# 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…

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 f0 = 50 cm and fe = 5 cm

For objective

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

vo = 200/3 cm

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

mo =-1/3

For eyepiece:

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

ue = -25/6 cm and

me = ve/ue

me =(-25)/(-25/6) = 6

(a) Magnification, m = m0 × me = -2

(b) Separation between objective and eyepiece.

L = v0 + |ue | = 200/3 + 25/6 = 425/6 = 70.83 cm