Question

An astronomical telescope uses two lenses of powers 10 dioptres and 1 dioptre. If the final image of a distant object is formed at infinity, find the length of the telescope.

Answer 1

Hint: The objective has a larger focal length than the eyepiece. Since power is inversely proportional to the focal length, the power of the objective will be lesser. The length of the telescope is the sum of the focal lengths of the objective lens and the eyepiece.

Formula used:
The power of a lens with a focal length $f$ is given by, $P = \dfrac{1}{f}$ . Its unit is dioptre.
The length of a telescope is given by, $L = {f_o} + {f_e}$ where ${f_o}$ denotes the focal length of the objective and ${f_e}$ denotes the focal length of the eyepiece.

Complete step by step answer:
Given, the powers of the two lenses of the telescope are 10 dioptres and 1 dioptre. We know that power of a lens is inversely proportional to its focal length. So, more power implies a lesser focal length. The objective lens has a larger focal length than the eyepiece.
Hence, the power of the eyepiece will be 1 dioptre and that of the objective will be 10 dioptres.
The power of the eyepiece is given by, $P = \dfrac{1}{{{f_e}}} = 10{\text{ dioptres}}$ .
Then the focal length of the eyepiece will be ${f_e} = \dfrac{1}{{10}} = 0.1{\text{m}} = 10{\text{cm}}$ .
The power of the objective is given by, $P = \dfrac{1}{{{f_o}}} = 1{\text{ dioptre}}$ .
Then the focal length of the eyepiece will be ${f_0} = \dfrac{1}{1} = 1{\text{m}} = 100{\text{cm}}$ .
Step 4: Find the length of the telescope.
The length of the telescope $L = {f_o} + {f_e}$ ------- (1)
where ${f_o}$ is the focal length of the objective and ${f_e}$ is the focal length of the eyepiece.
Substituting the values for ${f_0} = 100{\text{cm}}$ and ${f_e} = 10{\text{cm}}$ in equation (1), we get $L = 100 + 10 = 110{\text{cm}}$ .

Therefore, the length of the telescope is 110 cm.

Note:
It is given that the final image of the distant object is formed at infinity. This implies that the telescope is in normal adjustment.