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 ......

Options

A. 5 cm
B. 10 cm
C. 1 cm
D. 2 cm

Answer

A, C

Explanation

For objective `f_0=200` 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