Q: A telescope has an objective of focal length 50 cm and an eyepiece of focal length 5 cm. The least distance of distinct vision is 25 cm. The telescope is focused for distinct vision on a scale 2 m away from the objective. Calculate

(a) magnification produced

(b) separation between objective and eye piece,

Sol: Given f_{0} = 50 cm and f_{e} = 5 cm

For objective

$\large \frac{1}{v_o} – \frac{1}{-200} = \frac{1}{50} $

v_{o} = 200/3 cm

$\large m_o = \frac{v_o}{u_o} = \frac{200/3}{-200} $

m_{o} =-1/3

For eyepiece:

$\large \frac{1}{-25} – \frac{1}{u_e} = \frac{1}{5} $

u_{e} = -25/6 cm and

m_{e} = v_{e}/u_{e}

m_{e} =(-25)/(-25/6) = 6

(a) Magnification, m = m_{0} × m_{e} = -2

(b) Separation between objective and eyepiece.

L = v_{0} + |u_{e} | = 200/3 + 25/6 = 425/6 = 70.83 cm