In a compound microscope,the objective of focal length ${f_o}$ and the eyepiece of focal length ${f_e}$ are placed at a distance $L$ such that $L$ equals:

$A$ microscope has an objective of focal length $1 \ cm$ and an eye-piece of focal length $6 \ cm$. If the tube length is $30 \ cm$ and the image is formed at the least distance of distinct vision,what is the magnification produced by the microscope? Take $D = 25 \ cm$.

In a compound microscope,the focal length of the objective lens is $1.2 \, cm$ and the focal length of the eyepiece is $3.0 \, cm$. When the object is kept at $1.25 \, cm$ in front of the objective,the final image is formed at infinity. The magnifying power of the compound microscope is:

Magnifying power of a simple microscope is (when final image is formed at $D = 25 \ cm$ from eye)

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

Magnification of a compound microscope is $30$. The focal length of the eyepiece is $5 \, cm$ and the image is formed at the near point (distance of distinct vision) of $25 \, cm$. The magnification of the objective lens is: