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An object placed at 30 cm fron the pole of convex lens focal lenth 20 cm determine the image distance magnification and natureof image

Asked by sreenu.animator 5th January 2016, 1:51 PM
Answered by Expert
Answer:

Hi Sreenu,

begin mathsize 12px style Object space distance comma space straight u equals negative 30 space cm space left parenthesis object space is space on space the space left space side space of space lens right parenthesis Focal space length comma space straight f equals space plus 20 space cm space left parenthesis It space is space straight a space convex space lens right parenthesis Using space the space lens space formula colon 1 over straight v minus 1 over straight u equals 1 over straight f on space substituting space vvalues space we space get comma 1 over straight v minus fraction numerator 1 over denominator negative 30 end fraction equals 1 over 20 1 over straight v equals 1 over 20 minus 1 over 30 equals 1 over 60 straight v equals plus 60 space cm Thus space image space is space formed space at space straight a space distance space 60 space cm space from space the space convex space lens. space Plus space sign space for space image space distance space shows space that space the space image space is space formed space on space the space right space side space of space the space convex space lens. Hence comma space image space is space real space and space inverted. Magnification comma space straight m equals straight v over straight u equals fraction numerator 60 over denominator negative 30 end fraction equals negative 2 end style

Answered by Expert 5th January 2016, 6:00 PM
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