The diameter of the moon is 3.5 times 10^3 km and its distance from the earth is 3.8 times 10^5 km. IT is seen through a telescope having focal lengths of objective and eye piece as 4m and 10cm respectively. Calculate (a) magnifying power of telescope, (b) length of telescope tube and ( c) angular size of image of moon.

The diameter of the moon is 3.5 times 10^3 km and its distance from th

1.	The diameter of the moon is 3.5 times 10^3 km and its distance from the earth is 3.8 times 10^5 km. IT is seen through a telescope having focal lengths of objective and eye piece as 4m and 10cm respectively. Calculate (a) magnifying power of telescope, (b) length of telescope tube and ( c) angular size of image of moon.
Answer» SOLUTION :For normal adjustment (a) `\|M\|=f_0/f_e= ( 4 times 100)/10=40` B) `L=f_0+f_e=400+10=410 cm=4.10 m` As the angle subtended by moon on the objective of telescope `theta_0= (3.5 times 10^3)/(3.8 times 10^5)= 3.5/3.8 times 10^-2 rad` and as `\|M\|=\|theta/theta_0\|` the angular size of final image `\|theta\|=\|M\| times theta_0= 40 times 3.5/3.8 times 10^-2=0.3684 rad` i.e., `\|theta\|=0.368 times 180^@/PI =21^@`