Question

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Answer 1

It is given that,

The distance of the object(u) = 7 cm

The distance of the image(v) = (9+7) cm

                                               = 16 cm

Let the focal length be 'f' and the magnificent of the image be, 'm'

We know that,

1/f = (1/u) +(1/v)

⇒ 1/f = (1/7) + (1/16)

⇒ 1/f = (16+7)/112 [∵ The L.C.M. of the denominators is 112]

⇒ 1/f = 23/112

⇒ f = 112/23

⇒ f = 4.86 cm

So, the focal length of the mirror is 4.86 cm.

We know that,

m = -v/u

   = -16/7

  = -2.28

So, the magnificent of the image is -2.28

As the magnificent is negative so the image formed by the mirror is real.

Answer 2

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Refer the attachment.