Physics Topics are also essential for space exploration, allowing scientists to study phenomena such as gravitational waves and cosmic rays.
What are the Types of Astronomical Telescopes? What are the Advantages and Disadvantages of Refracting Telescopes?
Objects situated at large distances are not visible distinctly with the naked eye as these objects subtend very small visual angles at our eyes. The telescope makes it possible by forming an image that subtends a greater angle as compared to the original object. The Image formed is a virtual image.
Telescopes are mainly of two types:
1. Astronomical telescope: It is used to see astronomical objects such as planets, stars etc. Here the final image is virtual and inverted relative to the object. Sometimes the astronomical telescope is also called simply as telescope.
2. Terrestrial telescope: It is used to see distant objects situated on’ the surface or near the surface of the earth. Here the final image is virtual and erect relative to the object.
From the constructional point of view there is not much difference between an astronomical telescope and a terrestrial telescope. Terrestrial telescope has a provision of erecting the final image, which astronomical telescope does not have.
According to construction, astronomical telescopes are of two types:
- Refracting telescope: In this type of telescope, the objective is made of a single lens or a combination of lenses.
- Reflecting telescope: In this type of telescope, a large concave or paraboloidal mirror is used as the objective.
Now-a-days many different types of telescopes are in use, such as—radio telescope, infrared telescope, X-ray telescope, high-energy particle telescope, gravitational wave telescope etc.
Refracting Astronomical Telescope
In this instrument two convex lenses are mounted in a tube so as to have a common axis [Fig.]. Here, the objective O has a
large focal length and a large aperture. Comparatively, the eyepiece E has a very small focal length and a very small aperture. The aperture of the eyepiece is taken almost equal to that of the pupil of the eye so that all the refracted rays from the eyepiece enter the eye. The distance between the objective and the eyepiece in the tube can be adjusted by a screw.
Working principle: Fig. shows that the rays from a distant object are incident on the objective. As the object is situated at a large distance the rays from it may be taken as parallel. So a real, inverted and very diminished Image (FP) compared to the object is formed at the focal plane of the objective.
This image acts as an object for the eyepiece which then forms the final image. Rays from the Image FP diverge and fall on the eyepiece. The eyepiece is so placed that F becomes its first principal focus. Hence, the rays after refraction emerge as parallel rays. To adjust the eyepiece In the proper position is called focus-sing. After focussing, one can easily observe a virtual and largely magnified image through the eyepiece. Since, the first image is inverted, the virtual image is also inverted with respect to the object. This type of focussing of the telescope is called focussing for Infinity and this adjustment is called normal adjustment.
Again, the astronomical telescope may be used in such a way that the final virtual image is formed at the near point of the eye i.e., at the least distance of distinct vision (Fig.). For this, the eye piece is moved a little distance towards the objective so that the image PQ1 formed by the objective comes within the focal length of the eyepiece. Then the eyepiece forms a magnified virtual image pq at the near point of the eye. This image pq is erect with respect to PQ1 but inverted with respect to the original object. This type of focussing is called focussing for distinct vision.
Magnification: Magnification of an astronomical or terrestrial telescope is defined as the ratio of the angle subtended by the final image at the eye to the angle subtended by the object. As the object is situated at a large distance the angle subtended by the object at the naked eye is taken to be equal to the angle subtended at the objective. As the eye is placed very close to the eyepiece the angle subtended by the final image at the eye and the angle subtended at the eyepiece are virtually equal.
In case of Focussing for infinity: According to the Fig., angular magnification,
m \(=\frac{\text { angle subtended at the eye by the final image }}{\text { angle subtended at the eye by the object }}\)
= \(\frac{\beta}{\alpha}\) = \(\frac{\angle F E P}{\angle F O P}\) = \(\frac{\tan \angle F E P}{\tan \angle F O P}\) [If θ is small then tanθ ≈ θ]
∴ m = \(\frac{\frac{F P}{E F}}{\frac{F P}{O F}}\) = \(\frac{O F}{E F}\) = \(\frac{f_o}{f_e}\) ….. (1)
[fo = focal length of the objective and fe = focal length of the eyepiece]
So it is seen that for large magnification, the focal length of the objective should be large and that of the eyepiece should be small.
Again, with the help of similar triangles it can be proved,
\(\frac{f_o}{f_e}\) \(=\frac{\text { diameter of the objective }}{\text { diameter of the eye-piece }}\)
∴ m \(=\frac{\text { diameter of the objective }}{\text { diameter of the eve-piece }}\) …. (2)
In case of focussing for distinct vision: According to the Fig., angular magnification,
m = \(\frac{\beta}{\alpha}\) = \(\frac{\tan \beta}{\tan \alpha}\) = \(\frac{\frac{P Q_1}{E Q_1}}{\frac{P Q_1}{O Q_1}}\) = \(\frac{O Q_1}{E Q_1}\) = \(\frac{f_o}{E Q_1}\) …. (3)
[OQ1 = fo as the objective forms the image PQ1 of the distant object at its focal plane.]
Now, PQ1 is the object for refraction In the eyepiece and pq is its image.
∴ Object distance, u = EQ1
As the image pq is formed at the least distance of distinct vision,
Image distance, v = Eq = D
Therefore, according to the equation of lens,
\(\frac{1}{E q}\) – \(\frac{1}{E Q_1}\) = –\(\frac{1}{f_e}\) or, \(\frac{1}{E Q_1}\) = \(\frac{1}{E q}\) + \(\frac{1}{f_e}\) = \(\frac{1}{D}\) + \(\frac{1}{f_e}\)
We get from the equation (3),
m = f0(\(\frac{1}{D}\) + \(\frac{1}{f_e}\)) = \(\frac{f_o}{f_e}\)(\(\frac{f_e}{D}\) + 1) …… (4)
In case of focussing for infinity as the final image is formed at infinity, so Eq = D → ∞.
So, from equation (4) we get,
m = \(\frac{f_o}{f_e}\), which supports equation (1)
Hence, from equation (1) and (4) it cari be observed that in case of focussing for distinct vision, magnification of telescope increases in the ratio (\(\frac{f_e}{D}\) + 1) : 1. If the magnification is very large, the brightness of the image decreases. To increase the brightness of the image necessary arrangement has to be made for allowing light from the object to enter the instrument. So, the aperture of the objective of this type of telescope is made as large as possible I.e., the objective is generally of a large diameter. Moreover if the aperture of the objective is large, it is possible to examine the object minutely i.e., the resolving power of the instrument increases,
To remove or reduce the defects of all sorts of abberation, instead of a single lens for each of the objective and the eye-piece a combination of lenses should be used.
Length of the tube of the astronomical telescope: By the term length of the tube’ of an astronomical telescope we mean the distance between the objective and the eyepiece. If the length of the tube be L, then in case of focussing for infinity,
L = fo + fe …. (5)
in case of focussing for distinct vision,
L = OQ1 + EQ1 = f0 + \(\frac{1}{\frac{1}{D}+\frac{1}{f_e}}\) = f0 + \(\frac{D f_e}{D+f_e}\) …… (6)
Limitations of refracting astronomical telescope:
i) As only lens is used in refracting astronomical telescope, so this instrument has defects due to chromatic abberation. For this the final image becomes hazy. It is a disadvantage to scrutinise the object in view. It is to be noted that, no lens can converge the rays of all colous in a single point. For this reason, no image formed by lens becomes distinct, i.e., during the formation of image there remains some error all the time. This error is called chromatic aberration.
ii) To get bright images of distant objects like stars, planets etc., the aperture of objective lens is to be made large. But it is extremely difficult and expensive to make an objective lens of very large aperture.
iii) Lens is transparent. That is why, fixing of such a large lens (as on objective) rigidly in a right place is very difficult.
Reflecting Astronomical Telescope
In reflecting astronomical telescope (also called a reflector) concave mirror is used in place of objective lens. It was discovered in 17th century. It is free from chromatic aberration. Besides, it is easier to construct an objective of larger aperture and its construction cost is also less. Mirrors being not transparent, it is possible to fix a large mirror, as an objective, rigidly in the right place. For this reason, in astronomical studies, most of the telescopes used are of reflecting type. In this regard it is to be noted that reflecting astronomical telescope is one of the most important instruments used in scientific research.
Reflecting astronomical telescopes are of different types, namely—
- Gregorian telescope,
- Newtonian telescope,
- Cassegrain telescope.
The designs of modern day telescopes are mainly based on these three types.
Gregorian telescope: Scottish astrophysicist and mathematician James Gregory had published the design of this tele-scope first in his book in 1663. Based on this, British physicist Robert Hook made the telescope on 1673.
Construction: Gregorian telescope consists of two concave mirrors, out of which the primary mirror is parabolic and the secondary one is elliptical in nature. Parallel rays from very long distant object fall on the primary mirror and getting reflected from it meet at its focus [Fig.].
The secondary mirror is placed beyond the focus of the primary mirror in a such a manner that, rays after meeting at focus incident on secondary mirror as divergent rays. Then the rays get reflected from it (the secondary mirror) and finally emerge out through a small hole made in the primary mirror. Thus one can view a magnified erect image through the eyepiece placed at the back of the primary mirror.
Advantage: At the focus of the primary mirror a small but distinct real image of the object under observation is formed. As a result, the final virtual image becomes much bright and distinct.
Newtonian telescope:
Newton made this telescope in 1668. Ii was the first usable Gregorian reflector.
Construction: In this telescope, objective M1 is a concave mirror [Fig.] of large focal length and big aperture.
This mirror is placed at one end of a tube of large diameter and other end of the tube is focused towards distant object to watch. A plane mirror M2 is placed between concave lens and its focus. It is inclined at an angle 45° to the axis of mirror M1. The eyepiece E consisting of a convex lens of small focal length and small aperture is placed in an adjacent tube.
Formation of image: The parallel rays coming from a distant object AB are incident on the mirror M1. After reflection by M1, the reflected converging rays are reflected again by plane mirror M2 and a small real image A‘B’ is formed between M2 and E. This image A’B’ acts as an object for eyepiece E. A magnified virtual image A”B” of A‘B’ is formed at the least distance of distinct vision or at infinity. If A‘B’ is on focus plane of eyepiece E, the final image is formed at infinity.
Magnification: Magnification of telescope,
m \(=\frac{\text { angle subtended in eye by the final image }}{\text { angle subtended in eye by the object }}\)
If the final image is formed at infinity, then it can be shown that,
m = –\(\frac{f_o}{f_e}\)
Here, f0 focal length of concave mirror M1 which is used as an objective; fe = focal length of eyepiece
Now, f0 = \(\frac{R}{2}\); R = radius of curvature of the objective.
Therefore, magnification of reflecting telescope will be increased by increasing the radius of curvature of concave mirror, used as objective.
It is to be noted that, distorted images are formed due to rays emerged from the edge of the lens or reflected from the edge of the mirror. Hence the images become erroneous. Such errors are called spherical aberration.
To avoid such spherical aberration, paraboloid mirror is used in modern reflecting telescope.
Advantages:
- The image formed by reflecting telescope is brighter than the image formed by refracting telescope. This is because, in reflecting telescope, no loss of light takes place by reflection and absorption at lens surfaces.
- In reflecting telescope, the use of a parabolic mirror removes spherical aberration and chromatic aberration of the image.
- The concave mirror used in reflecting telescope is less expensive than the lens used in a refracting telescope.
Disadvantages:
- Alignment of different parts is required every time it is used.
- There is a central obstruction due to the secondary mirror in the light path, causing the light to be scattered in all directions. Hence, contrast of the image decreases.
- At the time of observation of star through reflecting telescope, If the eye of the observer is positioned near the edge of the field of view, then stars look like comets. This type of defect is known as coma.
Cassegrain reflecting telescope:
Laurent Cassegrain had published the design of this telescope in 1672.
Construction: In Fig., a Cassegrain reflecting telescope is shown.
In this telescope, the objective is a large parabolic mirror O with a hole in its centre. The aperture of this mirror is large. A small convex hyperbolic mirror M is placed in front of the objective. A convex lens E acts as an eyepiece is placed back of the objective. In a small size telescope, the focal length of the eyepiece is large.
Formation of image: The parallel rays coming from a distant object are incident on the objective O and after reflection from the objective, the rays are Incident on the convex mirror M. In absence of convex mirror M, the reflected rays from the objective would meet at its principle focus Fo. Hence, in presence of convex mirror M, the rays coming from the objective are reflected again and a real and inverted image is formed at the point F. This inverted image is seen through the eyepiece E.
In the world, presently many observatories are in function, among which a very notable one is— Roque de los Muchachos observatory of La palma, Spain. Its ‘Gran Telescopio Canarias’ is one of the most developed telescope of the world. Presently it is the largest reflecting telescope with unit aperture in the world. Its effective aperture is 10.4 in and focal length 16.5 m. It is situated at the top of a volcano at the height of 2267 m from sea level.
India can boast of having an observatory at highest altitude. It is located at the height of 4572 m from sea level at Hanle, Ladakh. Only In this part of the world, sky remains clear almost 250 days in night.
Numerical Examples
Example 1.
The focal lengths of the eyepiece and the objective are 10 cm and 200 cm respectively. If someone wants to observe moon with naked eye through this telescope then what should be the distance between the objective and the eyepiece?
Solution:
If someone wants to observe moon with naked eyes then the telescope should be focussed at infinity
For focussing at infinity, length of the tube is
L = fo + fe = 200 + 10 = 210 cm
Hence, distance between the two lenses is 210 cm.
Example 2.
The length of the tube of an astronomical telescope is 44 cm and its angular magnification is 10. What is the focal length of its objective?
Solution:
Magnification,
m = \(\frac{f_o}{f_e}\) or, fe = \(\frac{f_o}{m}\)
Length of the tube,
L = f0 + fe = f0 + \(\frac{f_o}{m}\) = f0(1 + \(\frac{1}{m}\)) = f0(\(\frac{m+1}{m}\))
or f0 = L \(\frac{m}{m+1}\) = 44 × \(\frac{10}{10+1}\) = 40 cm
Example 3.
A small astronomical telescope has an objective of focal length 50 cm and an eyepiece of focal length 5 cm. It is focussed at the sun and the final image is formed at a distance of 25 cm from the eyepiece. If the diameter of the sun subtends an angle of at the centre of the objective calculate the angular magnification of the instrument and the actual size of the image.
Solution:
Here, f0 = 50 cm, fe = 5 cm and D = 25 cm.
Angular magnification of the instrument,
m = \(\frac{f_o}{f_e}\)(1 + \(\frac{f_e}{D}\)) = \(\frac{50}{5}\)(1 + \(\frac{5}{25}\)) = 10 × \(\frac{6}{5}\) = 12
If the angle subtended at the eyepiece by the final image be β and the angle subtended at the centre of the objective by the sun be α, then
m = \(\frac{\beta}{\alpha}\)
or, β = mα = (12 × \(\frac{32}{60}\))° = \(\frac{32}{5}\) × \(\frac{\pi}{180}\)rad
If the actual size of the image be I, then
β = \(\frac{I}{D}\)
or, I = βD = \(\frac{32}{5}\) × \(\frac{\pi}{180}\) × 25 = \(\frac{8}{9}\)π = 2.793 cm
Example 4.
If the focal lengths of the objective and the eyepiece of an astronomical telescope be 100 cm and 20 cm respectively, calculate the angular magnification of the instrument. If a house of height 60 m situated at a distance of 1 km be observed by the instrument determine the height of the image formed by the objective.
Solution:
Angular magnification of the telescope,
m = \(\frac{f_o}{f_e}\) = \(\frac{100}{20}\) = 5
For the objective,
f0 = 100 cm, u = -1 km = -105 cm
According to the equation of lens we have,
\(\frac{1}{v}\) + \(\frac{1}{10^5}\) = \(\frac{1}{100}\) or, \(\frac{1}{v}\) = –\(\frac{1}{10^5}\) + \(\frac{1}{100}\)
or v = 100.1 cm
Magnification = \(\frac{v}{u}\) = \(\frac{100.1}{10^5}\) = \(\frac{1}{10^3}\) \(=\frac{\text { size of the image }}{\text { size of the object }}\)
or, \(\frac{1}{10^3}\) \(=\frac{\text { size of the image }}{60 \times 100}\)
∴ size of the image = \(\frac{6000}{1000}\) = 6 cm
Example 5.
Distance between the earth and the moon is 386242.56 km and diameter of the moon is 3218.69 km. If focal length of the objective of a telescope is 0.018288 km then what is the diameter of real Image of the moon formed by the objective?
Solution:
Moon is at a large distance from earth so image will be formed at focal plane of objective.
Hence, diameter of real image of the moon is 1.52 × 10-4 km.
Example 6.
The focal lengths of the objective and the eyepiece of a compound microscope are 2 m and 5 cm respectively. The distance between the two lenses is 20 cm. Final image is formed at 25 cm distance from the objective calculate the distance between eyepiece and final image. [HS’11]
Solution:
Here f0 = 2 m, fe = 5 cm
Distance between two lens (L) = 20 cm
Distance between final image and objective (u) = 25 cm
∴ Distance between final image and eyepiece is,
D = l + L = 25 + 20 = 45 cm