Terrestrial Telescope

A terrestrial telescope is a vital optical instrument designed for observing objects on Earth at great distances, providing an upright and magnified view. Unlike astronomical telescopes, which invert images, terrestrial telescopes incorporate an additional lens or prism system to ensure the final image is right-side up. These telescopes are commonly used in activities such as bird watching, nature study, surveillance, and even sports viewing, offering a closer look at distant scenes with clear detail. In everyday life, they enhance our ability to explore and appreciate the natural world, monitor wildlife, and enjoy scenic landscapes from afar. By bringing distant objects into sharp focus, terrestrial telescopes bridge the gap between our immediate surroundings and the broader environment, enriching our visual experience and understanding of the world around us. In this article, we will discuss the concept of a Terrestrial telescope and provide examples for better understanding.

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This Story also Contains

1. What is a Terrestrial Telescope?
2. Solved Examples Based on Terrestrial Telescope
3. Summary

What is a Terrestrial Telescope?

A refracting telescope has inverting lenses or an eyepiece that presents an erect image. A telescope for use on earth rather than for making astronomical observations. Such telescopes contain an additional lens or prism system to produce an erect image.

The erection of an image can be made by introducing a third lens between the objective and the eye-piece of the telescope. This modified telescope is known as the "Terrestrial Telescope" whose magnifying power is just equal to the magnification of an astronomical telescope but it just gives an erect image.

The terrestrial telescope contains three lenses as compared to the astronomical telescope. It is also known as the spyglass. As an astronomical telescope forms an inverted image of the object so, the main difference between the astronomical and terrestrial telescope is the erection of the final image with respect to the object. The third lens of short focal length f is placed at 2f which forms an inverted image of the object. This image serves as the object for the eyepiece. The lens placed in the centre of the telescope which actually erects the image is called the Erecting lens. The resolving power of the telescope can be given by the relations as follows:

$\begin{gathered}M=-\frac{f_o}{f_e} \times(-1)=\frac{f_o}{f_e} \\ L=f_o+f_e+4 f\end{gathered}$

Where,

$f_o=$ Focal length of the objective lens
$f_e=$ Focal length of the eye-piece lens
$f=$ Focal length of the lens placed between objective and eye-piece

Magnification at D,
$
m_D=\frac{f_0}{f_e}\left(1+\frac{f_e}{D}\right)
$

Magnification at infinity
$
m_{\infty}=\frac{f_0}{f_e}
$

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Solved Examples Based on Terrestrial Telescope

Example 1: In a terrestrial telescope, the focal length of the objective is 90cm, and of eye lens is 6cm. If the final image is at 30cm, then the magnification will be

1) 9

2) 18

3) 30

4) 36

Solution:

The magnification of a terrestrial telescope can be calculated as :

$m=\frac{f_o}{f_e}\left(1+\frac{f_e}{D}\right) \Rightarrow m=\frac{90}{6}\left(1+\frac{6}{30}\right) \Rightarrow m=18$

Hence, the answer is the option (2).

Example 2: A terrestrial telescope is made by introducing an erecting lens of focal length f between the objective and eyepiece lenses of an astronomical telescope. This causes the length of the telescope tube to increase by an amount equal to:

1) f

2) 2f

3) 3f

4) 4f

Solution:

$L=f_o+f_e+4 f$

In a terrestrial telescope, an erecting lens is introduced between the objective and eyepiece lenses to produce an upright image. This erecting lens essentially acts as an additional optical component that modifies the overall path of light through the telescope.

The focal length of the erecting lens is f. When this lens is added, the total length of the telescope increases by twice the focal length of the erecting lens. This is because the erecting lens forms an intermediate image at its focal length, and then this image is again projected to the eyepiece lens, requiring an additional distance equal to its focal length. Thus, the length of the telescope tube increases by 2f.

Hence, the answer is option (2).

Example 3: The magnifying power of a telescope with a tube length of 60 cm is 5. What is the focal length (in cm) of its eyepiece?

1) 10

2) 40

3) 30

4) 20

Solution:

Let the focal length of the objective as $f_0$ and the focal length of the eyepiece as $f_e$
Magnifying power, $M=\frac{f_0}{f_c}$
Tube length, $L=f_0+f_e$
Given: Magnifying power $=5$
$
\begin{aligned}
& \therefore f_0=5 \\
& \therefore f_0=5 f_e \\
& \text { and } \mathrm{L}=\mathrm{f}_0+\mathrm{f}_{\mathrm{c}}=60 \\
& 5 \mathrm{f}_{\mathrm{e}}+\mathrm{f}_{\mathrm{e}}=60 \quad\left[\because \mathrm{f}_0=5 \mathrm{f}_{\mathrm{e}}\right] \\
& 6 \mathrm{f}_{\mathrm{e}}=60 \\
& \mathrm{f}_{\mathrm{e}}=\frac{60}{6}=10 \mathrm{~cm}
\end{aligned}
$

Hence, the answer is option (1).

Example 4: In a reflecting telescope, a secondary mirror is used to:

1) Make chromatic aberration zero

2) Reduce the problem of mechanical support

3) Move the eyepiece outside the telescopic tube

4) Remove spherical aberration

Solution:

To move the eyepiece outside the telescopic tube.

Hence, the answer is option (3).

Example 5: The focal lengths of the objective and eye lenses of a telescope are respectively 200 cm and 5 cm. The maximum magnifying power of the telescope will be:

1) 40
2) 48
3) 60
4) 100

Solution:

Magnifying power is maximum when the image is formed at the least distance of distinct vision.$M=-\frac{f_0}{f_e}\left(1+\frac{f_e}{D}\right)=-\frac{200}{5}\left(1+\frac{5}{25}\right)=-48$

Hence, the answer is the option (2).

Summary

A terrestrial telescope, featuring an additional lens or prism system for an upright image, is used for Earth-bound observations like bird watching and surveillance. It consists of three lenses, including an erecting lens that inverts and then re-erects the image for clear, magnified viewing. Solved examples illustrate the calculations for determining magnification and the effects of introducing an erecting lens, enhancing understanding of the telescope’s functionality and practical applications.