Question

An object 6cm high is placed at a distance of 8cm in front of concave mirror of focal length 16cm. Find the position, nature and size of the image formed

Answer 1

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Answer 2

Answer:

Since it is a concave mirror, the given values are as follows:

Height of the object = 6cm ; Object distance (u) = - 8cm ; Focal length (f) = - 16cm.

It can be clearly found from the above values that the object is placed between the focus and the pole of the mirror. When an object is placed between focus and pole, the resulting image formed will be virtual, erect, enlarged and the image will be formed behind the mirror.

Therefore image specifications are as follows:

By mirror formula:  $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$

Therefore, on substituting the values of u and f we get,

$\frac{1}{v} - \frac{1}{8} = -\frac{1}{16}$

$\frac{1}{v} = -\frac{1}{16} + \frac{1}{8}$

$\frac{1}{v} = \frac{1}{16}$

$v= 16cm$

v = 16 cm

Position of the image :   16 cm ; Behind the mirror

Magnification (m) = - v/u

On substituting the values of u and v, we get

$m = - (-\frac{16}{8} )$

m = + 2

Image height ($h_{i}$ ) = m * Height of the object ( $h_{o}$)

$h_{i}$ = 2 * 6 = 12 cm

Height of the image = 12cm

Nature of the image : Virtual and erect

Size of the image : Magnified and enlarged image

Therefore, the Image distance is 16cm

                        Magnification is +2

                        Height of the image is 12cm

                        Position of the image is Behind the mirror

                        Nature of the image : Virtual and erect

                        Size of the image : Magnified and enlarged image

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