Question

The magnifying power of an astronomical telescope for normal adjustment is 10 and the length of the telescope is 110 cm. The magnifying power of the telescope when the image is formed at the least distance of distinct vision for normal eye is
A). 12
B). 14
C). 16
D). 18

Answer 1

Hint: To solve this question, we need to first analyze the data given in the question. We need to have an idea of the least distance of distinct vision in humans to calculate the magnifying power of a telescope.

Complete step by step solution:
We are given that the magnification power is 10 and the length of the telescope is 110 cm. We can consider D to be a variable representing the least distance of distinct vision which in this case is 25 cm. Then the magnification power at that distance D is to be found out. We know that the magnification power is equal to the focal length of the eyepiece divided by the focal length of the objective lens as is given in the equation below.
$\text{M = }\dfrac{{{\text{f}}_{\text{o}}}}{{{f}_{\text{e}}}}$
Where,
M = Magnifying power
${{\text{f}}_{\text{o}}}$ = focal length of objective lens
${{f}_{\text{e}}}$ = focal length of eyepiece
Therefore, we can write from this equation that
$\therefore \text{ }{{\text{f}}_{\text{o}}}\text{ = 10}\times {{\text{f}}_{\text{e}}}$
We know,
$\text{L = }{{\text{f}}_{\text{o}}}\text{ + }{{\text{f}}_{\text{e}}}\text{ = 110 cm}$
We can assume the focal length of an eyepiece as 10 cm.
We can also determine the focal length of an objective lens as 100 cm since, $\text{L = }{{\text{f}}_{\text{o}}}\text{ + }{{\text{f}}_{\text{e}}}\text{ = 110 cm}$.
With the formula,
${{\text{M}}_{\text{D}}}\text{ = }\dfrac{{{\text{f}}_{\text{o}}}}{{{\text{f}}_{\text{e}}}}\left( 1\text{ + }\dfrac{{{\text{f}}_{\text{e}}}}{\text{D}} \right)$
We can easily calculate the magnifying power of the lens at least distance of distinct vision D.
We know that the least distance of distinct vision is 25cm,
$\therefore \text{ D = 25 cm}$
Hence,
${{\text{M}}_{\text{D}}}\text{ = }\dfrac{{{\text{f}}_{\text{o}}}}{{{\text{f}}_{\text{e}}}}\left( 1\text{ + }\dfrac{{{\text{f}}_{\text{e}}}}{\text{D}} \right)$
$\text{= }\left( \text{1 + }\dfrac{10}{25} \right)\text{ }\times \text{ 10}$
$=\text{ }\dfrac{35\text{ }\times \text{ 10}}{25}$
$=\,\text{ }14$
Therefore, the calculated result for the magnifying power at least distance of distinct vision by a telescope of length 10 cm is 14.
Hence, the correct option is Option B.

Note: Here we are taking the assumption of focal length of eyepiece as 10 cm in case of normal adjustment as is in the question. The least distance of distinct vision is the distance at which someone with ‘normal’ vision can normally look comfortably. In human beings, this distance is 25 cm.