Magnifying Glass or Simple Microscope

The image of a distant tree is formed on a screen which is kept at a distance of $20 \mathrm{cm}$ from a convex lens. Find the focal length of the lens.

Explain auxillary plane method to calculate focal length of a convex lens.

In an experiment a convex lens of focal length 15cm is placed coaxially on an optical bench in front of a convex mirror at a distance of 5cm from it. It is found that an object and its image coincide, if the object is placed at a distance of 20cm from the lens. The focal length of the convex mirror is: 

Which of these is not an application of a simple microscope?

Why simple microscope is used in forensic examination?

How does a simple microscope depend on the focal length of the lens used?

What will be the focal length of a convex lens, which produces a virtual image at a distance of $30\mathrm{cm}$ of an object placed $15\mathrm{cm}$ in front of it.

List the uses of a simple microscope.

An object is placed at a distance $u$ from a simple microscope of focal length $f$. The angular magnification obtained depends

In a compound microscope, the objective and eyepiece have focal length of $0.95 \mathrm{cm}$ and $5 \mathrm{cm}$, respectively and are kept at a distance of $20 \mathrm{cm}$. The final image produced by this compound microscope is at a distance of $25 \mathrm{cm}$ from eyepiece. The total magnification is

In $\mathrm{a}$ compound microscope the focal length of objective lens is $1.2 \mathrm{cm}$ and focal length of eye piece is $3.0 \mathrm{cm}.$ When object is kept at $\text{1.25 cm}$
in front of objective, final image is formed at infinity. Magnifying power of the compound microscope should be :

For which of the following colour, the magnifying power of a microscope will be maximum?

Microscope is an optical instrument which

A compound microscope consists of an objective lens with focal length $20 \mathrm{c}\mathrm{m}$ the magnification will be

The objective lens and eye piece lens of a compound microscope have focal lengths 1.5 cm and 5 cm respectively. The object is placed at a distance 1.8 cm. What is its magnification?  ( Take the least distance of distinct vision = 24 cm)

A compound microscope having magnifying power $35$ with its eye-piece of focal length $10 \mathrm{cm}$. Assume that the final image is at least distance of distinct vision then the magnification produced by the objective is

A compound microscope has an eye piece of  focal length 10 cm and an objective of focal length of 4 cm. Calculate the magnification, if an object is kept at a distance of 5 cm from the objective then final image is formed at the least distance of distinct vision

The focal lengths of the objective and eye piece of a compound microscope are $1\mathrm{ }\mathrm{c}\mathrm{m}$ and $5\mathrm{ }\mathrm{c}\mathrm{m}$ respectively. An object is placed at a distance of $1.1\mathrm{ }\mathrm{c}\mathrm{m}$ from objective. If final image is formed at least distance of distinct vision 25 cm, the magnifying power is -

The focal length of the objective and the eye-piece of a microscope are $1 \mathrm{cm}$ and $5 \mathrm{cm}$ respectively. If the magnifying power for the relaxed eye is $45$, then the length of the tube is

In compound microscope, magnifying power is $95$ and the distance of the object from objective lens is $\frac{1}{3.8}$$\mathrm{cm}$. The focal length of the objective lens is $\frac{1}{4}$$\mathrm{cm}$. What is the magnification of eyepiece?