Question

An object is at a distance of 15 cm in front of a concave mirror of focal length 30 cm. Find the nature and position of the image?

Answer 1

Hey mate.
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Given,

Object Distance , u = -15 cm

Focal length , f = -30 cm

Nature of the image = ?

Image distance , v = ?

By applying Mirror Formula , we have

1/f = 1/v + 1/u

=> 1/f - 1/u = 1/v

=> u - f / f × u = 1/v

=> f × u / u - f = v

=> -30 (-15) / -15- (-30) = v

=> 30 = v

Thus , The image is located at 30 cm in front of the concave mirror.

Now ,

Magnification , m = -v / u

= -30 / -15

= 2

Since , the Magnification is (+) ve , the image is virtual.

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Answer 2

Hey there !!

Distance of the object from pole of Concave mirror , u = -15

Focal length , = -30

Image distance , v = ?

By mirror formula ,
$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$
putting given value
$\frac{1}{v} = \frac{ - 1}{ 30} + \frac{1}{15} \\ \\ = \frac{ - 1 + 2}{30} \\ = 30 \: cm$
since it is in +,ve sign therefore it located beyond the mirror by distance 30 cm

Magnification =
$\frac{ - v}{u} \\ \\ = \frac{ - 30}{15} \\ \\ = 2$
Since + ve therefore image is magnified , Erect , virtual

Hope this help u !!