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JEE Physics Focus Reading Time: 14 min 2 Key Formulas

The Physics of a Simple Microscope: Magnifying Power and Why It Works

Complete derivation of magnifying power formula for convex lens with ray diagrams and JEE applications.

2
Key Cases
100%
JEE Relevance
4+
Ray Diagrams
18min
Practice Time

Why Understanding Simple Microscope is Crucial

The simple microscope (single convex lens) forms the foundation of optical instruments in JEE Physics. Key concepts include:

  • Magnifying Power Definition - Angular magnification concept
  • Image Formation - Virtual, erect, and magnified images
  • Lens Formula Application - Direct application of $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
  • Visual Angle Concept - Fundamental to understanding magnification

🔍 What is a Simple Microscope?

A simple microscope uses a single convex lens of short focal length to magnify small objects.

  • Object placed between focus and optical center
  • Produces virtual, erect, and magnified image
  • Used as a magnifying glass
  • Foundation for compound microscopes
[Ray Diagram: Object between F and O]

Object placed between focus and lens

Case 1 Essential

Normal Adjustment (Image at Infinity)

$$M = \frac{D}{f}$$

Where $D$ = Least distance of distinct vision (25 cm), $f$ = focal length

📐 Derivation:

Step 1: Magnifying power is defined as:

$M = \frac{\text{Visual angle with instrument}}{\text{Visual angle without instrument}}$

Step 2: Without instrument (maximum angle when object at D):

$\alpha = \frac{h}{D}$

Step 3: With microscope (image at infinity, object at focus):

$\beta = \frac{h}{f}$

Step 4: Therefore, magnifying power:

$M = \frac{\beta}{\alpha} = \frac{h/f}{h/D} = \frac{D}{f}$

Ray Diagram (Normal Adjustment):

[Parallel rays emerging from lens - image at infinity]

Object at focus F, image at infinity

🎯 JEE Application Example:

Problem: A simple microscope has focal length 5 cm. Find its magnifying power for normal adjustment.

Solution: Using the formula with D = 25 cm:

$M = \frac{D}{f} = \frac{25}{5} = 5$

The microscope magnifies objects 5 times when image is at infinity.

Case 2 Important

Relaxed Vision (Image at Distance D)

$$M = 1 + \frac{D}{f}$$

Maximum magnification when image is at least distance of distinct vision

📐 Derivation Using Lens Formula:

Step 1: Lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$

Step 2: For relaxed vision: $v = -D$ (virtual image)

$\frac{1}{f} = \frac{1}{-D} - \frac{1}{u}$

Step 3: Solve for u:

$\frac{1}{u} = -\frac{1}{D} - \frac{1}{f} = -\left(\frac{1}{D} + \frac{1}{f}\right)$

$u = -\frac{Df}{D + f}$

Step 4: Linear magnification $m = \frac{v}{u}$:

$m = \frac{-D}{-\frac{Df}{D+f}} = \frac{D+f}{f} = 1 + \frac{D}{f}$

Step 5: For small angles, angular magnification = linear magnification:

$M = 1 + \frac{D}{f}$

Ray Diagram (Relaxed Vision):

[Rays appearing to come from point at distance D]

Virtual image formed at distance D from eye

🎯 JEE Application Example:

Problem: Same microscope (f=5 cm) used for relaxed vision. Find magnification.

Solution: Using the relaxed vision formula:

$M = 1 + \frac{D}{f} = 1 + \frac{25}{5} = 1 + 5 = 6$

Higher magnification achieved when image is at least distance D.

🚀 Problem-Solving Strategies

Key Points to Remember:

  • Normal adjustment: $M = \frac{D}{f}$ (eye relaxed)
  • Relaxed vision: $M = 1 + \frac{D}{f}$ (maximum magnification)
  • D = 25 cm for normal human eye
  • Shorter focal length → Higher magnification

JEE Exam Tips:

  • Always specify which case is being considered
  • Use sign convention consistently
  • Remember D is positive (25 cm)
  • Practice numericals with different focal lengths

📊 Comparison: Normal vs Relaxed Vision

Parameter Normal Adjustment Relaxed Vision
Image Position At infinity At distance D
Eye Condition Relaxed Strained
Magnifying Power $\frac{D}{f}$ $1 + \frac{D}{f}$
Usage Prolonged viewing Maximum magnification

Advanced Applications Available

Includes compound microscope, astronomical telescope, and JEE Advanced level problems

📝 Quick Self-Test

Try these JEE-level problems to test your understanding:

1. A simple microscope of focal length 10 cm is used. Calculate magnifying power for both cases.

2. Why is the magnifying power for relaxed vision greater than for normal adjustment?

3. Derive the formula $M = 1 + \frac{D}{f}$ using visual angle approach.

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