Question

An object is placed at a distance of 50 cm from a concave lens of focal length 20 cm find the position of real image

Answer 1

$\large\underline{\bigstar \: \: {\sf Given-}}$

➩ In concave lens -

• Object is placed at (u) = -50cm
• Focal Lenght (f) = -20cm

Sign convention -

Object and focal Lenght are always at left side of lens therefore they are taken as negative (-)

$\large\underline{\bigstar \: \: {\sf To \: Find -}}$

• Position of image

$\large\underline{\bigstar \: \: {\sf Formula \: Used -}}$

★ Lens formula -

$\bullet\underline{\boxed{\sf \dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}}}$

$\large\underline{\bigstar \: \: {\sf Solution-}}$

$\implies{\sf \dfrac{1}{-20}=\dfrac{1}{v}-\left(\dfrac{1}{-50}\right)}$

$\implies{\sf \dfrac{1}{-20}=\dfrac{1}{v}+\dfrac{1}{50} }$

$\implies{\sf \dfrac{1}{v}=\dfrac{1}{-20}-\dfrac{1}{50}}$

$\implies{\sf \dfrac{1}{v}=\dfrac{50-(-20)}{-20×50} }$

$\implies{\sf \dfrac{1}{v}=\dfrac{70}{-1000}}$

$\implies{\bf v=-14.2\:cm}$

$\large\underline{\bigstar \: \: {\bf Position \; of \; image -}}$

• Image is between Focus and Optical centre
• Virtual , erect diminished , toward the object

Answer 2

$\mathtt{\huge{ \fbox{Solution :)}}}$

By sign convention ,

Object distance (u) = -50 cm

Focal length (f) = -20 cm

We know that ,

The formula of lens is given by

$\large \mathtt{ \fbox{ \frac{1}{f} = \frac{1}{v} - \frac{1}{u}} }$

Thus ,

$\sf \hookrightarrow - \frac{1}{20} = \frac{1}{v} - ( - \frac{1}{50} ) \\ \\ \sf \hookrightarrow \frac{1}{v} = - \frac{1}{50} - \frac{1}{20} \\ \\ \sf \hookrightarrow \frac{1}{v} = \frac{ (- 20 - 50)}{1000} \\ \\ \sf \hookrightarrow v = - \frac{1000}{70} \\ \\ \sf \hookrightarrow v = - 14.2 \: \: cm$

The negative sign of v indicates the image is in left side of lens

Hence , the position of image is 14.2 cm