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Q. Find the position of final image from first lens.If the focal length of each lens is $10\, cm$.Physics Question Image

JIPMERJIPMER 2019Ray Optics and Optical Instruments

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

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