Astronomical Telescope

What is an Astronomical Telescope?

An astronomical telescope is an optical instrument which is used to see the magnified image of distant heavenly bodies like stars, planets, satellites galaxies etc. An astronomical telescope works on the principle that when an object to be magnified is placed at a large distance from the objective lens of a telescope, a virtual, inverted and magnified image of the object is formed at the least distance of distinct vision from the eye held close to the eyepiece.

An astronomical telescope consists of two convex lenses: an objective lens O and an eyepiece E. The focal length fo of the objective lens of an astronomical telescope is large compared to the focal length fe of the eyepiece. The aperture of objective lens O is large as compared to that of the eyepiece so that it can receive more light from the distant object and form a bright image of the distant object. Both the objective lens and the eyepiece are fitted at the free ends of two sliding tubes, at a suitable distance from each other.

The ray diagram to show the working of the astronomical telescope is shown in the figure. A parallel beam of light from a heavenly body such as stars, planets or satellites falls on the objective lens of the telescope. The objective lens forms a real, inverted and diminished image A’B’ of the heavenly body. This image (A’B’) now acts as an object for the eyepiece E, whose position is adjusted so that the image lies between the focus fe’ and the optical centre $C_2$ of the eyepiece. Now the eyepiece forms a virtual, inverted and highly magnified image of the object at infinity. When the final image of an object is formed at infinity, the telescope is said to be in ‘normal adjustment.

- $f_{\text {objective }}>f_{\text {eyelens }}$ and $d_{\text {objective }}>d_{\text {eye lens }}$
- The intermediate image is real, inverted and small.
- The final image is virtual, inverted and highly magnified.
- Magnification: $m_D=-\frac{f_0}{f_e}\left(1+\frac{f_e}{D}\right)$ and $m_{\infty}=-\frac{f_0}{f_e}$
- Length: $L_D=f_0+u_e$ and $L_{\infty}=f_0+f_e$

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Solved Examples Based on Astronomical Microscope

Example 1: The separation L between the objective (f=0.5cm) and the eyepiece (f=5cm) of a compound microscope is 7cm. The angular magnification produced by this microscope when eye is least strained is:

1) -5

2) -10

3) -15

4) -20

Solution:

$
m=-\frac{v_o}{u_o} \cdot \frac{D}{u_e}
$
wherein
$v_o$ and $u_o$ is the distance from the objective.
$u_e$ Distance from the eyepiece.

Maximum magnification
$
m=-\frac{v_o}{u_0}\left(1+\frac{D}{f_e}\right)
$

Since the eye is least strained hence the final image will form at infinity. In such a case, an image formed by the object should form at the focus of the eyepiece.

$
\begin{aligned}
& v_0=7 \mathrm{~cm}-5 \mathrm{~cm}=2 \mathrm{~cm} \quad f_0=0.5 \mathrm{~cm} \\
& \frac{1}{u}=\frac{1}{v_0}-\frac{1}{f_0}=\frac{1}{2}-\frac{1}{0.5}=\frac{-3}{2} \\
& \text { or } u=\frac{-2}{3} \mathrm{~cm}
\end{aligned}
$

Angular magnification
$
m=\frac{v_0}{u_0} \cdot \frac{D}{f_e}=\frac{-25}{5}=-15 \mathrm{~cm}
$

Hence, the answer is the option (3).

Example 2: In a compound microscope, the focal length of the objective lens is 1.2 cm and the focal length of the eyepiece is 3.0 cm. When an object is kept at 1.25 cm in front of the objective, the final image is formed at infinity. The magnifying power of the compound microscope should be :

1) 200

2) 100

3) 400

4) 150

Solution:

Compound Microscope

$
m=-\frac{v_o}{u_o} \cdot \frac{D}{u_e}
$
wherein
$v_o$ and $u_o$ is distance from the objective.
$u_e$ Distance from the eyepiece.

Maximum magnification
$
m=-\frac{v_o}{u_0}\left(1+\frac{D}{f_e}\right)
$
$
\begin{aligned}
& \quad f_e=3 \mathrm{~cm} \\
& f_0=1.2 \mathrm{~cm} \\
& u_0=1.25 \mathrm{~cm}, v_e=\infty \\
& \Rightarrow \frac{1}{V_0}=\frac{1}{f_0}+\frac{1}{u_0}=\frac{1}{1.2}-\frac{1}{1.25} \\
& V_0=30 \mathrm{~cm} \\
& \qquad \frac{v_0}{u_0} \cdot \frac{D}{f_e}=\frac{30}{1.25} \times \frac{25}{3}=200 \\
& \text { Magnification }
\end{aligned}
$

Hence, the answer is the option (1).

Example 3: The focal length of the objective and the eyepiece of the compound microscope are 2cm and 3cm respectively. The distance between the objective and the eyepiece is 15 cm. The final image formed by the eyepiece is at infinity the distance (in cm) of the object and image produced by the objective, measured from the objective lens respectively.

1) 2.4 and 12

2) 2.4 and 15

3) 2.3 and 3

4) 2.3 and 12

Solution:

Length of the compound microscope

$L= v_{o}+u_{e}$

$v_{o}=$ Image distance from the objective.

$u_{e}=$ Object distance from the eyepiece

$f_{0}$ = 2 cm $f_{e}$ = 3cm

l = 15cm

The final image is formed at infinity hence, the image formed by the objective is at the focal point of the eyepiece

$
\begin{aligned}
u_e & =f_e=3 \mathrm{~cm} \\
v_o & =l-f_e=12 \mathrm{~cm}
\end{aligned}
$

For objective:
$
\begin{array}{ll}
v_o=12 c m & f_o=2 \mathrm{~cm} \\
\frac{1}{v_0}-\frac{1}{u_0}=\frac{1}{f_0} \\
\frac{1}{12}-\frac{1}{u_0}=\frac{1}{2} &
\end{array}
$
or
$
\frac{1}{u_0}=\frac{1}{12}-\frac{1}{2}=\frac{1-6}{12}
$
or,
$
u_0=-2.4 \mathrm{~cm} \quad v_0=12 \mathrm{~cm}
$

Hence, the answer is the option (1).

Example 4: If we used a magnification of 375 from a compound microscope of tube length 150mm and an objective of focal length 5mm, the focal length of the eyepiece should be close to :
1) 12mm

2) 33mm

3) 22mm

4) 2mm

Solution:

The magnification is given by

$ \begin{aligned}
& M=\frac{L}{f_0}\left(1+\frac{D}{f_e}\right) \\
\Rightarrow & 375=\frac{150}{5}\left(1+\frac{25}{f_e}\right) \\
\Rightarrow & f_e=22 \mathrm{~mm}
\end{aligned}$

Hence, the answer is option (3).

Example 5: A compound microscope consists of an objective lens of focal length 1 cm and an eyepiece of focal length 5 cm with a separation of 10 cm . The distance between an object and the objective lens, at which the strain on the eye is minimum is $\frac{n}{40} \mathrm{~cm}$ The value of $n$ is ____.

1) 50

2) 100

3) 150

4) 200

Solution:

Image by objective is formed at the focus of eye-piece
$\therefore$ For objective, $v=5, u, f=1 \mathrm{~cm}$
$
\begin{array}{r}
\frac{1}{5}-\frac{1}{u}=\frac{1}{1} \Rightarrow \frac{1}{5}-1=\frac{1}{u} \\
\therefore \quad|u|=\frac{5}{4} \mathrm{~cm} \Rightarrow|u|=\frac{50}{40} \mathrm{~cm} \\
\therefore \quad n=50
\end{array}
$

Hence, the answer is option (1).

Summary

An astronomical telescope, consisting of two convex lenses—an objective and an eyepiece—enables the magnified viewing of distant celestial objects. The large focal length of the objective lens and its wide aperture allows it to gather ample light, forming a real, inverted image that is further magnified by the eyepiece into a virtual, highly magnified image. The telescope’s magnification and length depend on the focal lengths of its lenses. Solved examples illustrate calculations for determining magnification and focal lengths, enhancing understanding of the telescope’s functionality and applications in astronomy.