Question

If an object is placed at a distance of 50cm. From a concave lens of focal length 20cm. Find position, nature, size of the image

Answer 1

$\textbf{\underline{\underline{According\:to\:the\:Question}}}$

★Given :-

u = -50 cm (Distance of the object)

f = -20 cm (Focal length)

${\boxed{\sf\:{To\;Find}}}$

v = ? (Distance of object)

${\boxed{\sf\:{Using\;Lens\;Formula}}}$

$\tt{\rightarrow{\sf\:{\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}}}$

$\tt{\rightarrow{\sf\:{\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}}}$

$\tt{\rightarrow{\sf\:{\dfrac{1}{v}=\dfrac{1}{(-20)cm}+\dfrac{1}{(-50)cm}}}$

$\tt{\rightarrow{\sf\:{\dfrac{1}{v}=\dfrac{-7}{100cm}}}$

$\tt{\rightarrow{\sf\:{v= -\dfrac{100}{7}}}$

= -14.3 cm

$\text{Here\;we\;get}$

$\boxed{Image\;will\;be\;formed\;on\;same\;side\;at\; distance\;of\;14.3 cm\;from\;the\;lens}$

$\text{And\;Also}$

${\boxed{v\;is\;negative\;so\;image\;will\;be\;erect\;and\;virtual}$