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Question

Physics Question on Ray optics and optical instruments

Find the position of final image from first lens.If the focal length of each lens is 10 cm.

A

40 cm

B

50 cm

C

45 cm

D

55 cm

Answer

50 cm

Explanation

Solution

Refraction from lens A,
From lens formula,
1VA1uA=1fA\frac{1}{V_{A}}-\frac{1}{u_{A}}=\frac{1}{f_{A}}
1VA=1fA+1uA\frac{1}{V_{A}}=\frac{1}{f_{A}}+\frac{1}{u_{A}}
1VA=110140\frac{1}{V_{A}}=\frac{1}{10}-\frac{1}{40}
VA=403V_{A}=\frac{40}{3} cm
Refraction from lens B,
v is object for lens B,
uB=30403=503u_{B}=30-\frac{40}{3}=\frac{50}{3} cm
So, 1vB1uB=1fB\frac{1}{v_{B}}-\frac{1}{u_{B}}=\frac{1}{f_{B}}
1vB+350=110\frac{1}{v_{B}}+\frac{3}{50}=\frac{1}{10}
1vB=110350\frac{1}{v_{B}}=\frac{1}{10}-\frac{3}{50}
1vB=250\frac{1}{v_{B}}=\frac{2}{50}
vB=25v_{B}=25 cm
Refraction from lens C,
v is object for lens C,
u = 30 - 25 = 5 cm
1vC=1fC+1uC1vC=110151vC=110\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}
vc=10v_{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