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WAEC Physics 2011 Paper

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Question : 30

Total: 49

The focal length of the eye-piece of an astronomical telescope is f₁, while the focal length of the object is f₂. Fro normal adjustment, the angular magnification is given by

$\;\frac{f_2}{f_1}$
1 - $\;\frac{f_2}{f_1}$
$\;\frac{f_2}{f_1}$ - 1
$\;\frac{f_1}{f_2}$

Solution:

Concept:

For an astronomical telescope in normal adjustment, angular magnification is the ratio of the objective focal length to the eyepiece focal length.

Explanation:

The eyepiece focal length is

f_1

.

The objective focal length is

f_2

.

In normal adjustment, the final image is formed at infinity.

For this adjustment, angular magnification

M

is given by:

M = \frac{f_{\text{objective}}}{f_{\text{eyepiece}}}

Substitute

f_{\text{objective}} = f_2

and

f_{\text{eyepiece}} = f_1

:

M = \frac{f_2}{f_1}

Answer:

A.

\frac{f_2}{f_1}

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