Question

The angular magnification of a telescope is 30. If all the light which passes through objective lens of diameter 80 cm also passes through eyepiece lens then find out diameter of eyepiece lens
A. $\dfrac{1}{4}cm$
B. $\dfrac{3}{8}cm$
C. $\dfrac{8}{3}cm$
D. $\dfrac{1}{5}cm$

Answer 1

Hint: To solve this question, we will use the formula of angular magnification of a telescope which is given by, magnification = $\dfrac{{{\text{diameter of objection}}}}{{{\text{diameter of eyepiece}}}}$

Complete answer:
A telescope is an optical instrument made up of two lenses and used to observe distant heavenly objects.
Given that,
Diameter of objective lens = 80 cm.
Magnification of telescope = 30.
We know that, magnification of a telescope is given by,
Magnification = $\dfrac{{{\text{diameter of objection}}}}{{{\text{diameter of eyepiece}}}}$
Putting the values, we will get
$ \Rightarrow 30 = \dfrac{{80}}{{{D_e}}}cm$
$ \Rightarrow {D_e} = \dfrac{{80}}{{30}}cm$
$ \Rightarrow {D_e} = \dfrac{8}{3}cm$
Hence, we can say that the diameter of the eyepiece lens is $\dfrac{8}{3}cm$

So, the correct answer is “Option C”.

Note:
A telescope having high magnifying power produces apparently an enlarged image and hence enables to see the finer details of the distinct object clearly. When in normal adjustment, the magnifying power of a telescope is also given by, $M = - \dfrac{{{f_o}}}{{{f_e}}}$, where ${f_o}$ is the focal length of objective and ${f_e}$ is the focal length of the eye-piece.