Astronomical Telescope: Bringing the Stars Closer - Working and Formula
Master the working principle, ray diagram, and derivation of magnifying power in normal adjustment for JEE Physics.
Why Astronomical Telescope is Crucial for JEE
The astronomical telescope is a fundamental optical instrument that frequently appears in JEE Physics questions. Understanding its working and formulas helps in:
- Direct formula-based questions in JEE Main
- Derivation and conceptual questions in JEE Advanced
- Ray diagram drawing and analysis
- Comparative studies with other optical instruments
Basic Construction of Astronomical Telescope
| Component | Function | Specification |
|---|---|---|
| Objective Lens | Large aperture, large focal length converging lens | $f_o$ = 100-200 cm |
| Eyepiece | Small aperture, small focal length converging lens | $f_e$ = 5-10 cm |
| Tube | Holds both lenses coaxially | Length ≈ $f_o + f_e$ |
🎯 Key Points:
• Objective lens has large aperture to gather more light from distant objects
• Objective lens has large focal length to produce real, inverted image
• Eyepiece acts as a simple microscope to view the image formed by objective
• Final image is inverted - not suitable for terrestrial observations
Working Principle in Normal Adjustment
📐 Ray Diagram - Normal Adjustment:
Working Steps:
Step 1: Parallel rays from distant object fall on objective lens
Step 2: Objective forms real, inverted image at its focal plane
Step 3: This image lies at the focus of eyepiece
Step 4: Eyepiece acts as magnifying lens, forms virtual image at infinity
Step 5: Final image is inverted, magnified, and at infinity
Magnifying Power in Normal Adjustment
📐 Derivation of Magnifying Power:
Step 1: Magnifying power is defined as:
$M = \frac{\text{Angle subtended by final image at eye}}{\text{Angle subtended by object at unaided eye}}$
Step 2: When final image is at infinity:
$\beta = \frac{h}{f_e}$ (angle subtended by final image)
Step 3: Angle subtended by object at unaided eye:
$\alpha = \frac{h}{f_o}$
Step 4: Therefore, magnifying power:
$M = \frac{\beta}{\alpha} = \frac{h/f_e}{h/f_o} = \frac{f_o}{f_e}$
✓ Final Formula: $M = -\frac{f_o}{f_e}$ (negative sign indicates inverted image)
🎯 JEE Application Example:
Problem: An astronomical telescope has objective of focal length 100 cm and eyepiece of focal length 5 cm. Find magnifying power in normal adjustment.
Solution: Using the formula:
$M = \frac{f_o}{f_e} = \frac{100}{5} = 20$
The telescope magnifies 20 times. Negative sign indicates inverted image.
Length of Telescope in Normal Adjustment
📐 Explanation:
• In normal adjustment, final image is at infinity
• Image formed by objective lies at its focus ($f_o$ from objective)
• This image acts as object for eyepiece at its focus ($f_e$ from eyepiece)
• Therefore, distance between lenses = $f_o + f_e$
• This is the minimum length of telescope tube
🚀 Problem-Solving Strategies
Key Points to Remember:
- Normal adjustment = Final image at infinity
- $M = f_o/f_e$ (ignore negative sign for magnitude)
- Length $L = f_o + f_e$ in normal adjustment
- Image is always inverted in astronomical telescope
JEE Exam Tips:
- Practice ray diagram drawing
- Remember special cases and variations
- Understand when to use which formula
- Compare with terrestrial telescope
Advanced Concepts Available
Includes telescope when final image is at near point, resolving power, and JEE Advanced level problems
📝 Quick Self-Test
Try these JEE-level problems to test your understanding:
1. An astronomical telescope has magnifying power 10 in normal adjustment. If focal length of eyepiece is 5 cm, find focal length of objective.
2. Draw a labeled ray diagram of astronomical telescope in normal adjustment.
3. Why is the image formed by astronomical telescope inverted? Is this a disadvantage for astronomical observations?
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