Question

An object of length 17 m is placed at a distance of 14 m the image is formed 14 m away from the mirror.
Find the focal length of the mirror, and magnificant of the image. Say wheather it is real or virtual image.
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Answer 1

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The magnification is also the ratio of the distances lens- to-image and lens-to-object. Let's not worry about where on a lens barrel you would measure those distances. For objects well in front of the lens the lens to image distance will be a little larger than the focal length.

Answer 2

Answer:

It is stated that,

The length of the object (O) = 17 m

The distance of the object (u) = 14 m

The distance of the formed image (v) = (14+14)m

                                                              = 28 m

Let, the magnificant of the image be, 'm'

The focal length be, 'f'

We know that,

Magnificant of the image = - v/u

                                          = -28/14

                                          = -2

The magnificant of the image is -2

As the magnificant of the image is negative so, its a real image.

We know that

1/ focal length = (1/distance of the object) + (1/distance of the formed image)

⇒ 1/f = (1/u) + (1/v)

⇒ 1/f = (1/14) + (1/28)

⇒ 1/f = (1+1)/28 [∵ The L.C.M. of 14 and 28 is 28]

⇒ 1/f = 2/28

⇒ 1/f = 1/28

⇒ f = 14 m

So, the foal length of the mirror is 14 m