Question

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

• A. 6
• B. 7.75
• C. 9.25
• D. 11

Answer 1

Answer: (D)

For, objective,
$u=-2\, cm ,\, f=1.5\, m$
By $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$, we have,
$\frac{1}{v}=\frac{1}{1.5}-\frac{1}{2} v=6\, cm$
For eyepiece,
$v=25\, cm ,\, f=6.25\, cm$
Using $\frac{1}{v}-\frac{1}{u}=\frac{1}{f^{\prime}}$, we get
$\frac{1}{u}=\frac{1}{-25}-\frac{1}{6.25}$
$u=-5\, cm$

So, distance between two lenses $= 11$ cm.