A Cassegrain telescope uses two mirrors as shown in Fig. 9.30. Such a telescope is built with the mirrors 20mm apart. If the radius of curvature of the large mirror is 220mm and the small mirror is 140mm, where will the final image of an object at infinity be?

Explanation

`rArr` Here, radius of curvature of concave mirror of objective,
`R_1 = 220 mm`
`therefore` Focal length
`f_1 = (R_1)/(2) = (220)/(2) = 110 mm =11 cm `
Radius of curvature of secondary convex mirror,
`R_2 = 140` mm
and focal length
`f_2 = (R_2)/(2) = (140)/(2) = 70 mm = 7 cm `
`rArr` Distance between two mirrors
d= 20 mm = 2 cm
When parallel rays from object at infinity reflect from objective mirror and intersect at `f_1` they also incident on secondary mirror, hence, object distance, `u = f_1-d`
`therefore u =11 -2 = 9 cm `
`rArr` Distance between two mirrors
d= 20 mm = 2cm
When parallel rays from object at infinity reflect from objective mirror and intersect at `f_1` , they also incident on secondary mirror , hence object distance `u = f_1 - d`
`therefore u = 11 -2 = 9 cm `
`rArr` From mirror equation `(1)/(f_2) = (1)/(u) + (1)/(v)`
`(1)/(v) = (1)/(f_2) - (1)/(u) = (1)/(7) - (1)/(9) = (2)/(63)`
`therefore v= (63)/(2) = 31.5 cm `
At right side of convex mirror at 31.5 cm .