Question

Q8. An object is placed in front of a concave mirror of radius of curvature 15cm at a
distance of (a) 10cm. and (b) 5cm. Find the position, nature and magnification of the
image in each case.
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Answer 1

Answer:

Givens:

Object distance = 10 cm

Object distance = u = -10 cm 

Radius of curvature = R = -15 cm

Focal length = f

f = 2R

f = R/2 = -15/2 cm

We know that :

1/v + 1/u = 1/f

v = image distance

1/v = 1/f - 1/u

    = 1 / (-15/2) - (-1/10)

    = -2/15 + 1/10

    = -4/30 + 3/30

    = -1/30

1/v = -1/30

v = -30 cm

Image distance = v = -30 cm

The position of image is 30 cm in front of mirror.

We know that ,

magnification , m  = -v/u

                            = -(-30)/-10

                            = -3

Nature : This negative sign indicates that the image is real ,and inverted.

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ii) Object distance = 5 cm

Object distance  = u = -5cm

Focal length = -15/2 cm

1/v = 1/f - 1/u

     = -2/15 - (-1/5)

    = -2/15 + 1/5

    = 1/15

v = 15 cm

Image distance = v = 15 cm

m = -v/u

   = -15/-5

   = 3

Nature of image : The positive sign of magnification indicates that the image is virtual , erect and magnified 

Explanation:

Hope this helps.