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ExamsJEE AdvancedPhysics

The optical axis of a thin equiconvex lens is the x-axis. A point object at (-40 cm, 1 cm) forms an image at (50 cm, -2 cm). Determine the x-coordinate of the lens.

  1. x = 20 cm
  2. x = -30 cm
  3. x = -10 cm
  4. origin

Correct answer: x = -10 cm

Solution

Transverse magnification = (image height)/(object height) = -2/1 = -2 (negative because image is inverted). m = v/u. Object is at x=-40, lens at x=a: u = -40-a (measured as signed distance, taking sign convention: distances measured from lens). Image at x=50: v = 50-a. m = v/u: (50-a)/(-40-a) = -2 => 50-a = -2*(-40-a) = 80+2a => 50-a = 80+2a => -30 = 3a => a = -10. So lens is at x = -10 cm.

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