A small astronomical telescope has an objective of focal length 50 cm and an eyepiece of focal length 5 cm . It is focussed at the sun and the final image is formed at a distance of 25 cm from the eyepiece . If the diameter of the sun subtends an angle of 32' at the centre of the objective calculate the angular magnification of the instrument and the actual size of the image.

Explanation

Here, `f_o=50cm, f_e=5 cm and D=25 cm` .
Angular magnification of the instrument,
`m=(f_o)/(f_e)(1+(f_e)/D)=(50)/5 (1+5/(25))=10xx6/5 =12`
If the angle subtended at the eyepiece by the final image be `beta` and the angle subtended at the centre of the objective by the sum be `alpha` , then
`m=(beta)/(alpha)`
or, `beta =malpha =(12xx(32)/(60))^@ =(32)/5 xx(pi)/(180) rad`
If the actual size of the image be I , then,
`beta =I/D`
or, I=`beta D =(32)/5xx(pi)/(180)xx25 =8/9 (pi)=2.793 cm `