1.

The focal length of the objective and the eyepiece of a telescope are 50 cm and 5 cm respectively. If the telescope is focussed for distinct vision on a scale distant 2 m from its objective, then its magnifying power will be :

Answer»

`-4`
`-98`
`+8`
`-2`

Solution :Given : `f_(0)=50cm, f_(E)=5cm`
`d=25cm, u_(0)=-200cm`
Magnification `M=?`
As `(1)/(v_(0))-(1)/(u_(0))=(1)/(f_(0))`
`rArr (1)/(v_(0))=(1)/(f_(0))+(1)/(u_(0))=(1)/(50)-(1)/(200)=(4-1)/(200)=(3)/(200) or v_(0)=(200)/(3)CM`
Now `v_(e)=d=-25cm`
From, `(1)/(v_(e))-(1)/(u_(e))=(1)/(f_(e))`
`-(1)/(u_(e))=(1)/(f_(e))-(1)/(v_(e))=(1)/(5)+(1)/(25)=(6)/(25) or,""v_(e)=(-25)/(6)cm`
Magnification `M=M_(0)xxM_(e)`
`=(v_(0))/(u_(0))xx(v_(e))/(u_(e))=(-200//3)/(200)xx(-25)/(-25//6)=-(1)/(3)xx6=-2`


Discussion

No Comment Found

Related InterviewSolutions