In This post you will get answer of Q.
Find the position of final image from first lens.If the focal length of each lens is 10 cm.
Solution:
Refraction from lens A,
From lens formula,
$frac{1}{V_{A}}-frac{1}{u_{A}}=frac{1}{f_{A}}$
$frac{1}{V_{A}}=frac{1}{f_{A}}+frac{1}{u_{A}}$
$frac{1}{V_{A}}=frac{1}{10}-frac{1}{40}$
$V_{A}=frac{40}{3}$ cm
Refraction from lens B,
vA is object for lens B,
$u_{B}=30-frac{40}{3}=frac{50}{3}$ cm
So, $frac{1}{v_{B}}-frac{1}{u_{B}}=frac{1}{f_{B}}$
$frac{1}{v_{B}}+frac{3}{50}=frac{1}{10}$
$frac{1}{v_{B}}=frac{1}{10}-frac{3}{50}$
$frac{1}{v_{B}}=frac{2}{50}$
$v_{B}=25$ cm
Refraction from lens C,
vB is object for lens C,
uC = 30 – 25 = 5 cm
$frac{1}{v_{C}}=frac{1}{f_{C}}+frac{1}{u_{C}}Rightarrow frac{1}{v_{C}}=frac{1}{10}-frac{1}{5}Rightarrow frac{1}{v_{C}}=frac{-1}{10}$
$v_{c}=-10$ cm
Final image distance from lens C is 10 cm towards lens B. So the final image distance from lens A is,
= 20 + 30 = 50 cm

Solution:
Refraction from lens A,
From lens formula,
$frac{1}{V_{A}}-frac{1}{u_{A}}=frac{1}{f_{A}}$
$frac{1}{V_{A}}=frac{1}{f_{A}}+frac{1}{u_{A}}$
$frac{1}{V_{A}}=frac{1}{10}-frac{1}{40}$
$V_{A}=frac{40}{3}$ cm
Refraction from lens B,
vA is object for lens B,
$u_{B}=30-frac{40}{3}=frac{50}{3}$ cm
So, $frac{1}{v_{B}}-frac{1}{u_{B}}=frac{1}{f_{B}}$
$frac{1}{v_{B}}+frac{3}{50}=frac{1}{10}$
$frac{1}{v_{B}}=frac{1}{10}-frac{3}{50}$
$frac{1}{v_{B}}=frac{2}{50}$
$v_{B}=25$ cm
Refraction from lens C,
vB is object for lens C,
uC = 30 – 25 = 5 cm
$frac{1}{v_{C}}=frac{1}{f_{C}}+frac{1}{u_{C}}Rightarrow frac{1}{v_{C}}=frac{1}{10}-frac{1}{5}Rightarrow frac{1}{v_{C}}=frac{-1}{10}$
$v_{c}=-10$ cm
Final image distance from lens C is 10 cm towards lens B. So the final image distance from lens A is,
= 20 + 30 = 50 cm