Question

The objective lens and the eyepiece lens of a telescope are 1.0 m apart. The telescope has an angular magnification of 45. Part A Find the focal length of the eyepiece. Express your answer with the appropriate units. fef e = nothing nothing Request Answer Part B Find the focal length of the objective. Express your answer with the appropriate units. fof o = nothing nothing

Answer 1

Answer:

(A) 0.0217 m

(B) 0.9765 m

Explanation:

Distance between the objective lens and the eye piece, d = 1.0 m

Angular magnification, m = 45

Now,

(A) To calculate the focal length of objective:

$\frac{f_{o}}{f_{e}} = 45$

where

$f_{ob}$ = focal length of object

$f_{ey}$ = focal length of eye piece

Thus

$f_{ob} = 45f_{ey}$              (1)

$f_{ob} + f_{ey} = d$

$f_{ob} + f_{ey} = 1.0$

From eqn (1):

$45f_{ey} + f_{ey} = 1.0$

$f_{ey} = \frac{1.0}{46} = 0.0217\ m$

(B) To calculate the focal length of eye piece:

From eqn (1):

$f_{ob} = 45\times 0.0217 = 0.9765\ m$

Answer 2

Based on the data provided, the focal length of the eyepiece is 2 cm while the focal length of the objective is 98 cm.

What is the focal length of a lens?

The focal length of a lens is the distance between the principal focus and the centre of the lens.

Magnification of a lens is given by the formula below:

• magnification = Fo/Fe

where:

• Fo is focal length of objective lens
• Fe is focal length of eyepiece

From the data provided;

angular magnification = 45

Fo/Fe = 45

Fo = 45 × Fe

Also;

Fe + Fo = 1.0

Fe = 1 - Fo

Substituting in the previous equation

Fo = 45 × (1 - Fo)

Fo = 45 - 45Fo

46Fo = 45

Fo = 45/46

Fo = 0.98 m = 98 cm

From Fe = 1 - Fo

Fe = 1 - 0.98

Fe = 0.02 m = 2 cm

Therefore, the focal length of the eyepiece is 2 cm while the focal length of the objective is 98 cm.

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