Astronomical Telescope

It is a tool that helps us see faraway things in space—like stars, planets, moons, and galaxies. It works by collecting light from these distant objects and making them look bigger and clearer so we can see more details. There are two main types of telescopes: refracting telescopes, which use lenses to bend and focus the light, and reflecting telescopes, which use mirrors to gather and direct the light. Some modern telescopes combine both lenses and mirrors, and those are called catadioptric telescopes.

1.0Definition of Telescopes and its Types

• It is  an optical instrument used to observe distant objects by collecting and magnifying light or other forms of electromagnetic radiation. Telescopes are primarily used in astronomy to view celestial bodies like stars, planets, and galaxies, but they are also used in other fields such as navigation, surveillance, and photography.

2.0Astronomical Telescope Working

Astronomical Telescope is a refracting type telescope used to see heavenly bodies like stars, planets and satellites.

Construction: It consists of two converging lenses mounted coaxially at the outer ends of two sliding tubes.

1. Objective:It’s a convex lens with a large focal length and a wide aperture, designed to face distant objects and form bright images by gathering plenty of light.
2. Eyepiece:It is a convex lens with a small focal length and aperture, facing the eye. The eyepiece aperture is kept small to ensure all telescope light enters the eye for clear vision.

Working:

Final Image at Least Distance of Distinct Vision:

• A parallel light beam from a distant object hits the objective lens, forming a real, inverted, and smaller image (A’B’) at its focal plane. This image is then magnified by the eyepiece, placed so that A’B’ lies within its focal length. The final image (A’’B’’) is magnified, inverted, and seen clearly at the least distance of distinct vision.

Magnifying Power:It's the ratio of the angle formed by the final image at the eye (Near Point) to the angle formed by a distant object at the objective, which is nearly the same as at the eye.

$\begin{array}{rl}& \angle A^{\prime }O{B}^{\prime }=\alpha \\ & \angle A^{\prime \prime }E{B}^{\prime \prime }=\beta \end{array}$

$\begin{array}{rl}& \text{ Magnifying Power }(m)=\frac{\tan \beta }{\tan \alpha }=\frac{\beta }{\alpha }=\frac{A^{\prime }{B}^{\prime }/B^{\prime }E}{A^{\prime }{B}^{\prime }/O{B}^{\prime }}=\frac{O{B}^{\prime }}{B^{\prime }E}\\ & O{B}^{\prime }=+f_o(\text{ Focal length of objective })\\ & B^{\prime }E=-u_e=\text{ distance of }{A}^{\prime }{B}^{\prime }\text{ from the eyepiece acting as an object }f\text{ or it }\\ & m=-\frac{f_o}{u_e}\end{array}$

For Eyepiece

$\begin{array}{rl}& u=-u_e\text{ and }v=-D\\ & \frac{1}{v}-\frac{1}{u}=\frac{1}{f}\\ & \frac{1}{-D}-\frac{1}{-u_e}=\frac{1}{f_e}\\ & \frac{1}{u_e}=\frac{1}{f_e}+\frac{1}{D}=\frac{1}{f_e}\left(1+\frac{f_e}{D}\right)\\ & m=-\frac{f_o}{f_e}\left(1+\frac{f_e}{D}\right)\end{array}$

For large magnifying power $f_o\gg f_e$ The negative sign for magnifying power indicates that the final image formed is real and inverted.

Normal Adjustment – Final Image at Infinity

In normal adjustment, parallel light from a distant object enters the objective lens and forms a real, inverted, and diminished image at its focal plane. The eyepiece is adjusted so this image lies at its focus, resulting in the final image forming at infinity. This final image is inverted and highly magnified.

Magnifying Power (Normal Adjustment)

It’s the ratio of the angle formed by the concluding image at the eye to the angle formed by a distant object at the objective, which is nearly the same as at the eye.

$\begin{array}{rl}& \angle A^{\prime }O{B}^{\prime }=\alpha \\ & \angle A^{\prime }E{B}^{\prime }=\beta \\ & \text{ Magnifying Power }(m)=\frac{\tan \beta }{\tan \alpha }=\frac{\beta }{\alpha }=\frac{A^{\prime }{B}^{\prime }/B^{\prime }E}{A^{\prime }{B}^{\prime }/O{B}^{\prime }}=\frac{O{B}^{\prime }}{B^{\prime }E}\\ & O{B}^{\prime }=+f_o\left(\text{ Distance of }{A}^{\prime }{B}^{\prime }\text{ from objective }\right)\\ & B^{\prime }E=-u_e=\text{ distance of }{A}^{\prime }{B}^{\prime }\text{ from the eyepiece }\\ & m=-\frac{f_o}{f_e}\end{array}$

A negative sign for magnifying power indicates that the final image formed is real and inverted.

3.0Terrestrial Telescope

•  A terrestrial telescope is a type of refracting telescope used to view distant earthly objects in an erect position. To achieve this, it uses an extra convex lens (called the erecting lens) placed between the objective and the eyepiece.
• The objective lens first forms a real, inverted, and diminished image (A'B') of the distant object. The erecting lens, placed at twice its focal length from this image, produces a real, inverted, and same-size image (A"B') that is now erect relative to the original object.
• Finally, the eyepiece is adjusted so that this image lies at its focus, forming a highly magnified, erect image at infinity.
• Since the erecting lens doesn't magnify, the angular magnification is the same as in an astronomical telescope when the concluding image is at infinity.
• When the image is formed at Near Point, $m=\frac{f_o}{f_e}\left(1+\frac{f_e}{D}\right)$

Drawbacks of Terrestrial Telescope

1. It is longer than an astronomical telescope. In normal adjustment, its length is $f_o+4f+f_e$, where f is the focal length of the erecting lens.
2. Image brightness is reduced due to extra reflections from the erecting lens surfaces.

4.0Reflecting Telescope

Newtonian Reflecting Telescope

• The first reflecting telescope, built by Isaac Newton in 1668, uses a large concave mirror (made of a copper-tin alloy) as the objective. It reflects light to a flat mirror, which directs it to the eyepiece, forming a virtual, erect, and magnified image of distant objects.
• Light from a distant star strikes the objective mirror. Before converging at the focusF , a 45° inclined plane mirror reflects the rays sideways to an eyepiece, placed perpendicular to the main axis. The eyepiece forms a highly magnified, virtual, and erect image of the distant object.

5.0Cassegrain Reflecting Telescope

• This telescope uses a large concave paraboloidal mirror with a central hole as the primary mirror and a small convex secondary mirror near its focus. Light from a distant object reflects off the primary mirror, then off the secondary mirror before reaching focus, and finally converges near the hole.
• The eyepiece, placed near this hole, views the final image. Since the image at the primary focus is inverted, and the secondary mirror forms an erect image relative to it, the final image remains inverted with respect to the original object.
• For viewing at the least distance of distinct vision, the eyepiece is adjusted accordingly. Let f_o be the focal length of the objective and $f_e$ that of the eyepiece.
• For the final image formed at the least distance of distinct vision, $m=\frac{f_o}{f_e}\left(1+\frac{f_e}{D}\right)$
• For the final image formed at infinity, $m=\frac{f_o}{f_e}=\frac{R/2}{f_e}$

6.0Advantages of Reflecting Telescopes

7.0Essential Details Telescopes

(1) Astronomical Telescope

(1) Used to see heavenly bodies.

(2) $f_{\text{objective }}>f_{\text{eye lens }}\text{ and }{d}_{\text{objective }}>d_{\text{eye len }}$

(3) The intermediate image is real, inverted and small.

(4) The final image is virtual, inverted and small.

(5)Magnification, $m=-\frac{f_o}{f_e}\left(1+\frac{f_e}{D}\right)\text{ and }{m}_{\infty }=-\frac{f_o}{f_e}$

(6) Length: $L_D=f_o+u_e=f_o+\frac{f_{e}D}{F_o+D}\text{ and }{L}_{\infty }=f_o+f_e$

(2) Terrestrial Telescope

(1) Used to see far off objects on the earth.

(2) It consists of three converging lenses : objective, eye lens and erecting lens.

(3) Its final image is virtual erect and smaller.

(4)Magnification: $m_D=\frac{f_o}{f_e}\left(1+\frac{f_e}{D}\right)\text{ and }{m}_{\infty }=\frac{f_o}{f_e}$

(5) Length: $L_D=f_o+4f+u_e=f_o+4f+\frac{f_{e}D}{F_e+D}\text{ and }{L}_{\infty }=f_o+4f+f_e$

(3) Galilean Telescope

(1) It is also a terrestrial telescope but of much smaller field of view.

(2) The Objective is a converging lens while the eye lens is a diverging lens.

(3)Magnification: $m_D=\frac{f_o}{f_e}\left(1-\frac{f_e}{D}\right)\text{ and }{m}_{\infty }=\frac{f_o}{f_e}$

(4) Length: $L_D=f_o-u_e\text{ and }{L}_{\infty }=f_o-f_e$

Note:

The least distance (d) between two objects that a telescope can just resolve is

$d=\frac{r}{R.P}\text{ where }r=\text{ space of objects from telescope. }$

(4) Binocular

When two telescopes are mounted side-by-side to be viewed by both eyes simultaneously, the setup is called a binocular. Binoculars use reflecting prisms to shorten the tube length and produce bright, upright images without lateral inversion. Viewing through binoculars gives two slightly different angles of the same object, creating a combined 3D image with depth perception.          

Few important points

(1) Since the magnifying power is negative, astronomical telescope images are inverted—left becomes right and upside down. However, this doesn’t impact observations because most celestial objects are symmetrical.

(2) If the objective and eyepiece lenses are swapped, the telescope won’t work as a microscope, and the object will appear very small.

(3) In a telescope, if field and eye lenses are interchanged magnification will change from $\left(\frac{f_o}{f_e}\right)\text{ to }\left(\frac{f_e}{f_o}\right)$ i.e., it will change from  m to 1/m i.e., will become $\left(\frac{1}{m^2}\right)$ times of its initial value.

(4) As magnification for normal setting as $\left(\frac{f_o}{f_e}\right)$ so to have large magnification,fomust be as large as practically possible and $f_e$ small.This is why in a telescope, the object is of large focal length while the eye piece is small.

(5)In a telescope, the field lens aperture is made as large as possible to boost resolving power, which is proportional to aperture size (D). A larger objective aperture also gathers more light, brightening the image, but it can increase aberrations, especially spherical.telescope $\propto \left(\frac{D}{\lambda }\right)$

(6) For a telescope with an increase in length of the tube, magnification decreases.

(7) When both the object and final image in a telescope are at infinity, then: $m=\frac{f_o}{f_e}=\frac{D}{d}$

(8)  If we are given four convex lenses having focal lengths $f_1>f_2>f_3>f_4$.For making a good telescope and microscope. We choose the following lenses respectively.

Telescope $f_1(o),f_4(e)$

Microscope $f_4(0),f_3(e)$

(9)If a parrot sits on a telescope’s objective, it won’t be visible when viewing distant objects like the moon, but it will slightly reduce image brightness by blocking some light and lowering the aperture.

8.0Solved Examples On Astronomical Telescope

Q-1.An astronomical telescope has an objective lens focal length of 50 cm and an eyepiece focal length of 5 cm. What is its total length when the final image forms at a near point.

Solution:

By using $L_D=f_o+u_e=f_o+\frac{f_{e}D}{F_o+D}=50+\frac{5\times 25}{(5+25)}=\frac{325}{6}\mathrm{cm}$

Q-2.An object subtends an angle of 2° at the eye when viewed directly. If it is observed through an astronomical telescope with an objective focal length of 60 cm and an eyepiece focal length of 5 cm, what angle does the image subtend at the eye through the telescope?

Solution:

By using $\frac{\beta }{\alpha }=\frac{f_o}{f_e}\Rightarrow \frac{\beta }{20}=\frac{60}{5}\Rightarrow \beta =24^{\circ }$

Q-3.An astronomical telescope has a magnifying power of 8, and the distance between the objective and the eyepiece is 54 cm. What are the focal lengths of the objective lens and the eyepiece, respectively?

Solution:

Given that $m_{\infty }=8and{L}_{\infty }=54$

By using $\left|m_{\infty }\right|=\frac{f_o}{f_e}\text{ and }{L}_{\infty }=f_o+f_e$

We get $f_o=6\mathrm{cm}\text{ and }{f}_e=48\mathrm{cm}$

Q-4.The diameter of the Moon is $3.5\times 10^3\mathrm{km}$, and its distance from Earth is $3.8\times 10^5\mathrm{km}$. If the Moon is observed through a telescope with an objective focal length of 4 m and an eyepiece focal length of 10 cm, what is the approximate angle subtended by the Moon at the eye?

Solution:

The angle formed by the moon at the telescope’s objective lens.

$\begin{array}{rl}& \alpha =\frac{3.5\times 10^3}{3.8\times 10^5}=\frac{3.5}{3.8}\times 10^{-2}\mathrm{rad}\\ & m=\frac{f_o}{f_e}=\frac{\beta }{\alpha }\Rightarrow \frac{400}{10}=\frac{\beta }{\alpha }\Rightarrow \beta =40\alpha \\ & \beta =40\times \frac{3.5\times 10^3}{3.8\times 10^5}\times \frac{180}{\pi }=20^{\circ }\end{array}$

Q-5.A telescope with a 10 cm objective lens observes two objects 1 km away using 5000 Å light. What is the closest separation between the objects that it can resolve?

Solution:

Suppose minimum distance between objects is x and their distance from telescope is r

So, resolving limit $d\theta =\frac{1.22\lambda }{a}=\frac{x}{r}$

$\Rightarrow x=\frac{1.22\lambda \times r}{a}=\frac{1.22\times \left(5000\times 10^{-10}\right)\times \left(1\times 10^3\right)}{(0-1)}=6.1\times 10^{-3}\mathrm{m}=6.1\mathrm{mm}$

Hence its order is ≈ 5 mm