An astronomical telescope has an angular magnification of magnitude 5 for distant objects. The separation between the objective and the eyepiece is 36 cm and final image is formed at infinity. The focal lengths of the objective and eyepiece are respectively:

Question

An astronomical telescope has an angular magnification of magnitude 5 for distant objects. The separation between the objective and the eyepiece is 36 cm and final image is formed at infinity. The focal lengths of the objective and eyepiece are respectively: (A) 20 cm, 16 cm (B) 50 cm, 10 cm (C) 30 cm, 6 cm (D) 45 cm, - 9 cm

Tardigrade Admin · Accepted Answer

Given f0+fe=36 ?(i) and m=5=(f0/fe) ?(ii) Solving (i) and (ii), we get f0=30 cm and fe=6 cm

Q. An astronomical telescope has an angular magnification of magnitude 5 for distant objects. The separation between the objective and the eyepiece is 36 cm and final image is formed at infinity. The focal lengths of the objective and eyepiece are respectively:

20 cm, 16 cm

50 cm, 10 cm

30 cm, 6 cm

45 cm, - 9 cm

Given $f_{0} + f_{e} = 36$ ?(i) and $m = 5 = \frac{f _{0}}{f _{e}}$ ?(ii) Solving (i) and (ii), we get $f_{0} = 30 c m$ and $f_{e} = 6 c m$

Q. An astronomical telescope has an angular magnification of magnitude 5 for distant objects. The separation between the objective and the eyepiece is 36 cm and final image is formed at infinity. The focal lengths of the objective and eyepiece are respectively:

20 cm, 16 cm

50 cm, 10 cm

30 cm, 6 cm

45 cm, - 9 cm

Given f0​+fe​=36 ?(i) and m=5=fe​f0​​ ?(ii) Solving (i) and (ii), we get f0​=30cm and fe​=6cm

Given $f_{0} + f_{e} = 36$ ?(i) and $m = 5 = \frac{f _{0}}{f _{e}}$ ?(ii) Solving (i) and (ii), we get $f_{0} = 30 c m$ and $f_{e} = 6 c m$