Question

The magnifying power of the astronomical telescope for normal adjustment is 50. The focal length of the eyepiece is 2 cm. The required length of the telescope for normal adjustment is:
A. 100 cm
 B. 98 cm
C. 25 cm
D. 102 cm

Answer 1

Hint: A telescope is an optical device used to look at distant objects like a star, a planet, or a distant tree etc. The magnification power of this telescope is given as $m = \dfrac{{ - {f_ \circ }}}{{{f_e}}}$ . And the length of telescope in normal adjustment i.e., when both object and image are at infinity is given by${f_u} + {f_e} = L$ .
$\eqalign{
  & {\text{where }}m{\text{ is the magnification power of the telescope,}} \cr
  & {f_ \circ }{\text{ is the focal length of the objective,}} \cr
  & {f_e}{\text{ is the focal length of the eye piece,}} \cr
  & L{\text{ is the length of the telescope for normal adjustment}}{\text{.}} \cr} $
Using the above-mentioned formulas we can solve the question, by substituting the given values and thus finding the required ones.

Complete step by step answer:
An astronomical telescope is a refracting type telescope used to see heavenly bodies like stars, planets, satellites, etc. It consists of two converging lenses mounted coaxially at the outer ends of two sliding tubes.
Magnifying power in the normal adjustment 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 image and object are at infinity.
Mathematically,
$m = \dfrac{{{f_ \circ }}}{{{f_e}}} \cdots \cdots \cdots \cdots \left( 1 \right)$
When the beam of light is incident on the objective, it forms a real, inverted, and diminished image of the object in its focal plane. The eyepiece is so adjusted that the image is formed at infinity, and is highly magnified and inverted with respect to the object. This setup is known as the normal adjustment.
For normal adjustment separation between object and image, i.e. the length of the telescope at normal adjustment is:

${f_ \circ } + {f_e} = L \cdots \cdots \cdots \cdots \left( 2 \right)$
Given:
m=50 and ${f_e} = 2cm$
Substituting the given values in equation (2) & (3) we get:
$\eqalign{
  & {f_ \circ } + 2cm = L \cdots \cdots \cdots \left( 3 \right) \cr
  & {\text{and }}50 = \dfrac{{{f_ \circ }}}{{{f_e}}} \cdots \cdots \cdots \cdots \left( 4 \right) \cr} $
Solving equation (4) we get:
$\eqalign{
  & 50 = \dfrac{{{f_ \circ }}}{{{f_e}}} \cr
  & \Rightarrow 50{f_e} = {f_ \circ } \cr
  & {\text{But }}{f_e} = 2cm\left[ {given} \right] \cr
  & \Rightarrow {f_ \circ } = 50 \times 2cm \cr
  & \therefore {f_ \circ } = 100cm \cr} $
Substituting the calculated value of ${f_ \circ }$ in equation (3), we get:
$\eqalign{
  & {f_ \circ } + 2cm = L \cr
  & \Rightarrow 100cm + 2cm = L \cr
  & \therefore L = 102cm \cr} $
Therefore, the correct option is D. i.e., 102cm

Note:
Students usually get confused a lot between the formula of the image formed at the distance of least distinct vision and in the normal adjustment. In order to rise above the confusion, they should associate the meaning of these system settings to the obtained result. For example at normal adjustment, both the image and the object are at infinity.