Telescope : It is an optical instrument used to increase the visual angle of distant large objects.

**Astronomical Telescope :**

It consists of two converging lenses placed coaxially with objective having large aperture and a large focal length while the eye- piece having smaller aperture and focal length. The separation between eye- piece and objective can be varied.

Light from a distant object enters the objective and a real image is formed at the focal point of the objective. The eyepiece then further magnifies the image. In case of a telescope, magnification is defined in terms of angles subtended by the image and the object.

Magnifying power

$ \displaystyle M = \frac{Visual \;Angle \; with \; Instrument }{Visual \;Angle \; for \; unaided \; eye } $

$ \displaystyle M = \frac{\theta}{\theta_0} $

$ \displaystyle \theta_0 = \frac{y}{f_0} $

$ \displaystyle \theta = \frac{y}{-u_e} $

$ \displaystyle M = -\frac{f_0}{u_e} $

Length of Tube $ \displaystyle L = f_0 + u_e $

Now there are two possibilities :

Case (i) Final image is at infinity ( far point)

v = ∞ then u_{e} = f_{e}

$ \displaystyle M = -\frac{f_0}{f_e} $

Length of Tube $ \displaystyle L = f_0 + f_e $

Case (ii) : Final image is at D ( near point)

For Eye Lens ,

$ \displaystyle \frac{1}{v_e} – \frac{1}{u_e} = \frac{1}{f_e}$

$ \displaystyle \frac{1}{-D} – \frac{1}{- u_e} = \frac{1}{f_e}$

$ \displaystyle \frac{1}{u_e} = \frac{1}{f_e} + \frac{1}{D}$

$ \displaystyle \frac{1}{u_e} = \frac{1}{f_e}( 1+ \frac{f_e}{D}) $

$ \displaystyle M = -\frac{f_0}{f_e} ( 1 + \frac{f_e}{D}) $

### Also Read :

← Back Page |Next Page→