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. 46.0cm
B. 50.0cm
C. 54.0cm
D. 37.3cm

Answer

C

Explanation

(c) According to question,
Focal length of objective lens, `f_(o)=+40cm`
Focal length of eye-piece lens,`f_(e)= 4 cm`
Object distance for objective lens, `u_(o)= -200 cm`

Applying lens formula for objective lens,
`1/v-1/u=1/frArr1/v-1/((-200))=1/40`
`rArr1/v=1/40-1/200=(5-1)/200=4/200`
`rArrv=50cm`
Image will be form at first focus of eye-piece lens.
So, for normal adjustment distance between objectives and eye-piece lens (length of tube) will be,
`v+f_(e)rArr50+4rArr54`cm