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

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?

A jeweller examines a $1.0 \mathrm{cm}$ diamond through a small magnifying glass (called a loupe) having an angular magnification of $3.5$. The near point of her eye is $25.0 \mathrm{cm}$ and she wants the image to be located at the near point. The focal length of the loupe should be

A person uses a handheld "magnifying glass" of focal length $f=5 \mathrm{cm}$ to observe some very small print in a book. What is the typical angular magnification he can hope to get if the image is formed at infinity ?

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 student studying the similarities and differences between a camera and the human eye makes the following observations:
$\left(\mathrm{I}\right)$ Both the eye and the camera have convex lenses.
$\left(\mathrm{II}\right)$In order to focus, the eye lens expands or contracts while the camera lens moves forward or backward.
$\left(\mathrm{III}\right)$ The camera lens produces upside down real images while the eye lens produces only upright real images.
$\left(\mathrm{IV}\right)$ A screen in camera is equivalent to the retina in the eyes.
$\left(\mathrm{V}\right)$ A camera adjusts the amount of light entering in it by adjusting the aperture of the lens. In the eye the cornea controls the amount of light. The correct statements are :

Total magnification of compound microscope is $20$ while magnification for eye piece is $5$. If final image is formed at near point which is taken $20 \mathrm{cm}$ then focal length of eye piece is

The focal length of an objective of a telescope is 3 meter and diameter 15 cm. Assuming for a normal eye, the diameter of the pupil is 3 mm for its complete use, the focal length of eye piece must 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

Image formed on retina of eye is proportional to:

Electron microscope is better than optical microscope because of:

Circular part in the centre of retina is called

A astronomical telescope has objective and eyepiece of focal lengths $40 \mathrm{cm}$ and $4 \mathrm{cm}$ respectively. To view an object $200 \mathrm{cm}$ away from the objective, the lenses must be separated by a distance :