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 $\text{cm}$).

The angular magnification of a simple microscope can be increased by increasing

$A$ compound microscope consists of an objective lens of focal length $2.0 \, cm$ and an eyepiece of focal length $6.25 \, cm$ separated by a distance of $15 \, cm$. What is the magnifying power when the final image is formed at the least distance of distinct vision $(25 \, cm)$?

In a compound microscope,the intermediate image is

$A$ compound microscope produces a magnification of $24$. The focal length of the eyepiece is $5 \ cm$. The final image is formed at the least distance of distinct vision. The magnification produced by the objective is

$A$ card sheet divided into squares each of size $1 \, mm^{2}$ is being viewed at a distance of $9 \, cm$ through a magnifying glass (a converging lens of focal length $10 \, cm$) held close to the eye.
$(a)$ What is the magnification produced by the lens? How much is the area of each square in the virtual image?
$(b)$ What is the angular magnification (magnifying power) of the lens?
$(c)$ Is the magnification in $(a)$ equal to the magnifying power in $(b)$? Explain.