InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
Define magnifying power of a telescope. Write its expression. A small telescope has an objective lens of focal length 150 cm and an eye piece of focal length 5 cm. If this telescope is used to view a 100 m high tower 3 cm away, find the height of the final image when it is formed 25 cm away from the eye piece. |
|
Answer» Magnifying power of an astronomical telescope is defined as the ratio of the angle subtended at the eye by the final image of the angle subtended at the eye, by the object directly. The expression for magnifying power : `m=(f_(0))/(f_(e ))(1+(f_(e ))/(d))` (a) In normal adjustment `m= -(f_(0))/(f_(e ))` (b) When final image is formed at infinity `m=(f_(0))/(f_(e ))(1+(f_(e ))/(d)), f_(0)=150 cm implies f_(e )=5 cm`. Using `(1)/(f )=(1)/(v)-(1)/(u)` for objective. `(1)/(150)=(1)/(v)-(1)/(-3000) implies (1)/(v)=(1)/(150)-(1)/(3000)=(20-1)/(3000) implies v=(3000)/(19)` `m_(0)=(3000)/(19)xx(1)/(-3000)=-(1)/(19)` Using `(1)/(f)=(1)/(v)-(1)/(u)` for eye piece, `(1)/(5)=(1)/(-25)-(1)/(u) implies (1)/(u) =(1)/(-25)-(1)/(5) implies u=(1+5)/(-25)` `u=(-25)/(6) cm implies m_(e )=(v)/(u)` `m_(e )=(-25)/(-25)xx6=6` `therefore` Total magnification `=m_(0)xxm_(e )= -0.315` `h_(1)=0.315xx100=31.578`. |
|