If a 3cm tall object placed perpendicular to principal axis of a convex lens of focal length 15cm produces a real invented image of height 15cm- then its object distance -u- is - and image distance -v- is -u-18m-v-90mu-18cm-V-90cmy-18cm-v-90cmu-18cm-v-90cm

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Toppr Admin · Accepted Answer

H1-3cmh2-15cmm-h2h1-153-5As image is real and invertedm-x2212-5object distance is always negative-x2234- u-x2212-18cmand v-90cm

h1=3cmh2=15cmm=h2h1=153=5As image is real and invertedm=−5object distance is always negative∴ u=−18cmand v=+90cm

$h_{1} = 3 c m$
$h_{2} = 15 c m$
$m = \frac{h_{2}}{h_{1}} = \frac{15}{3} = 5$
As image is real and inverted
$m = - 5$
object distance is always negative
$∴$ $u = - 18 c m$
and $v = + 90 c m$