Question

A telescope has an objective of focal length `50 cm` and eye piece of focal length `5 cm`. The least distance of distinct vision is `25 cm`. The telescope is focussed for distinct vision on a scale `200 cm` away from the objective. Calculate
(i) the separation between objective and eye piece
(ii) the magnification produced.

Answer 1

Here, `f_0 = 50 cm, f_e = 5 cm`,
`d = 25 cm, u_0 = -200 cm, m = ? L = ?`
As `(1)/(v_0)-(1)/(u_0)=(1)/(f_0)`
`:. (1)/(v_0)=(1)/(f_0)+(1)/(u_0)=(1)/(50)-(1)/(200)=(4 - 1)/(200) = (3)/(200)`
`v_0 = (200)/(3) cm`
Linear magnification produced by objective lens,
`m_0 = (v_0)/(u_0) = -(200//3)/(200) = -(1)/(3)`
Now, `v_e = d = -25 cm`
From `(1)/(v_e) - (1)/(u_e)=(1)/(f_e)`
`(-1)/(u_e)=(1)/(f_e)-(1)/(v_e)=(1)/(5)+(1)/(25)=(5 + 1)/(25)`
`u_e = -(25)/(6) cm`
Linear magnification produced by eye piece
`m_e = (v_e)/(u_e) = (-25)/(-25//6) = 6`
Net magnification produced,
`m = m_0 xx m_e = -(1)/(3) xx 6 = -2`
Distance between objective and eye piece
`L = |v_0|+|u_e|= (200)/(3) + (25)/(6) = (400 + 25)/(6) = (425)/(6)`
=`70.8 cm`.