Question

An astronomical telescope, in normal adjustment position, has magnifying power 5. The distance between the objective and the eyepiece is 120 cm. Calculate the focal lengths of the objective and of the eyepiece.
A. 100cm,20m
B. 20m and 100cm
C. 100cm and 100cm
D. 20m and 20m

Answer 1

Magnifying power (m)= $\dfrac{{({f_0})}}{{({f_e})}}$ is the formula required to solve the sum. Two main considerations with an astronomical telescope are: (i) its light gathering power and (ii) it is resolving power. The light gathering power depends on the area of the objective. With larger diameters, fainter objects can be observed. Magnifying power in normal adjustment.

Complete step by step solution:
First, we write down all the values that have been provided,
Magnifying power (m)= 5
Magnifying power (m) can be also written as (m)= $\dfrac{{({f_0})}}{{({f_e})}}$
Distance between the objective which is written as $({f_0})$ and the eyepiece which is written as $({f_e}) = 120cm$
Which also can be written as,
$({f_0})$+$({f_e})$=120

We know that,
Magnifying power(m) of an astronomical telescope = the ratio of the focal length of the objective $({f_0})$to the focal length of the eye piece $({f_e})$
Since m=5 and m=$\dfrac{{({f_0})}}{{({f_e})}}$
We can write,
m=5=$\dfrac{{({f_0})}}{{({f_e})}}$
Now we can write,
5=$\dfrac{{({f_0})}}{{({f_e})}}$
$({f_0})$=5 \times $$$$({f_e})by simplification,
Now putting the value of $({f_0})$in the equation $({f_0}) + ({f_e}) = 120$ we get,
$5({f_e}) + ({f_e}) = 120$
Now by adding we get,
$6({f_e}) = 120$
We get,
$({f_e}) = \dfrac{{120}}{6}$ by further simplification
Now by dividing we get,
$({f_e}) = 20$

Now we have got the value of $({f_e})$so putting the value of $({f_e})$in the equation $({f_0}) = 120 \times ({f_e})$we get,
$({f_0}) = 5 \times 20$
= 100cm

Therefore, the correct option is A.

Note: Students need to understand the meaning of magnifying power. It is defined as the ratio of the angle subtended at the eye by the final image as seen through the telescope to the angle subtended at the eye by the object seen directly, when both the image and the object lie at infinity. They need to understand where the magnifying power is negative or positive and thus, therefore students often go wrong in this step.