An astronomical telesope has objective and eyepiece of focal lengths `40cm` and `4 cm` respectively. To view an object `200cm` away from the objective, the lenses must be separated by a distance `:`

Options

A. `46.0 cm`
B. 50 cm
C. `54.0 cm`
D. 37.3 cm`

Answer

C

Explanation

Here, `f_0 = 40 cm, f_e = 4 cm, u_0 = -200 cm`
As `(1)/(v_0)-(1)/(u_0)=(1)/(f_0)`
`:. (1)/(v_0)=(1)/(f_0)+(1)/(u_0)`
`= (1)/(40)-(1)/(200) = (4)/(200) = (1)/(50)`
`v_0 = 50 cm`
In normal adjustment, distance between objective and eye piece `=v_0 + f_e = 50 + 4 - 54 cm`.