Optical Instruments - Extending Human Vision
The Movie Hook ๐ฌ
Remember how Interstellar uses telescope shots to view distant planets? Or the microscope scenes in Ant-Man showing quantum realm? Or how Tony Stark’s glasses in Spider-Man have augmented reality? All these instruments - telescopes, microscopes, eyeglasses - use the same optics principles we’ll master today!
The Big Picture
Optical instruments extend our vision beyond natural limits:
- Microscopes: See the very small (cells, bacteria)
- Telescopes: See the very far (stars, planets)
- Human eye: Nature’s perfect camera
- Cameras: Capturing light permanently
Connection Alert: Uses Thin Lens, Reflection, and Diffraction.
1. The Human Eye
Structure
Components:
- Cornea: Most refraction happens here (n โ 1.38)
- Aqueous humor: Liquid between cornea and lens
- Lens: Fine-tunes focus (variable focal length!)
- Retina: Light-sensitive screen (like camera sensor)
- Optic nerve: Transmits signals to brain
Focal length range:
- Relaxed (far objects): f โ 2.5 cm
- Maximum accommodation (near point): f โ 2.3 cm
Key Parameters
Near point (D): Closest distance for clear vision
- Normal eye: D = 25 cm (standard value for JEE!)
- Symbol: D or d
Far point: Farthest distance for clear vision
- Normal eye: Infinity
- Myopic eye: Finite distance
Range of vision: Distance between near point and far point
Power of Accommodation
Accommodation: Eye’s ability to change focal length by changing lens shape
Power change:
$$\boxed{\Delta P = P_{\text{near}} - P_{\text{far}}}$$For normal eye viewing at 25 cm to infinity:
$$P_{\text{near}} = \frac{100}{2.3} \approx 43.5 \text{ D}$$ $$P_{\text{far}} = \frac{100}{2.5} = 40 \text{ D}$$ $$\Delta P \approx 4 \text{ D}$$Defects of Vision
1. Myopia (Near-sightedness)
Problem: Can see near objects clearly, but not distant objects
- Far point at finite distance (not infinity)
- Eye too long OR lens too strong
Correction: Concave lens (diverging)
$$\boxed{P = -\frac{100}{d}}$$Where d = far point distance in cm
Example: Far point at 50 cm โ P = -100/50 = -2 D
2. Hypermetropia (Far-sightedness)
Problem: Can see distant objects clearly, but not near objects
- Near point > 25 cm
- Eye too short OR lens too weak
Correction: Convex lens (converging)
$$\boxed{P = \frac{100}{25} - \frac{100}{d}}$$Where d = actual near point in cm
Example: Near point at 50 cm โ P = 4 - 2 = 2 D
3. Presbyopia
Problem: Loss of accommodation with age
- Both near and far vision affected
- Lens becomes less flexible
Correction: Bifocal lens
- Upper part: for distant vision
- Lower part: for near vision
4. Astigmatism
Problem: Unequal curvature of cornea in different directions
- Blurred vision at all distances
Correction: Cylindrical lens
2. Simple Microscope (Magnifying Glass)
Setup
- Single convex lens
- Object placed between F and optical center
- Forms virtual, erect, magnified image
Magnifying Power
When image at near point (D = 25 cm):
$$\boxed{M = 1 + \frac{D}{f}}$$When image at infinity (relaxed eye):
$$\boxed{M = \frac{D}{f}}$$Where:
- M = magnifying power (angular magnification)
- D = near point distance = 25 cm
- f = focal length of lens (in cm)
Memory Trick: “Define Magnification with Focal length” (M = D/f or 1 + D/f)
Comparison
| Condition | Formula | Use |
|---|---|---|
| Image at D (25 cm) | M = 1 + D/f | Maximum magnification |
| Image at โ | M = D/f | Relaxed viewing (comfortable) |
JEE Trick: Unless specified, assume image at infinity (relaxed eye)!
3. Compound Microscope
Setup
Two convex lenses:
- Objective lens (O): Short focal length (f_o)
- Eyepiece lens (E): Short focal length (f_e)
Working:
- Object just beyond f_o of objective
- Forms real, inverted, magnified image
- This acts as object for eyepiece
- Eyepiece acts as simple magnifier
- Final image: virtual, inverted (relative to object), highly magnified
Magnifying Power
When final image at D (25 cm):
$$\boxed{M = \frac{v_o}{u_o} \times \left(1 + \frac{D}{f_e}\right)}$$When final image at infinity (normal adjustment):
$$\boxed{M = \frac{v_o}{u_o} \times \frac{D}{f_e}}$$For small focal lengths (approximation):
$$\boxed{M \approx \frac{L \cdot D}{f_o \cdot f_e}}$$Where:
- L = tube length (distance between f_o and f_e)
- v_o = image distance from objective
- u_o = object distance from objective
Length of Microscope
$$\boxed{L = v_o + u_e}$$For normal adjustment (image at โ):
$$u_e = f_e$$So:
$$\boxed{L = v_o + f_e}$$Total magnification:
$$M = m_o \times M_e$$Where:
- m_o = magnification by objective = v_o/|u_o|
- M_e = magnification by eyepiece = D/f_e (at infinity)
4. Astronomical Telescope
Setup
Two convex lenses:
- Objective (O): Large focal length (f_o), large aperture
- Eyepiece (E): Small focal length (f_e)
Working:
- Collects light from distant object
- Objective forms real, inverted image at its focus
- Eyepiece magnifies this image
- Final image: virtual, inverted, magnified
Magnifying Power
Normal adjustment (image at infinity):
$$\boxed{M = -\frac{f_o}{f_e}}$$Negative sign indicates inverted image.
When image at D (25 cm):
$$\boxed{M = -\frac{f_o}{f_e}\left(1 + \frac{f_e}{D}\right)}$$Length of Telescope
Normal adjustment:
$$\boxed{L = f_o + f_e}$$When image at D:
$$\boxed{L = f_o + u_e}$$Where u_e can be found using lens formula for eyepiece.
Resolving Power
Rayleigh’s criterion: Two point sources are just resolved when central maximum of one coincides with first minimum of other.
For telescope:
$$\boxed{R = \frac{D}{1.22\lambda}}$$Where:
- D = diameter of objective
- ฮป = wavelength of light
Minimum angular separation:
$$\boxed{\theta_{\min} = \frac{1.22\lambda}{D}}$$Key insight: Larger diameter โ better resolution!
5. Terrestrial Telescope
Problem with astronomical telescope: Image is inverted (upside down)
Solution: Add erecting lens system
- Additional convex lens between objective and eyepiece
- OR use prisms (like in binoculars)
Length: L = f_o + 4f + f_e
Where f = focal length of erecting lens
Magnification: Same as astronomical telescope (M = -f_o/f_e)
6. Galilean Telescope
Different design:
- Objective: Convex lens (long f_o)
- Eyepiece: Concave lens (short |f_e|)
Advantage: Erect image (no inversion!)
Magnifying power:
$$\boxed{M = \frac{f_o}{|f_e|}}$$Length:
$$\boxed{L = f_o - |f_e|}$$Disadvantage: Smaller field of view
Application: Opera glasses, Galileo’s original telescope
7. Reflecting Telescope
Why use mirrors?
- No chromatic aberration
- Can make larger diameter (better resolution)
- Lighter than lenses for same aperture
Types:
- Newtonian: Plane mirror deflects light to side
- Cassegrain: Convex secondary mirror
- Prime focus: Direct focus (for large telescopes)
Advantage over refracting:
- Only one surface to polish
- Can support from back
- No chromatic aberration
The Formula Cheat Sheet
OPTICAL INSTRUMENTS ESSENTIALS:
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SIMPLE MICROSCOPE:
Image at โ: M = D/f
Image at D: M = 1 + D/f
COMPOUND MICROSCOPE:
Normal: M = (v_o/u_o) ร (D/f_e)
Approx: M โ LD/(f_o ร f_e)
Length: L = v_o + f_e
TELESCOPE:
Magnification: M = -f_o/f_e
Length: L = f_o + f_e
Resolving: R = D/(1.22ฮป)
EYE DEFECTS:
Myopia: P = -100/d (concave)
Hypermetropia: P = 100/25 - 100/d (convex)
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Common Traps & Mistakes
Trap #1: Magnification Signs
โ Wrong: Forgetting negative sign in telescope magnification โ Right: M = -f_o/f_e (negative means inverted)
Trap #2: Normal Adjustment
โ Wrong: Assuming image always at D = 25 cm โ Right: “Normal adjustment” means final image at infinity!
Trap #3: Microscope vs Telescope
โ Wrong: Using telescope formula for microscope โ Right: Microscope: M โ LD/(f_of_e), Telescope: M = f_o/f_e
Trap #4: Eye Correction
โ Wrong: Using positive power for myopia โ Right: Myopia needs diverging (concave) lens โ negative power
Practice Problems
Level 1: JEE Main Warmup
Problem 1.1: A simple microscope has focal length 5 cm. Find magnifying power when image is formed at (a) infinity (b) 25 cm from eye.
Solution
Given: f = 5 cm, D = 25 cm
(a) Image at infinity:
$$M = \frac{D}{f} = \frac{25}{5} = 5$$(b) Image at 25 cm (near point):
$$M = 1 + \frac{D}{f} = 1 + \frac{25}{5} = 1 + 5 = 6$$Answers:
- (a) M = 5 (relaxed viewing)
- (b) M = 6 (maximum magnification)
Problem 1.2: A person can see clearly up to 50 cm. What power lens should be used to correct this defect?
Solution
Far point = 50 cm (should be infinity) This is myopia (near-sightedness)
Correction: Concave lens to make object at infinity appear at 50 cm
$$P = -\frac{100}{d} = -\frac{100}{50} = -2 \text{ D}$$Answer: -2 D (concave lens)
Level 2: JEE Main/Advanced
Problem 2.1: A compound microscope has objective of focal length 2 cm and eyepiece of focal length 5 cm. The distance between them is 20 cm. Find magnifying power when final image is at infinity.
Solution
Given: f_o = 2 cm, f_e = 5 cm, Length = 20 cm, D = 25 cm
For image at infinity, eyepiece must have object at its focus:
$$u_e = f_e = 5 \text{ cm}$$Length of microscope:
$$L = v_o + |u_e|$$ $$20 = v_o + 5$$ $$v_o = 15 \text{ cm}$$Using lens formula for objective:
$$\frac{1}{v_o} - \frac{1}{u_o} = \frac{1}{f_o}$$ $$\frac{1}{15} - \frac{1}{u_o} = \frac{1}{2}$$ $$\frac{1}{u_o} = \frac{1}{15} - \frac{1}{2} = \frac{2-15}{30} = -\frac{13}{30}$$Magnification by objective:
$$m_o = \frac{v_o}{|u_o|} = \frac{15}{30/13} = \frac{15 \times 13}{30} = 6.5$$Magnification by eyepiece (at infinity):
$$M_e = \frac{D}{f_e} = \frac{25}{5} = 5$$Total magnification:
$$M = m_o \times M_e = 6.5 \times 5 = 32.5$$Answer: M = 32.5
Problem 2.2: An astronomical telescope has objective of focal length 100 cm and eyepiece of focal length 5 cm. Find: (a) Magnifying power in normal adjustment (b) Length of telescope (c) Distance between objective and eyepiece when final image is at 25 cm from eye
Solution
Given: f_o = 100 cm, f_e = 5 cm, D = 25 cm
(a) Magnifying power (normal = image at โ):
$$M = -\frac{f_o}{f_e} = -\frac{100}{5} = -20$$Magnitude: |M| = 20
(b) Length in normal adjustment:
$$L = f_o + f_e = 100 + 5 = 105 \text{ cm}$$(c) When image at D = 25 cm:
For eyepiece, v_e = -25 cm (virtual image on same side)
Using lens formula:
$$\frac{1}{v_e} - \frac{1}{u_e} = \frac{1}{f_e}$$ $$\frac{1}{-25} - \frac{1}{u_e} = \frac{1}{5}$$ $$\frac{1}{u_e} = -\frac{1}{25} - \frac{1}{5} = -\frac{1+5}{25} = -\frac{6}{25}$$ $$u_e = -\frac{25}{6} = -4.17 \text{ cm}$$Length:
$$L = f_o + |u_e| = 100 + 4.17 = 104.17 \text{ cm}$$Answers:
- (a) M = 20 (inverted)
- (b) L = 105 cm
- (c) L = 104.17 cm (slightly less than normal adjustment)
Level 3: JEE Advanced
Problem 3.1: A person has near point at 100 cm and far point at 200 cm. What power lenses should be used for: (a) Reading at 25 cm (b) Seeing distant objects clearly
Solution
Near point = 100 cm (should be 25 cm for reading) Far point = 200 cm (should be infinity for distant objects)
(a) For reading (correct near point):
Lens should make object at 25 cm appear at 100 cm (person’s near point)
Using lens formula:
$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$u = -25 cm (object), v = -100 cm (image on same side, virtual)
$$\frac{1}{-100} - \frac{1}{-25} = \frac{1}{f}$$ $$\frac{1}{f} = -\frac{1}{100} + \frac{1}{25} = \frac{-1+4}{100} = \frac{3}{100}$$ $$f = \frac{100}{3} = 33.33 \text{ cm} = 0.333 \text{ m}$$ $$P = \frac{1}{0.333} = +3 \text{ D}$$(convex)
(b) For distant objects:
Lens should make object at infinity appear at 200 cm
u = -โ, v = -200 cm
$$\frac{1}{-200} - \frac{1}{-\infty} = \frac{1}{f}$$ $$\frac{1}{f} = -\frac{1}{200}$$ $$f = -200 \text{ cm} = -2 \text{ m}$$ $$P = \frac{1}{-2} = -0.5 \text{ D}$$(concave)
Answers:
- (a) +3 D (convex, for reading)
- (b) -0.5 D (concave, for distance)
Person needs bifocals!
Problem 3.2: The diameter of the objective of a telescope is 0.1 m. What is the resolving power for light of wavelength 5000 ร ?
Solution
Given: D = 0.1 m, ฮป = 5000 ร = 5000 ร 10โปยนโฐ m = 5 ร 10โปโท m
Resolving power:
$$R = \frac{D}{1.22\lambda} = \frac{0.1}{1.22 \times 5 \times 10^{-7}}$$ $$R = \frac{0.1}{6.1 \times 10^{-7}} = \frac{10^{-1}}{6.1 \times 10^{-7}} = \frac{10^6}{6.1} = 1.64 \times 10^5$$Answer: R = 1.64 ร 10โต
This means telescope can resolve two stars separated by angle:
$$\theta_{\min} = \frac{1}{R} = \frac{1}{1.64 \times 10^5} = 6.1 \times 10^{-6} \text{ rad}$$Quick Revision Cards
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โ OPTICAL INSTRUMENTS FACTS โ
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HUMAN EYE:
๐ฏ Normal near point: D = 25 cm
๐ฏ Normal far point: Infinity
๐ฏ Myopia: Concave lens (P < 0)
๐ฏ Hypermetropia: Convex lens (P > 0)
SIMPLE MICROSCOPE:
๐ฏ M = D/f (at โ), M = 1+D/f (at D)
COMPOUND MICROSCOPE:
๐ฏ Two convex lenses (short f)
๐ฏ M โ LD/(f_o ร f_e)
๐ฏ Final image: inverted
TELESCOPE:
๐ฏ M = -f_o/f_e (astronomical)
๐ฏ L = f_o + f_e (normal adjustment)
๐ฏ R = D/(1.22ฮป) (resolving power)
Cross-Topic Connections
- Thin Lens: Lens formula for all instruments
- Reflection: Reflecting telescopes
- Diffraction: Resolving power limitation
- Wave Optics: Rayleigh criterion
Final Tips
- Normal adjustment = Final image at infinity (most comfortable)
- Simple microscope: M = D/f (easy to remember!)
- Telescope: Objective has LONG f, eyepiece has SHORT f
- Microscope: Both have SHORT f
- Myopia โ Concave, Hypermetropia โ Convex
Pro Tip: In JEE, unless stated otherwise, assume normal adjustment (image at โ)!
Next up: Wave Optics - Huygens principle and wavefronts!