Question

. A 6 cm object is placed perpendicular to the principal axis of a concave mirror of
focal length 15 cm. The distance of the object from the mirror is 10 cm. Find the
position, size and nature of the image formed.​

Answer 1

Answer:

Here you go ...........

Explanation:

object distance (u) = -10cm

focal length of the mirror (f) = -15cm

height of object = 6cm

We know that 1/f = 1/u + 1/v   (mirror formula)

                     1/-15 = 1/-10 + 1/v

                     -1/15 = -1/10 + 1/v

                     1/10 - 1/15 = 1/v

                     15-10/150 = 1/v

                     5/150 = 1/v

                     1/30 = 1/v

                     v = 30cm = image distance

Magnification = -v/u = (height of image) / (height of object)

                         -30/-10 = (height of image) /  6

                          3 =  (height of image) /  6

                           height of image = 18cm

   Height of the image is greater than the height of the object.

  So the image is magnified. Since the image distance is positive, the image is formed behind the mirror.

So the image is virtual, erect and is magnified