A point source S is placed in front of a concave mirror of radius of curvature 50 cm.A plane mirror M is placed 200 cm away from the concave mirror and an opaque object prevents direct light from the source on plane mirror.What should be the distance of S from concave mirror to have a image of S at S itself. Draw a neat diagram to explain the ans.

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nice question dude!! take the image formed by the concave mirror to be the object for plane mirror and the image formed by plane mirror is formed at initial position of S. use focal length for plane mirror is infinity...so v=-u..
sry bt its nt dude its mimansha bt due to sme inconvenience this site has kept my name mimansh in shrt i'm a girl
can u plz explain me better through this image

Answers

2015-01-12T02:10:04+05:30

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Suppose  we place Source S at the center C of curvature of concave mirror CM.  Then the real image is formed at C itself.    Then there is no role of plane mirror in this.  So in this case  S is placed at 50 cm from CM.

  If plane mirror is to be involved  then, the Source S is placed between the Focus F (25 cm) and C.  So the rays form an image  beyond the position of plane mirror M.  That is rays reflected from concave mirror CM trying to converge behind the plane mirror M, get reflected again.  The relation we have here is that , the distance of the image (to be formed by concave mirror) behind plane mirror is equal to the distance of the image in front of  plane mirror.

Let V (positive value) be the distance of formation of image from concave mirror, if there is no plane mirror.  Then (V - 200) will be the distance of  this image behind plane mirror.  Then the real image will be formed after reflection from plane mirror M at  (V-200) infront of  M.  So  actual image is at  200 - (V-200) = 400 - V  cm  from the concave mirror.

That is the source is placed at  u = 400 - V  cm from  CM.

sign convention :  u = - (400-V)     v = - V
  f = - 25 cm

1/v + 1/u = 1/f  = -1/25
   1/(400 - V)  + 1/V = 1/25
 400 / (400 V - V²) = 1/25
  10,000 - 400 V + V²  = 0

     V = 200 + 100√3 cm  and 200 - 100√3 cm  -- ignore this as it is less than 200 cm
      V =  373.20 cm
  hence  u = 400- V = 26.80 cm

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thanks a lot
can u teach me sign conventions
horizontal distances are measured always from the pole of mirror. vertical distances always from the axis of the mirror. Distances against the direction of the rays coming from object towards mirror -- are negative. distances in the direction of rays from object towards mirror -- are positive..
heights upward from axis (in the same direction as object) are positive. Heights against the direction of the height of objects are negative.. So erect image height is positive. height of an Inverted image is negative.