Optical Instruments

For a simple microscope in normal adjustment (image at infinity): $m = \frac{D}{f}$

Where:

• D is the least distance of distinct vision (typically 25 cm)
• f is the focal length of the microscope lens

Key insight: Magnifying power increases as focal length decreases, which is why microscopes use lenses with very short focal lengths.

Note: This represents angular magnification - the ratio of the angle subtended by the image when viewed through the microscope to the angle subtended by the object at the near point.

For compound microscope:

Normal adjustment (image at infinity): $m = \frac{v}{u}\frac{D}{f_e} \approx -\frac{l}{f_o}\frac{D}{f_e}$

Image at near point: $m = \frac{v}{u}\left(1+\frac{D}{f_e}\right) \approx -\frac{l}{f_o}\left(1+\frac{D}{f_e}\right)$

Where:

• l = tube length
• f₀ = focal length of objective
• f₉ = focal length of eyepiece
• D = least distance of distinct vision

The difference (the added term D/f₉) occurs because:

• At near point, the eyepiece forms a virtual image at distance D
• This produces greater angular magnification due to increased angular subtension
• The eye must accommodate to view this closer image

The sign difference in magnifying power relates directly to image orientation:

Astronomical telescope:

• Uses two converging lenses (positive focal lengths)
• Magnifying power: $m = -\frac{f_o}{f_e}$ (negative)
• Produces inverted final image
• Negative sign occurs because angles β and α are on opposite sides of axis

Galilean telescope:

• Uses converging objective and diverging eyepiece
• Magnifying power: $m = -\frac{f_o}{f_e}$ (positive, since f₉ is negative)
• Produces erect final image
• Angular subtension maintains same sign because rays don't cross axis

The diverging eyepiece in Galilean design intercepts rays before they form a real image, preserving orientation but reducing field of view.

Accommodation process:

• Ciliary muscles control lens curvature and effective focal length
• Relaxed muscles → maximum focal length → distant objects
• Strained muscles → decreased focal length → closer objects
• Near point (D): closest distance for clear vision

Relationship to age:

• Young eyes (under 10): near point as close as 7-8 cm
• Normal adult: near point around 25 cm
• Elderly: near point shifts to 1-2 m or more

Age-related changes:

• Presbyopia: hardening of lens with age reduces accommodation ability
• Ciliary muscles cannot effectively increase lens curvature
• Near point gradually recedes away from eye

Correction:

• Converging lenses (reading glasses, bifocals)
• Focal length selected to form virtual image of close objects at person's near point
• Formula: $\frac{1}{f} = \frac{1}{D_{desired}} - \frac{1}{D_{actual}}$
• Creates an optical system that compensates for diminished accommodation

When viewing objects:

• With naked eye at near point: maximum visual angle θ₀ = h/D (where h is object height, D is near point distance ~25 cm)
• With simple microscope (normal adjustment): angle θ = h/f (where f is focal length)

The magnifying power equals θ/θ₀ = D/f

This shows that using a converging lens with focal length less than the eye's near point distance creates greater angular magnification, making objects appear larger by increasing the visual angle.

Key insight: The magnification increases as focal length decreases, with maximum theoretical magnification of D/f.

The angle θ' is crucial for understanding angular magnification in a compound microscope:

• θ' represents the angle subtended by the first image at the eyepiece
• This angle determines how large the final image appears to the eye
• Angular magnification (m) compares θ' to the angle θo subtended by the object when viewed at the near point

Key relationships:

• θ' = h'/ue (where h' is first image height, ue is distance from eyepiece)
• θo = h/D (where h is object height, D is near point distance)
• Angular magnification: m = θ'/θo = (h'/ue)/(h/D) = (h'/h)(D/ue)

Since h'/h = v/u (linear magnification by objective), the total magnification becomes: m = (v/u)(D/ue)

The larger θ' becomes through proper positioning of lenses, the greater the magnifying power of the microscope.

• The magnifying power formula is: $m = -\frac{f_o}{f_e}$

Where:

• $f_o$ = focal length of the objective lens
• $f_e$ = focal length of the eyepiece
• The negative sign indicates the image is inverted

This represents the ratio of the angle subtended by the final image to the angle subtended by the object when viewed by unaided eye.

Example: A telescope with 100 cm objective focal length and 2 cm eyepiece focal length has magnifying power of -50, meaning objects appear 50 times larger but inverted.

• Best used when:

  1. Compact design is required (opera glasses, binoculars)
  2. Upright image is necessary (terrestrial viewing)
  3. Wide field of view isn't critical

• Key mathematical relationships:

  1. Length = fo + fe (where fe is negative, making it shorter than fo)
  2. For normal adjustment: magnification = -fo/fe
  3. Field of view is limited by: FOV ≈ |fe|/fo × (diameter of eyepiece)
  4. Resolving power: R = 1/Δθ = a/(1.22λ) where a is objective diameter

• Limitations:

  1. Lower magnification than comparable astronomical telescopes
  2. Narrower field of view due to diverging eyepiece
  3. Chromatic aberration increases with magnification
  4. Difficulty achieving high magnification while maintaining clarity

• Example application: Opera glasses typically use Galilean design with magnification of 2-4×

Simple Microscope (Magnifying Glass)

Magnifying power depends on final image position:

1. Image at infinity (normal adjustment):

   • Object placed at focal point
   • Magnifying power: $m = \frac{D}{f}$
   • Where D = near point distance, f = focal length

2. Image at near point:

   • Object placed closer than focal point
   • Using lens equation: $\frac{1}{u_o} = \frac{1}{f} - \frac{1}{D}$
   • Magnifying power: $m = \frac{D}{u_o} = 1 + \frac{D}{f}$
   • Higher magnification than normal adjustment by factor of 1

Compound Microscope

Magnifying power calculated as:

1. Image at infinity:

   • $m \approx -\frac{L}{f_o}\frac{D}{f_e}$
   • Where L = tube length, $f_o$ = objective focal length, $f_e$ = eyepiece focal length

2. Image at near point:

   • $m \approx -\frac{L}{f_o}\left(1+\frac{D}{f_e}\right)$
   • Higher magnification than normal adjustment

Key Principles

• Higher magnification always achieved with image at near point
• Trade-off: Near point viewing causes eye strain (accommodation required)
• Eyepiece position: Must move closer to objective when switching from infinity to near point viewing
• Negative sign: Indicates inverted final image in compound microscope

Example (Compound Microscope)

For $f_o = 8$ mm, $f_e = 25$ mm, L = 160 mm, D = 250 mm:

• Magnification (infinity): -200×
• Magnification (near point): -220×
• Eyepiece must move ~2.3 mm closer to objective for near point viewing

Uncorrected Vision (Natural Eye)

The nearest point of clear vision depends on:

$\frac{1}{u_{min}} = \frac{1}{f_{min}} - \frac{1}{v_o}$

Where:

• $u_{min}$ = minimum distance for clear vision (near point)
• $f_{min}$ = minimum focal length achievable by ciliary muscles
• $v_o$ = fixed distance from lens to retina

For normal vision, $u_{min}$ should be approximately 25 cm or less.

Corrective Lens Calculation

To determine the focal length ($f_{lens}$) needed to correct vision to a desired minimum distance:

1. The corrective lens must create a virtual image at the person's actual near point ($d_{near}$) when viewing an object at the desired distance ($u_{min}$)

2. Using the lens equation and accounting for virtual image formation: $\frac{1}{f_{lens}} = \frac{1}{u_{min}} - \frac{1}{d_{near}}$

3. Solving for the required focal length: $f_{lens} = \frac{u_{min} \times d_{near}}{d_{near} - u_{min}}$

Example: If a person's near point is 100 cm (presbyopia) and they want to read at 25 cm, the required focal length would be: $f_{lens} = \frac{25 \times 100}{100 - 25} = \frac{2500}{75} = 33.3 \text{ cm}$

Note: The corrective lens power in diopters would be $P = 1/f_{lens}$ (measured in meters), so $P = 1/0.333 = 3$ diopters.