Question

1) An object is placed at a distance of 4 cm from
lens of focal length 12cm find the
position and nature of the image
con cave​

Answer 1

Explanation:

u = -4 cm

f = -12 cm

Lens formula : 1/v – 1/u = 1/f

1/v – 1/-4 = 1/-12

1/v = - (1/12 ) – 1/4

1/v = -4/12

V = - 3cm

Image is formed 3 cm infront of the concave lens.

Image is virtual and erect.

Answer 2

Given-

• ☛Object distance= 4cm

• ☛Focal length = 12cm

To Find-

• ☛Image distance-?

Formula to be used-

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

where,

v= image distance

u= object distance

f= focal length

SoLution-

• ☛using lens formula-

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

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

 

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

 

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

 

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

 

$\bf{ \implies \dfrac{1}{v} = \dfrac{-1}{3} }$

 

$\bf{\implies v= -3}$

$\sf{\underline{\pink{\therefore Position\ of \ image \ is \ 3cm \ away }}}$

- shows Image formed infront of lens

Nature of the image-

☛Virtual

☛Erect