Question

The focal length of the objectives and the eyepiece of an astronomical telescope are `60cm`and`5cm` respectively .Calculate the magnifying power and the length of the telescope when the final image is formed at(i)infinity,(ii)least distance of distinct vision`(25cm)`

Answer 1

(i)When the final image is at infinity , then :
`m=-(f_0)/(f_e) = -60/5=-12`
and length of the telescope is `L=f_0+f_e=60+5=65 cm`
(ii)For least distance of dis tinct vision, the magnifying power is : The length of telescope in this position is :
`m=-(f_(0))/(f_(e))(1+(f_(e))/(D))=-(60)/(5)(1+(5)/(25))=-(12xx6)/(5)=-14.4`
Now,`(1)/(f_(e))=(1)/(v_(e))-(1)/(u_(e)) v_(e)` and `u_(e)` are both negative:
`:.(1)/(f_(e))=-(1)/(25)-(-(1)/(u_(e)))` or `(1)/(5)=-(1)/(25)+(1)/(u_(e))`
`(1)/(u_(e))=(1)/(25)+(1)/(5)` or `u_(e)=4.17cm`
The length of telescope in this position is:
`L=f_(0)+u_(e)=60+4.17=64.17cm`