A convex mirror of focal length 2 meter forms image of 2 cm . If the object is 10 meter away from mirror then find the position of

2 thoughts on “A convex mirror of focal length 2 meter forms image of 2 cm . If the object is 10 meter away from mirror then find the position of”

1. Answer:

   It is to be remembered that a convex mirror always forms the virtual, erect and diminished image.

   Given: Object distance, u=−10cm Focal length, f=15cm

   To find: Image distance, v

   From mirror formula :

   v

   1

   +

   u

   1

   =

   f

   1

   v

   1

   −

   10

   1

   =

   15

   1

   v

   1

   =

   15

   1

   +

   10

   1

   =

   150

   25

   v=6cm

   Hence image will be formed 6 cm beyond the mirror, i.e virtual magnification

   m=

   u

   −v

   =

   −10

   −6

   =0.6

   Hence it will be erect and diminished.

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2. Answer:

   Given = focal length,f = 2 cm (taken positive because forcal length of convex mirror is behind the mirror)

   image height, h’ = 2 cm

   object distance, u = -10 cm

   We know that,

   Using mirror formula,

   [tex] \boxed{ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} }[/tex]

   [tex] \boxed{ \frac{1}{2} = \frac{1}{v} – \frac{1}{10} }[/tex]

   [tex] \boxed{ \frac{1}{2} + \frac{1}{10} = \frac{1}{v} }[/tex]

   [tex] \boxed{ \frac{5 + 1}{10} = \frac{1}{v} }[/tex]

   [tex] \boxed{ \frac{1}{v} = \frac{6}{10} }[/tex]

   [tex] \boxed{v = \frac{10}{6} }[/tex]

   [tex] \boxed{v = 1.7 \: cm}[/tex]

   Linear magnification = -v/u

   => -1.7/-10

   => 0.17

   Since the magnification is positive hence the image is virtual and erect and the image size will be highly diminished.

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