
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
233.1k+ 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
JEE Main 2023 April 6 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 6 Shift 2 Question Paper with Answer Key

JEE Main 2023 (January 31 Evening Shift) Question Paper with Solutions [PDF]

JEE Main 2023 January 30 Shift 2 Question Paper with Answer Key

JEE Main 2023 January 25 Shift 1 Question Paper with Answer Key

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Key

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

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

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

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

Understanding Uniform Acceleration in Physics

Dual Nature of Radiation and Matter Class 12 Physics Chapter 11 CBSE Notes - 2025-26

Understanding Electromagnetic Waves and Their Importance

Understanding the Electric Field of a Uniformly Charged Ring

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

