Question

1. An object is placed at a distance of 4 cm from a concave lens of focal length 12 cm. Find the position and nature of the image.​

Answer 1

Given :

↗ Distance of object = 4cm

↗ Focal length = 12cm

↗ Type of lens : concave

To Find :

⟶ Position and nature of image.

SoluTion :

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

$:\implies\sf\:\dfrac{1}{v}-\dfrac{1}{(-4)}=\dfrac{1}{(-12)}$

$:\implies\sf\:\dfrac{1}{v}=-\dfrac{1}{12}-\dfrac{1}{4}$

$:\implies\sf\:\dfrac{1}{v}=\dfrac{-1-3}{12}$

$:\implies\sf\:\dfrac{1}{v}=\dfrac{-4}{12}$

$:\implies\boxed{\bf{v=-3\:cm}}$

✴ Nature of image :

• Virtual
• Erect

Answer 2

$\huge{\bold{\star{\fcolorbox{black}{lightgreen}{Answer}}}}⋆$

Position of image : -3 cm

Nature of image :

• Virtual
• Erect

$\huge{\bold{\star{\fcolorbox{black}{lightgreen}{Explanation}}}}⋆$

$\blue{\bold{\underline{\underline{Given}}}}$

• Distance of object = 4cm
• Lens(Type) = Concave
• Focal length = 12cm

$\blue{\bold{\underline{\underline{To \: Find}}}}$

• Position and nature of the image

$\blue{\bold{\underline{\underline{Formula}}}}$

${\boxed{\rm{\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}}}}$

$\blue{\bold{\underline{\underline{Solution}}}}$

Put the values,

$\dfrac{1}{v}-\dfrac{1}{(-4)}=\dfrac{1}{(-12)}$

$\implies \: \dfrac{1}{v}=-\dfrac{1}{12}-\dfrac{1}{4}$

$\implies \: \dfrac{1}{v}=\dfrac{-1-3}{12}$

$\implies \: \dfrac{1}{v}=\dfrac{-4}{12}$

$\implies v=-3\:cm$