Advertisements
Advertisements
प्रश्न
A lady cannot see objects closer than 40 cm from the left eye and closer than 100 cm from the right eye. While on a mountaineering trip, she is lost from her team. She tries to make an astronomical telescope from her reading glasses to look for her teammates. (a) Which glass should she use as the eyepiece? (b) What magnification can she get with relaxed eye?
Advertisements
उत्तर
Given:
For the left glass lens of the lady, we have:
v = – 40 cm and u = – 25 cm
The lens formula is given by
`1/v -1/u =1/f`
Putting the values, we get:
`1/f = 1/((-40)) - 1/((-25)) = 3/200`
`=> f = 200/3 = 66.6` cm
For the right glass lens of the lady, we have:
v = – 100 cm, u = – 25 cm
The lens formula is given by
`1/v -1/u =1/f`
Putting the values, we get:
`1/f = 1/((-100)) - 1/((-25)) =3/100`
`=> f = 100/3 = 33.3 cm `
(a) An astronomical telescope consists of two lenses: the objective lens having a large focal length and the eyepiece lens having a smaller focal length. So, she should use the right lens of focal length 33.3 cm as the eyepiece lens.
(b) With relaxed eye in normal adjustment,
f0 = 66.6 cm and fe = 33.3 cm
Magnification (m) in normal adjustment is given by
`m =f_0/f_e`
∴ m =`(66.6 text/"cm") /(33.3 text/" cm")` = 2
So, with the relaxed eye, she can get the magnification of 2.
APPEARS IN
संबंधित प्रश्न
- For the telescope is in normal adjustment (i.e., when the final image is at infinity)? what is the separation between the objective lens and the eyepiece?
- If this telescope is used to view a 100 m tall tower 3 km away, what is the height of the image of the tower formed by the objective lens?
- What is the height of the final image of the tower if it is formed at 25 cm?
Why should the objective of a telescope have large focal length and large aperture? Justify your answer.
Draw a labelled ray diagram of an image formed by a refracting telescope with the final image formed at infinity. Derive an expression for its magnifying power with the final image at infinity
State the condition under which a large magnification can be achieved in an astronomical telescope.
Define magnifying power of a telescope. Write its expression.
How is the working of a telescope different from that of a microscope?
The focal lengths of the objective and eyepiece of a microscope are 1.25 cm and 5 cm respectively. Find the position of the object relative to the objective in order to obtain an angular magnification of 30 in normal adjustment.
Draw a labeled ray diagram of a reflecting telescope. Mention its two advantages over the refracting telescope.
A Galilean telescope is 27 cm long when focussed to form an image at infinity. If the objective has a focal length of 30 cm, what is the focal length of the eyepiece?
Draw a labelled ray diagram showing the formation of an image by a refracting telescope when the final image lies at infinity.
A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm. Find the magnifying power of the telescope for viewing distant objects when
- the telescope is in normal adjustment,
- the final image is formed at the least distance of distinct vision.
Draw a ray diagram of astronomical telescope for the final image formed at infinity.
Read the following paragraph and answer the questions.
| A number of optical devices and instruments have been designed and developed such as periscope, binoculars, microscopes and telescopes utilising the reflecting and refracting properties of mirrors, lenses and prisms. Most of them are in common use. Our knowledge about the formation of images by the mirrors and lenses is the basic requirement for understanding the working of these devices. |
- Why the image formed at infinity is often considered most suitable for viewing. Explain
- In modern microscopes, multicomponent lenses are used for both the objective and the eyepiece. Why?
- Write two points of difference between a compound microscope and an astronomical telescope
OR
Write two distinct advantages of a reflecting type telescope over a refracting type telescope.
With the help of a ray diagram explain the working of a reflecting telescope.
The magnifying power of an astronomical telescope in normal adjustment is 2.9 and the objective and the eyepiece are separated by a distance of 150 cm. Find the focal lengths of the two lenses.
Read the passage given below and answer the question that follows.
| There are two types of optical instruments: Microscopes and Telescopes. Microscopes are used to magnify very tiny objects whereas telescopes are used to study distant objects. Both of them deploy convex lenses. In his telescope, Newton used a large parabolic mirror to collect light from the stars and reduce aberrations. |
- Rohit observed the launch of Chandrayan 3 with the help of an optical instrument. Name the instrument used by him.
- State any one advantage of a reflecting telescope over a refracting telescope.
- Which instrument is used to study the structure of a virus?
- What is the ability of an optical instrument to form enlarged images called?
-
What is the difference between a compound microscope and an astronomical telescope (refracting type), as far as their lenses are concerned?
Useful Constants & Relations:
| 1 | Charge of a proton | e | 1.6 × 10-19 C |
| 2 | Speed of light in vacuum | c | 3 × 108 ms-1 |
| 1 u = 931 MeV | |||
Assertion: An astronomical telescope has an objective lens having large focal length.
Reason: Magnifying power of an astronomical telescope varies directly with focal length of the objective lens.
A small telescope has an objective of focal length 140 cm and an eye piece of focal length 5.0 cm. The magnifying power of telescope for viewing a distant object?
