A telescope objective of focal length 1m forms a real image of the moon 0.92cm in diameter. Calculate the diameter of the moon taking its mean distance for the earth to be 38 times 10^4 km . IF the telescope uses an eyepiece of 5cm focal length. What would be the distance between the two lenses for (i) the final image to be formed at infinity (ii) the final image (virtual) at 25 cm from the eye.

A telescope objective of focal length 1m forms a real image of the moo

1.	A telescope objective of focal length 1m forms a real image of the moon 0.92cm in diameter. Calculate the diameter of the moon taking its mean distance for the earth to be 38 times 10^4 km . IF the telescope uses an eyepiece of 5cm focal length. What would be the distance between the two lenses for (i) the final image to be formed at infinity (ii) the final image (virtual) at 25 cm from the eye.
Answer» SOLUTION :`f_0=1m` object distance from the objective = distance of the moon from the earth `=3.8 times 10^5 km =3.8 times 10^8m`, image distance from the objective =FOCAL length of the objective =1 m, image size= image diameter=`0.92 times 10^-2m` object size= object diameter i.e., diameter of the moon=? We know that, `(Object diameter)/(Imag e diameter)= (Object distance)/(Imag e distance)` `(Diameter for moon)/(Im ag e diameter)= (3.8 times 10^8)/1` `therefore` Diameter of the moon =`3.8 times 10^8 times ` Image diameter `=3.8 times 10^8 times 0.92 times 10^-2 m=3.496 times 10^6m=3496km` (i) For NORMAL adjustment, the distance between the two lenses `f_0+f_a=100+5=105cm` (ii) FOr the final image at 25CM from the eye, the distance between the two lenses `=f_0+((Df_e)/(D+f_e))=100+ ((25 times 5)/(25+5))=104.2cm`