A astronomical telescope has objective and eyepiece of focal lenghts 40 cm 4 cm respectively. To view an object 200 cm away from the objective, the lenses must be seperated by a distance

Options

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

Answer

C

Explanation

According to question,
focal length of objective lens, `f_(o)=+40 cm`
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)/(f) rArr (1)/(v)-(1)/((-200))=(1)/(40)`
`rArr " " (1)/(v)=(1)/(40)-(1)/(200)=(5-1)/(200)=(4)/(200)`
`rArr " " v=50 cm`
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 )rArr 50+4rArr 54 cm`