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Focal lenghts of objective and eye-piece of telescope 200 cm and 2 cm respectively. If a building of 50 m height at 2 km away is observed from this, then height of image of objective will be ...... |
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Answer» 5 cm `u_0=-2xx10^5` cm `(1)/(v_0)-(1)/(u_0)=(1)/(f_0)` `THEREFORE (1)/(v_0)=(1)/(f_0)+(1)/(u_0)` `therefore (1)/(v_0)=(1)/(200)+(1)/(-2xx10^5)=(1)/(200)-(1)/(2xx10^5)` `(1)/(v_0)=(999)/(2xx10^5)` `therefore v_0=(2xx10^5)/(999)` cm Now `("dimensions of image")/("dimensions of object")=((v_0)/(u_0))` `therefore` Image DIMENSION`=((v_0)/(u_0))xx`Object dimension `=((2xx10^5)/(999)xx(1)/(2xx10^5))xx(50xx100)` `(5000)/(999) ~~5` cm |
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