Question

A telescope objective lens has a focal length of `100 cm` . When the final image is formed at the least distance of distinct vision, the distance between the lenses is `105 cm`. Calculate the focal length of eye piece and magnifying power of telescope.

Answer 1

Correct Answer - `6.25 cm ; 20`
Here, `f_0 = 100 cm , f_e = ?, m = ?`
When the final image is formed at the least distance of distinct vision, distance between objective and eye lens `= f_0 + |u_e| = 105 cm`
`:. u_e = 105 - f_0 = 105 - 100 = 5 cm`
For the eye piece, `u_e = -5 cm`,
`v_e = -d = -25 cm`
`:. (1)/(f_e)=(1)/(v_e)-(1)/(u_e)=(-1)/(25)+(1)/(5) = (4)/(25)`
`f_e = (25)/(4) = 6.25 cm`
`m = (f_0)/(f_e) ((1 + f_e)/(d)) = (100)/(6.25) (1 + (6.25)/(25)) = 20`.