An astronomical telescope in normal adjustment consists of an objective lens with a focal … — JAMB UTME

An astronomical telescope in normal adjustment consists of an objective lens with a focal length of $60\text{ cm}$ and an eyepiece with a focal length of $5\text{ cm}$ . What is the angular magnification produced by the telescope and the separation distance between the two lenses?

$12$ and $65\text{ cm}$ Cevap
B
$0.083$ and $65\text{ cm}$
C
$12$ and $55\text{ cm}$
D
$300$ and $65\text{ cm}$

Cevap

The angular magnification is

12

and the separation distance between the lenses is

65\text{ cm}

For an astronomical telescope in normal adjustment, the magnification is given by

M = \frac{f_o}{f_e} = \frac{60}{5} = 12

, and the length of the telescope tube (distance between lenses) is

L = f_o + f_e = 60 + 5 = 65\text{ cm}

Adım Adım Çözüm

Calculate the angular magnification (

M

) of the telescope

M = \frac{f_o}{f_e} = \frac{60}{5} = 12

In an astronomical telescope in normal adjustment, angular magnification is given by the ratio of the focal length of the objective lens to that of the eyepiece.

Calculate the separation distance (

L

) between the two lenses

L = f_o + f_e = 60 + 5 = 65\text{ cm}

When in normal adjustment, the final image is formed at infinity, so the distance between the objective lens and eyepiece equals the sum of their focal lengths.

Anahtar Kavram

Astronomical telescope in normal adjustment (magnification and lens separation)

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