Question

With neat diagram describe the construction and working of an astronomical telescope. Find its magnifying power.

Hint:

For a telescope, remember: Objective lens is "large" (large aperture, large focal length) to collect light and create a spread-out intermediate image. The eyepiece is "small" (small focal length) to act as a powerful magnifier for this image. The total magnification is essentially the ratio of their focal lengths.

Answer 1

Step 1: Construction:
An astronomical telescope is an optical instrument used to see magnified images of distant heavenly bodies like stars and planets. It consists of two convex lenses mounted coaxially at the ends of a tube.
1. Objective Lens: It is a convex lens of large focal length (\(f_o\)) and large aperture. Its function is to gather as much light as possible from the distant object and form a bright, real image.
2. Eyepiece (or Ocular): It is a convex lens of short focal length (\(f_e\)) and small aperture. It acts as a simple magnifier to view the intermediate image formed by the objective.
The distance between the lenses can be adjusted using a rack and pinion arrangement.
Step 2: Diagram and Working (Normal Adjustment):
In normal adjustment, the final image is formed at infinity for relaxed viewing.

Working:
1. Parallel rays of light from a distant object (at infinity) enter the objective lens.
2. The objective lens converges these rays to form a real, inverted, and highly diminished image (\(A'B'\)) at its second focal point (\(F_o\)).
3. For normal adjustment, the eyepiece is positioned such that this intermediate image \(A'B'\) lies at its first focal point (\(F_e\)).
4. Therefore, the intermediate image acts as an object for the eyepiece. Since the object is at the focal point of the eyepiece, the final rays emerge parallel from the eyepiece.
5. These parallel rays enter the observer's eye, which perceives them as coming from a highly magnified, virtual, and inverted image at infinity. The length of the telescope tube in this case is \(L = f_o + f_e\).
Step 3: Magnifying Power:
The magnifying power (\(M\)) of a telescope is defined as the ratio of the angle (\(\beta\)) subtended at the eye by the final image to the angle (\(\alpha\)) subtended at the eye by the object directly.
\[ M = \frac{\beta}{\alpha} \] Since the angles are small, we can approximate \(\alpha \approx \tan(\alpha)\) and \(\beta \approx \tan(\beta)\).
From the diagram, with the intermediate image \(A'B'\) of height \(h\):
The angle subtended by the object at the objective is \(\alpha\). From triangle \(\triangle A'B'O_1\) (where \(O_1\) is the optical center of the objective):
\[ \tan(\alpha) = \frac{A'B'}{O_1B'} = \frac{h}{f_o} \] The angle subtended by the final image at the eyepiece is \(\beta\). From triangle \(\triangle A'B'O_2\) (where \(O_2\) is the optical center of the eyepiece):
\[ \tan(\beta) = \frac{A'B'}{O_2B'} = \frac{h}{f_e} \] Now, we can find the magnifying power:
\[ M = \frac{\tan(\beta)}{\tan(\alpha)} = \frac{h/f_e}{h/f_o} \] \[ M = \frac{f_o}{f_e} \] This is the expression for magnifying power in normal adjustment.
Condition for Maximum Magnification:
For maximum magnification, the final image is formed at the least distance of distinct vision (\(D\)). In this case, the magnifying power is:
\[ M = \frac{f_o}{f_e} \left(1 + \frac{f_e}{D}\right) \]