Saved Bookmarks
| 1. |
(a) Define magnifying power of a telescope. (b) 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 100m high tower 3 x 10^(5)cm away , find the height of the final image when it is formed 25 cam away from the eye piece. |
|
Answer» Solution :(a) Magnifying power is the ratio of the angle subtended at the eye by the image to the angle subtended at the unaided eye by the object. Expression `:` `m = ( beta )/( alpha) = ( f_(0))/( f_(e ))` ` m = ( f_(0))/( f_(e )) = (1+ ( f_(e))/(D))` (b) USING ,the lens equation for objective lens , `:` `(1)/(f_(0))= ( 1)/( v_(0)) - ( 1)/( u_(0))` `RARR (1)/( 150) = ( 1)/( v_(0)) - ( 1)/( - 3 xx 10^(5))` `rArr (1)/(v_(0)) = ( 1)/( 150) - (1)/( - 3 xx 10^(5)) = ( 2000 - 1)/( 3 xx 10^(5))` `rArr v_(0) = ( - 3 xx 10^(5))/( 1999) cm` `~~ 150 cm` Hence, MAGNIFICATION due to the objective lens ` m_(0) = ( v_(0))/( u _(0) ) = ( 150 xx 0^(-2) m)/( 3000 m ) ` `~~ (10^(-2))/( 20) =0.05 xx 10^(-2)` Using lens FORMULA for eyepiece, `(1)/(f_(e )) = ( 1)/( v_(e )) - ( 1)/( u _(e ))` `rArr(1)/( 5) = ( 1)/( -25) - (1)/(u_(e ))` `rArr (1)/( u_(e )) = ( 1)/( -25) - ( 1)/( 5) = ( -1-5)/( 25)` `rArr u_(e ) = ( - 25)/( 6) cm` `:.` Magnification due to eyepiece,` m_(e ) = ( - 25)/(- (25)/(6)) = 6` Hence, total magnification`rArrm= m_(e ) xx m _(0)` `m = 6 xx 5 xx 10^9-4) \= 30 xx 10^(-4)` Hence, size of the final image `= 30 xx 10^(-4) xx 100 `m =30cm |
|