Define Astronomical Telescope , it's construction and principal. Expression for magnifying power ( Angular Magnification) For Distinct Vision Adjustment and For normal Adjustment
Construction: It consists of two achromatic convex lenses mounted co-axially in two metallic tubes. The lens facing the object (which is at infinity) is called objective lens. It has large aperture and large focal length. The other lens through which the image is observed is called eyepiece. It is of small aperture and has small focal length. The tube having eyepiece can be moved in and out of the tube holding objective lens with the help of rack and pinion arrangement.
Principle and Theory: Principle of astronomical telescope can be discussed by considering two extreme cases:
(a) When the final image is formed at least distance of distinct vision ( Distinct Vision Adjustment)
(b) When final image is formed at infinity (Normal adjustment).
Define magnifying power of an astronomical telescope. Derive expression for it ?
Magnifying power of an astronomical telescope is defined as the ratio of the angle subtended by the final image at the eye to the angle subtended by the object at the eye.
If $\alpha$ and $\beta$ be the angles subtended by the object and image with the eye respectively, then
$M.P. = \frac{\beta}{\alpha}... (i)$
The value of M.P. will depend upon the adjustments of the telescope i.e., distinct vision and normal adjustment.
Using $M.P. = \frac{\beta}{\alpha}$
where, $\alpha$ and $\beta$ being small angles, so using the relation $\theta = l/r$, we get
$\beta = \frac{A'B'}{C_2A'}$ and $\alpha = \frac{A'B'}{C_1A'} ... (ii)$
Hence eqn. (i) becomes,
$M.P. = \frac{A'B'}{C_2A'} \times \frac{C_1A'}{A'B'} = \frac{C_1A'}{C_2A'}...(iii)$
Now $C_1A' = f_o$ and $C_2A' = -u_e$ (using sign conventions)
Hence, eqn. (iii) becomes
$M.P. = -\frac{f_o}{u_e} ... (iv)$
(a) For distinct vision
For an eyepiece;
$\frac{1}{u} - \frac{1}{v} = \frac{1}{f_e}$
Here $u = -u_e$ and $v = -D$
$\frac{1}{-u_e} - \frac{1}{-D} = \frac{1}{f_e}$
$\frac{1}{u_e} = \frac{1}{D} + \frac{1}{f_e}$
$\frac{1}{u_e} = \frac{1}{f_e} \left( 1 + \frac{f_e}{D} \right) ... (v)$
Substituting the value of $\frac{1}{u_e}$ in eqn (iv), we get
$M.P. = -\frac{f_o}{f_e} \left( 1 + \frac{f_e}{D} \right) ... (vi)$
In this case, length of telescope,
$L = C_1f_o + f_o C_2 = f_o + u_e$.
But from (v),
$u_e = \frac{f_e D}{f_e + D}$
$L = f_o + \frac{f_e D}{f_e + D}... (vii)$
(b) For normal adjustment :
$u_e = f_e$
Hence from eqn. (iv), the magnifying power of a telescope in normal adjustment is given by
$M.P.= -\left(\frac{f_o}{f_e}\right)$
$|M.P.| = \frac{f_o}{f_e}$
In this case, the length of the telescope,
$\text{L} = \text{C}_1\text{A}'\text{A}'\text{C}_2$
$\text{L} = f_o + f_e$