
An astronomical telescope has an eyepiece of a focal-length of $5cm$. If the angular magnification in normal adjustment is 10 when the final image is at least distance of distinct vision ($25cm$) from the eyepiece, then angular magnification will be:
A) $10$
B) $12$
C) $50$
D) $60$
Answer
219.3k+ views
Hint: In order to solve this question you have to know the formula for the angular magnification for both the normal adjustment and when the final image is at least distance of distinct vision from the eyepiece. Also, remember all the concepts related to an astronomical telescope.
Formula used:
$m' = - \dfrac{{{f_o}}}{{{f_e}}}\left( {1 + \dfrac{{{f_e}}}{D}} \right)$
here ${f_e}$ is the focal length of the eyepiece of the given astronomical telescope
\[{f_o}\] is the focal length of the given astronomical telescope
$D$ is the least distance of distinct vision of the human eye
Complete step by step solution:
All the information given in the question are:
An astronomical telescope has an eyepiece of focal length, ${f_e} = 5cm$
Angular magnification in normal adjustments, $m = 10cm$
Final image is at least distance of distinct vision, $D = 25cm$
Firstly apply the formula for magnification for normal adjustment that is,
$m = - \dfrac{{{f_o}}}{{{f_e}}}$
On putting all the given values,
$10 = \dfrac{{{f_o}}}{5}$
On further solving, we have
$ \Rightarrow {f_o} = 50cm$
Now, when final image is at least distance of distinct vision from eyepiece, then angular magnification $m'$ is given by
$m' = - \dfrac{{{f_o}}}{{{f_e}}}\left( {1 + \dfrac{{{f_e}}}{D}} \right)$
Here ${f_e}$ is the focal length of the eyepiece of the given astronomical telescope
\[{f_o}\] is the focal length of the given astronomical telescope
$D$ is the least distance of distinct vision of the human eye
$ \Rightarrow m' = 10\left( {1 + \dfrac{5}{{25}}} \right)$
On further solving, we get the angular magnification as
$ \Rightarrow m' = 12$
Therefore, the correct option is (B).
Note: A astronomical telescope is an optical instrument used for observing far away distance. It has an object with an eyepiece with a short focal length and a large focal length to observe distant objects or celestial bodies. The ability of a telescope to magnify a distant object is known as its magnifying power. Mathematically, it is equal to the ratio of the focal length of the objective to the focal length of the eyepiece.
Formula used:
$m' = - \dfrac{{{f_o}}}{{{f_e}}}\left( {1 + \dfrac{{{f_e}}}{D}} \right)$
here ${f_e}$ is the focal length of the eyepiece of the given astronomical telescope
\[{f_o}\] is the focal length of the given astronomical telescope
$D$ is the least distance of distinct vision of the human eye
Complete step by step solution:
All the information given in the question are:
An astronomical telescope has an eyepiece of focal length, ${f_e} = 5cm$
Angular magnification in normal adjustments, $m = 10cm$
Final image is at least distance of distinct vision, $D = 25cm$
Firstly apply the formula for magnification for normal adjustment that is,
$m = - \dfrac{{{f_o}}}{{{f_e}}}$
On putting all the given values,
$10 = \dfrac{{{f_o}}}{5}$
On further solving, we have
$ \Rightarrow {f_o} = 50cm$
Now, when final image is at least distance of distinct vision from eyepiece, then angular magnification $m'$ is given by
$m' = - \dfrac{{{f_o}}}{{{f_e}}}\left( {1 + \dfrac{{{f_e}}}{D}} \right)$
Here ${f_e}$ is the focal length of the eyepiece of the given astronomical telescope
\[{f_o}\] is the focal length of the given astronomical telescope
$D$ is the least distance of distinct vision of the human eye
$ \Rightarrow m' = 10\left( {1 + \dfrac{5}{{25}}} \right)$
On further solving, we get the angular magnification as
$ \Rightarrow m' = 12$
Therefore, the correct option is (B).
Note: A astronomical telescope is an optical instrument used for observing far away distance. It has an object with an eyepiece with a short focal length and a large focal length to observe distant objects or celestial bodies. The ability of a telescope to magnify a distant object is known as its magnifying power. Mathematically, it is equal to the ratio of the focal length of the objective to the focal length of the eyepiece.
Recently Updated Pages
Chemical Equation - Important Concepts and Tips for JEE

JEE Main 2022 (July 29th Shift 1) Chemistry Question Paper with Answer Key

Conduction Explained: Definition, Examples & Science for Students

Analytical Method of Vector Addition Explained Simply

Atomic Size - Important Concepts and Tips for JEE

JEE Main 2022 (June 29th Shift 1) Maths Question Paper with Answer Key

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

Understanding Uniform Acceleration in Physics

Derivation of Equation of Trajectory Explained for Students

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Angle of Deviation in a Prism

Understanding Collisions: Types and Examples for Students

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Understanding Atomic Structure for Beginners

How to Convert a Galvanometer into an Ammeter or Voltmeter

Understanding Centrifugal Force in Physics

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Degree of Dissociation: Meaning, Formula, Calculation & Uses

