Question

A group of students are given a project for constructing a telescope and they were supplied with two biconvex lens of power 1 diopter and 0.1 dioptre.

1. Of the two lenses, which can be used as objective?

2. Draw the ray diagram for the formation of the image by a telescope.

3. Arrive at an expression for magnification of a telescope.

4. Prepare a notice/ label about the precaution to be taken while using the telescope and limitations of the telescope constructed.

Answer 1

1. Biconvex lens of power 0.1 dioptre.

2.

3. Magnification:

The magnifying power of a telescope is the ratio of the angle subtended by the image at the eye to the angle subtended by the object at the objective.

i.e. m = \(\frac{β}{\alpha}\)  …….(1)[from figure]

But from ∆CIM, tanα = \(\frac{IM}{IC}\), α = \(\frac{IM}{IC}\)

(For small values tan α ≈ α)

from ∆C¹ IM, tanβ = \(\frac{IM}{IC^1}\) ,β = \(\frac{IM}{IC^1}\)

substituting α and β in eq (1) we get

But IC = f_o (the focal length objective lens)

and IC¹ = f_e (the focal length eyepiece lens.)

∴  \(m =\frac{f_0}{f_e}\)

In this case the length of the telescope tube is (f_o + f_e).

Case 1 :

When the image formed by the objective is within the focal length of the eyepiece, Then the final image is formed at the least distant of distinct vision. In this case, magnifying power.

m = \(\frac{f_0}{f_e}\Big[1+\frac{f_e}{D}\Big]\)

4. 

• As magnifying power is negative, the final image in an astronomical telescope is inverted.
• To have large magnifying power, f₀ must be as large as possible and f_e must be as small as possible.
• As intermediate image is between the two lens, cross wire (ora measuring device) can be used.
• In normal setting of telescope, the final image is at infinity.