Question

In a compound microscope, the focal lengths of two lenses are $1.5 \, cm$ and $6.25 \, cm$ and an object is placed at $2 \, cm$ from objective and the final image is formed at $25 \, cm$ from eye lens. The distance between the two lenses is

• A. $6.00 \, cm$
• B. $7.75 \, cm$
• C. $9.25 \, cm$
• D. $11.00 \, cm$

Answer 1

Answer: (D)

$f_{0}=1.5 \, cm, \, f_{e}=6.25 \, cm, \, u_{0}=-2 \, cm, \, v_{e}=-D=-25 \, cm$
By objective lens $\frac{1}{f_{0}}=\frac{1}{v_{0}}-\frac{1}{u_{0}}$
$\frac{1}{1.5}=\frac{1}{v_{0}}-\frac{1}{- 2}$
$\Rightarrow $ $\frac{1}{v_{0}}=\frac{1}{1.5}-\frac{1}{2}$ or $v_{0}=6 \, cm$
By eyepiece $\frac{1}{f_{e}}=\frac{1}{v_{e}}-\frac{1}{u_{e}}$
$\frac{1}{6.25}=\frac{1}{- 25}-\frac{1}{- u_{e}}$
$\Rightarrow $ $\frac{1}{u_{e}}=\frac{1}{6.25}+\frac{1}{25}=\frac{4}{25}+\frac{1}{25}=\frac{1}{5}$
$u_{e}=5 \, cm$
Length of tube $=L=v_{0}+u_{e}=6.0cm +5.0cm$
$L=11 \, cm$