Question

An object is placed at a distance of 20 cm in front of concave mirror of focal length 10 cm. The image produced is?

Answer 1

Answer:

The image produced is real, inverted and same size.

Explanation:

Using the sign conventions for spherical mirrors,

Object distance,$u = -20 cm$

Focal length, $f = -10 cm$

Using the mirror formula,

$\begin{aligned}&\frac{1}{\mathrm{~V}}+\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}} \\&\Rightarrow \frac{1}{\mathrm{v}}+\frac{1}{-20}=\frac{-1}{10} \\&\Rightarrow \frac{1}{\mathrm{v}}=-\frac{1}{10}+\frac{1}{20}=\frac{-2+1}{20} \\&=\frac{-1}{20} \\&\mathrm{v}=-20 \mathrm{~cm}\end{aligned}$

The radius of curvature of the mirror is double the focal length (R = 2f). The image formed at the centre of curvature of the concave mirror is real, inverted and of the same size. For the given mirror, the center The object distance is equal to the centre of curvature of the concave mirror

Answer 2

Answer:

The image produced will be:

• real
• inverted
• same size as the object

Explanation:

1. The reflective surface of a concave mirror is curved inward and it is away from the source of light.
2. The image formed by a concave mirror shows different characteristics which depend on the distance between the object and the mirror.
3. By considering the sign conventions for spherical mirrors we will get,

• object distance, $u$ = $-20 \ cm$
• focal length, $f$ = $-10 \ cm$

Using the mirror formula we get,

$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$

$\frac{1}{v} + \frac{1}{-20} = \frac{-1}{10}$

$\frac{1}{v} = -\frac{1}{10} + \frac{1}{20} = \frac{-2+1}{20}$

$-\frac{1}{20}$

$v = -20 \ cm$

We know that the radius of curvature of a mirror is basically two times the focal length.

This means that, $R = 2f$

The final image formed will have the following characteristics:

• real
• inverted
• same size as the object

The object distance is equal to the center of curvature of the concave mirror.

Link:

https://brainly.in/question/9911939?referrer=searchResults

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