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Q.36 Physics · Section B ⭐ 360° Option Elimination Active
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NEET 2024 Physics Question 36 Solution

Topic: Ray Optics and Optical Instruments · Official NTA Booklet Code T1
NEET 2024 Physics Question 36: Magnifying Power of an Astronomical Telescope for Distant Objects

Question

A small telescope has an objective of focal length 140 cm and an eye piece of focal length 5.0 cm. The magnifying power of telescope for viewing a distant object is:

✓
Verified Against NTA Final Answer Key
Option (4) : 28
📚 Ray Optics and Optical Instruments Easy
📋 Given Parameters & Problem Setup
Objective focal length f_o = 140 cm
Eyepiece focal length f_e = 5.0 cm
Object Distant object (normal adjustment)
⚡

The 30-Second Exam Hall Shortcut: ⚡ Normal Adjustment Telescope Shortcut

⏱️ Save ~30 seconds

Magnifying power m = f_o / f_e = 140 / 5.0 = 28!

1 For viewing distant objects in normal adjustment (image formed at infinity), magnifying power of telescope is m = f_o / f_e.
2 Substitute given values: f_o = 140 cm and f_e = 5.0 cm.
3 m = 140 / 5.0 = 28.
4 Option (4) in under 3 seconds!
🛡️
Why You Can 100% Trust This Shortcut: Astronomical Telescope Magnifying Power Formula in Normal Adjustment (m = f_o / f_e).
📐

Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Formula for Magnifying Power of an Astronomical Telescope

For an astronomical refracting telescope viewing a distant celestial object under normal adjustment (relaxed eye, final image at infinity), the angular magnification m is given by the ratio of the focal length of the objective lens to that of the eyepiece lens:

m=fofem = \frac{f_o}{f_e}
Step 2

Substitution and Evaluation

Substituting f_o = 140 cm and f_e = 5.0 cm:

m=140 cm5.0 cm=28m = \frac{140\,\text{cm}}{5.0\,\text{cm}} = 28
⭐

360° Option Elimination Engine Key Feature

Diagnose every choice: Know exactly why the right option works and why distractors fail.

4-Option Deep Scan
1 (1)
✗ Incorrect
Verdict & Scientific Reason:

Arbitrary incorrect value.

⚠️
Student Fallacy (Why candidates pick this trap): Arithmetic miscalculation.
2 (2)
✗ Incorrect
Verdict & Scientific Reason:

Incorrect division.

⚠️
Student Fallacy (Why candidates pick this trap): Rough guessing.
3 (3)
✗ Incorrect
Verdict & Scientific Reason:

Incorrect.

⚠️
Student Fallacy (Why candidates pick this trap): Division slip.
4 (4)
✓ 100% Correct
Verdict & Scientific Reason:

Exact quotient: 140 / 5 = 28.

💡
Why this is airtight: None.
⚠️

NTA Trap Alert: Inverting Objective and Eyepiece

Students who confuse microscope and telescope formulas sometimes divide f_e by f_o, getting a fraction < 1.

🛡️ How to avoid the -1 mark: Telescope OBJECTIVE is always huge (large f_o) to gather light from distant stars; Eyepiece is small (small f_e). Hence m = f_o / f_e >> 1.
📖

Core NEET Syllabus Standard

Physics Core Optics · Ray Optics and Optical Instruments Standard Benchmark
Core Scientific Principle Tested:

"The magnifying power of an astronomical telescope in normal adjustment is m = f_o / f_e."

Syllabus Focus: Refracting Telescopes
✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗
Telescope: Big f_o over tiny f_e: 140 / 5 is 28 to see!

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⚖️ Legal & Fair Dealing Notice: Sourced from publicly released NEET 2024 question papers and NEET answer keys for private study, research, and non-commercial educational practice under Fair Dealing (Section 52(1)(a) of the Indian Copyright Act, 1957). Not affiliated with or endorsed by NTA or NMC.

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