InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
A telescope has an objective of focal length `50 cm` and eye piece of focal length `5 cm`. The least distance of distinct vision is `25 cm`. The telescope is focussed for distinct vision on a scale `200 cm` away from the objective. Calculate (i) the separation between objective and eye piece (ii) the magnification produced. |
|
Answer» Given `u=-200cm, f=50cm` For image `I_(1)` object formed by objective lens, `(1)/(f)=(1)/(v)-(1)/(u)` we have `(1)/(v)=(1)/(f)+(1)/(u)=(1)/(50)+(1)/(200)=(4-1)/(200)=(3)/(200)` `rArr v=+(200)/(3) cm` Also, magnification produced by objective lens `m_(0)=(v)/(u)=-(200//3)/(200)=(1)/(3)` Image `I_(1)` acts as an object for eye lens. Here, `v=-25cm, f=5 cm` `:. (1)/(f)=(1)/(v)-(1)/(u)` `rArr (1)/(u)=(1)/(v)-(1)/(f)=-(1)/(25)-(1)/(5)=-(1+5)/(25)` `:. u=-(25)/(6)cm` And magnification produced by eye lens, `m_(e)=(v)/(u)=(-25)/((-25//6))=6` a. The separation between objective and eyepiece `=|V|+|u|=(200)/(3)+(25)/(6)=(425)/(6)=70.73cm` b. Magnification produced, `m=m_(0)xxm_(e)=-(1)/(3)xx6=-2` The negative sign show that the final image is inverted. |
|