A 2 cm tall  object is placed perpendicular to the  principal axis of a  convex  lens of  focal length 10 cm. the distance  of the  object  from the  lens is 15 cm. find the  nature, position and size  of the  image

Asked by sisugsgsgsysyhshz | 9th Jun, 2017, 01:39: PM

Expert Answer:

begin mathsize 12px style For space the space object colon
straight h equals 2 space cm semicolon space straight f equals 10 space cm semicolon space straight u equals negative 15 space cm
From space lens space formula comma
1 over straight v minus 1 over straight u equals 1 over straight f
therefore 1 over straight v equals 1 over straight f plus 1 over straight u equals 1 over 10 plus fraction numerator 1 over denominator negative 15 end fraction equals 1 over 10 minus 1 over 15
therefore 1 over straight v equals fraction numerator 3 minus 2 over denominator 30 end fraction equals 1 over 30
therefore straight v equals 30 space cm
As space straight v space is space positive comma space the space image space is space real.
The space other space aspects space about space the space nature space canbe space determined space from space the space value space of space magnification.
straight m equals straight v over straight u equals fraction numerator straight h apostrophe over denominator straight h end fraction
therefore fraction numerator 30 over denominator negative 15 end fraction equals fraction numerator straight h apostrophe over denominator 2 end fraction
therefore straight h apostrophe equals negative 4
Hence comma space the space height space of space the space image space is space 4 space cm. space This space suggests space that space the space image space is space enlarged.
The space negative space sign space of space the space magnification space suggests space that space the space image space is space inverted.
Hence comma space the space image space is space real comma space inverted space and space enlarged. end style

Answered by Romal Bhansali | 9th Jun, 2017, 02:41: PM

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