Question

In telescope power of an objective and eye lens are respectively + 1.25 D and - 20 D, then for relaxed vision the length and magnification will be ______.

Options

• 21.25 cm & 16

• 75 cm & 20

• 75 cm & 16

• 8.5 cm & 21.25

Answer 1

In telescope power of an objective and eye lens are respectively + 1.25 D and - 20 D, then for relaxed vision the length and magnification will be 75 cm & 16.

Explanation:

f_o = `1/1.25` = 0.8, f_e = `1/(- 20)` = - 0.05

∴ |L| = |f_o| + |f_e|         [For telescope]

|L| = 0.8 - 0.05

|L| = 0.75 m = 75 cm

|m| = `(|"f"_"o"|)/("f"_"e") = 0.8/0.05` = 16