Question

an object is placed at a distance of 30 cm from a convex lens of focal length 20 cm . find the position of image .is the image is virtual or real.​

Answer 1

Answer:

the position image is 60cm and it is real

Answer 2

$\bf \: \underline{Given} :$

$\sf\implies Object \: distance \:(u) = -30 \: cm$

$\sf \implies Focal \: length \: of \: the \: lens \: (f) = 20 \: cm$

$\bf \: \underline{To \: find} :$

$\sf \implies Position \: of \: the \: mirror = \: ?$

$\bf \: \underline{\underline{Solution}}$

$\: { \implies \boxed{ \red{\sf \: \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f} }}}$

$\sf \: \underline{Where},$

$\sf \implies\sf \: u = Object \: distance \: from \: mirror$

$\sf \implies\sf \: f = Focal \: length\: of \: mirror$

$\sf \implies\sf \: v = image \: distance \: from \: mirror$

$\sf \implies \dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u}$

$\sf \implies \dfrac{1}{v} = \dfrac{1}{20} - \dfrac{1}{ - 30}$

$\sf \implies \dfrac{1}{v} = \dfrac{3}{60} - \dfrac{2}{ 60}$

$\sf \implies \dfrac{1}{v} = \dfrac{3 - 2}{60}$

$\sf \implies \dfrac{1}{v} = \dfrac{1}{60}$

$\red{\sf \implies v = 60 \: cm }$

$\underline{ \boxed{ \sf \pink{Hence, the \: position \: of \: the \: image = 60 \: cm}}}$