Question

Option B is correct

I want full explanation ​

Answer 1

Given :

Distance of object = -10 cm

Focal length of lens A = 20 cm

Focal length of lens B = 5 cm

Distance between two lenses = 5 cm

To Find :

Position of final image.

Solution :

❖ First of all we need to find position of first image which is formed by lens A.

Applying lens formula for lens A,

[Note : Focal length of convex lens is taken positive]

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

$\sf:\implies\:\dfrac{1}{v}-\dfrac{1}{(-10)}=\dfrac{1}{20}$

$\sf:\implies\:\dfrac{1}{v}=\dfrac{1}{20}-\dfrac{1}{10}$

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

$\bf:\implies\:v=-20\:cm$

♦ First image $\bf{(I_1)}$ will behave as an object for lens B.

In this case, distance of object will be,

➙ u' = v + d

where d is the distance between two lenses.

➙ u' = (-20) + (-5)

➙ u' = -25 cm

Applying lens formula for lens B,

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

$\sf:\implies\:\dfrac{1}{v'}-\dfrac{1}{(-25)}=\dfrac{1}{5}$

$\sf:\implies\:\dfrac{1}{v'}=\dfrac{1}{5}-\dfrac{1}{25}$

$\sf:\implies\:\dfrac{1}{v'}=\dfrac{5-1}{25}$

$\sf:\implies\:v'=\dfrac{25}{4}$

$:\implies\:\underline{\boxed{\bf{\orange{v'=6.25\:cm\:from\:B}}}}$