetic energy of the electron is represented as Ek when the electrodes are positioned at a distance of "d" apart.

Our goal is to determine the kinetic energy when they are separated by a distance of d/3.

K.E = ½mv²

It’s important to note that the mass remains constant; only velocity varies.

Additionally,

K.E = Work done by the electron

K.E = F × d

K.E = W = ma × d

Assuming constant acceleration

Hence, m and a are fixed,

therefore,

K.E is directly related to d

Thus, as d increases, K.E increases, and conversely, when d decreases, K.E decreases.

Consequently,

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With K.E_1 equating to E_k

and d_1 being d

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This leads to K.E_2 = K.E_1 / d_1 × d_2

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1 month ago

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