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At what distance from a concave mirror of focal length 10 cm should an object 2 cm long be placed in order to get an erect image?¿?¿

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Answer 1

Answer:-

Object should be placed at a distance of 6.667 cm in front of the mirror.

Solution :-

concave mirror , f is negative ,  f = -10 cm

   u = ?

 h = 2 cm 

 erect image  so  h' is positive,  h' = 6 cm

     so m = h'/h = 3

         as  m = -v/u = 3  => v = -3 u

   mirror equation  is  1/v + 1/u = 1/f

       1/(-3u) + 1/u = 1/(-10)

          2/(3u) = -1/10

            u = -20 /3 cm = -6.67 cm    

    and  v = -3 u = 20 cm

           v is positive,  so image is behind the mirror and is virtual.

  In concave mirror , an erect image is obtained when the object is between the pole and the focus. it is magnified and is virtual.  So  you can verify from values of u, v and f.

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Answer 2

Answer:

Given that focal length, f= -10cm

Given that focal length, f= -10cmMagnification, m= Image height/ Object height= 6/2 = 3

Given that focal length, f= -10cmMagnification, m= Image height/ Object height= 6/2 = 3Also magnification= - v/u

Given that focal length, f= -10cmMagnification, m= Image height/ Object height= 6/2 = 3Also magnification= - v/u3= -v/u

Given that focal length, f= -10cmMagnification, m= Image height/ Object height= 6/2 = 3Also magnification= - v/u3= -v/uv= -3u

Given that focal length, f= -10cmMagnification, m= Image height/ Object height= 6/2 = 3Also magnification= - v/u3= -v/uv= -3uAs 1/u + 1/v = 1/f

Given that focal length, f= -10cmMagnification, m= Image height/ Object height= 6/2 = 3Also magnification= - v/u3= -v/uv= -3uAs 1/u + 1/v = 1/f1/u + 1/-3u = 1/ -10

Given that focal length, f= -10cmMagnification, m= Image height/ Object height= 6/2 = 3Also magnification= - v/u3= -v/uv= -3uAs 1/u + 1/v = 1/f1/u + 1/-3u = 1/ -10Therefore u= -6.667

Given that focal length, f= -10cmMagnification, m= Image height/ Object height= 6/2 = 3Also magnification= - v/u3= -v/uv= -3uAs 1/u + 1/v = 1/f1/u + 1/-3u = 1/ -10Therefore u= -6.667So object should be placed at a distance of 6.667 cm in front of the mirror.

Hope it helps

Mark brainliest