Optical Instruments

An astronomical telescope uses two lenses of powers 10 D and 1 D.Its magnifying power in normal adjustment would be:

The expression for magnifying power of compound microscope is

The expression for astronomical telescope magnifying power is

The focal lengths of the objective and the eye piece of a compound microscope are $2.0 \mathrm{cm}$ and $3.0 \mathrm{cm},$ respectively. The distance between the objective and the eye piece is $15.0 \mathrm{cm}.$ The final image formed by the eye piece is at infinity. The two lenses are thin. The distance in $\mathrm{cm}$ of the object and the image produced by the objective, measured from the objective lens, are respectively

A convex lens $\mathrm{A}$ of focal length $20 \mathrm{cm}$ and a concave lens $\mathrm{B}$ of focal length $5 \mathrm{cm}$ are kept along the same axis with a distance $\mathrm{d}$ between them. If a parallel beam of light falling on $\mathrm{A}$ leaves $\mathrm{B}$ as a parallel beam, then $\mathrm{d}$ is equal to _________ $\mathrm{cm}$.

In a reflecting telescope, a secondary mirror is used to:

A microscope has an objective of focal length $1.5 \mathrm{cm}$ and an eye-piece of focal length $2.5 \mathrm{cm}$. If the distance between objective and eye-piece is $25 \mathrm{cm}$ is the approximate value of magnification produced for relaxed eye?

Tube length of a compound microscope is equal to $30 \mathrm{cm}$. The focal length of objective and eyepiece is equal to $2 \mathrm{cm}$ and $4 \mathrm{cm}$ respectively. The magnification produced by the compound microscope when the final image is produced at infinity, is approximately
 

A normal human eye can see an object making an angle of $1.8°$ at the eye. What is the approximate height of object which can be seen by an eye placed at a distance of $1 \mathrm{m}$ from the object?

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We can see that in normal eye the things taking part in refraction are the cornea, aqueous humour, crystalline eye lens and the vitreous humour. Let us assume that we consider only $2$ things namely the cornea and the eye lens. The power of the cornea is fixed and equal to $40 D$. The ciliary muscles can change the power of the eye lens which varies between $10 D$ to $20 D$. Calculate the least distance of distinct vision for such an eye.

Which of the following is correct regarding microscope and telescope?

In an astronomical telescope, $f_0=20 \mathrm{cm}$ and $f_e=5 \mathrm{cm}$, then find out the maximum magnification of a distant object. Take least distance of clear vision, $D=25 \mathrm{cm}$.

What is the nature of the intermediate image in a compound microscope?

A certain far-sighted person cannot see objects closer to the eye than$100 \mathrm{cm}$. The power of lens which will enable him to read at a distance of $25 \mathrm{cm}$ will be 

An astronomical telescope has a converging eyepiece of focal length $5 \mathrm{cm}$ and an objective of focal length $80 \mathrm{cm}$ . When the final image is formed at the least distance of distinct vision$\left(25 \mathrm{cm}\right)$, the separation between the two lenses is

An observer looks at a tree of height $15 \mathrm{m}$ with a telescope of magnifying power $10$. To him the tree appears

A presbyopic patient can see objects only when they lie between $50 \mathrm{cm}$ and $200 \mathrm{cm}$ from the eye. Then the focal length of the correcting lens required while reading is 

A myopic patient uses