An object is placed on a metre scale at 8 cm mark was focused on a white screen placed at 92 cm mark, using a converging lens placed on the scale at 50 cm mark.; (i)Find the focal length of the converging lens.; (ii)Find the position of the image formed if the object is shifted towards the lens at a position of 29.0 cm.; (iii)State the nature of the image formed if the object is further shifted towards the lens.(All India - 2013)
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Position of the object = 8 cm (Given)
Position of the lens = 50 cm (Given)
Position of the mage = 92 cm (Given)
Thus,
Object distance = 50-8
= 42 cm = u
Image distance = 92-50
= 42 cm = v
Focal length = 1/v - 1/u = 1/f
= 1/42 - 1/(-42) = 1/f
= 1/42 + 1/42 = 1/f
= 1/f = 2/42
= 1/21
= f = 21cm
(ii) If the object is shifted to 29cm
New object distance = 50-29
= 21 cm
Focal Length -
= 1/v - 1/(-21) = 1/21
= 1/v = 1/21 -1/21 = 0
= v = ∞
(iii) If the object is further shifted towards the lens, the object will be between focus and the optical center making a virtual, erect and a magnified image.
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