Question

An object is placed at a distance 50 cm from a concave lens of focal length 20 cm. Find the nature and position of the image​

Answer 1

$Given:-$

• $f = -20\:cm$
• $u = -50\:cm$

$Formula \: Applied:-$

• $\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

$To\:Find:-$

• Nature & position of image

$Solution:-$

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

$\implies\:\:\dfrac{1}{-20}+\dfrac{1}{-50}$

$\implies\:\:\dfrac{-5-2}{100}$

$\implies\:\:\dfrac{-7}{100}$

$\implies\:\:\dfrac{-100}{7}$

$\implies\:\:{\boxed{\red{v = -14.29\:cm}}}$

★ Image distance is negative indicates that it is a virtual and erect image.