Optical Instruments

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Optical Instruments — Key Facts

• The Microscope: An instrument used to magnify very small objects. The angular magnification (or magnifying power) $M$ of a compound microscope is given by: $$M = \frac{v_o}{u_o} \times \frac{D}{f_e}$$ where $v_o$ = image distance from objective, $u_o$ = object distance from objective, $D$ = near point distance (25 cm), and $f_e$ = focal length of eyepiece.

• The Astronomical Telescope: Used to view distant objects like stars and planets. Its angular magnification in normal adjustment is: $$M = \frac{f_o}{f_e}$$ where $f_o$ = focal length of objective and $f_e$ = focal length of eyepiece.

• The Camera: A converging lens forms a real, inverted, diminished image on a light-sensitive film or sensor. The lens equation applies: $$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$$

• The Human Eye: Acts like a camera. The least distance of distinct vision is the near point, conventionally taken as $D = 25$ cm. The eye can focus on objects at different distances by changing the curvature of the lens — this is accommodation.

• Defects of Vision:

  • Myopia (short sight): image forms in front of retina; corrected by concave (diverging) lens.
  • Hypermetropia (long sight): image would form behind retina; corrected by convex (converging) lens.
  • Presbyopia: loss of accommodation with age; corrected by bifocal lenses.
  • Astigmatism: different focal lengths in different planes; corrected by cylindrical lenses.

⚡ WAEC Exam Tip: Know the difference between linear magnification $m = \frac{v}{u}$ (ratio of image size to object size) and angular magnification $M$ (ratio of angular size of image to angular size of object viewed at the near point). WAEC questions frequently ask for the magnifying power of microscopes and telescopes — be sure to identify which formula applies!

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Optical Instruments — WAEC WASSCE Study Guide

Compound Microscope

A compound microscope consists of two converging lenses: an objective lens (near the object) with a short focal length and an eyepiece (near the eye) with a longer focal length. The object is placed just beyond the focal length of the objective lens, producing a magnified real, inverted image. This image acts as the object for the eyepiece, which functions as a magnifier, producing a further magnified virtual image.

The total magnifying power is the product of the linear magnification of the objective and the angular magnification of the eyepiece:

$$M = m_o \times m_e = \frac{v_o}{u_o} \times \left(1 + \frac{D}{f_e}\right)$$

In normal adjustment (final image at infinity), the eyepiece term simplifies to $\frac{D}{f_e}$.

Astronomical Telescope

In a Keplerian (astronomical) telescope, both lenses are convex. For a relaxed eye (image at infinity, normal adjustment):

$$M = -\frac{f_o}{f_e}$$

The negative sign indicates the image is inverted. If the final image is formed at the near point instead:

$$M = -\frac{f_o}{f_e}\left(1 + \frac{f_e}{D}\right)$$

Astronomical vs Terrestrial Telescope

A terrestrial telescope adds an erecting lens between the objective and eyepiece to produce an upright image. The Galileo telescope uses a diverging eyepiece and produces an upright image directly, but with a narrower field of view.

The Eye as an Optical Instrument

The eye forms an image on the retina using the principles of a converging lens. The ciliary muscles adjust the lens curvature to change its focal length — this is accommodation. The retina contains light-sensitive cells (rods and cones) that transmit signals to the brain via the optic nerve.

The power of accommodation decreases with age, leading to presbyopia. The near point recedes from about 10 cm in childhood to 25 cm at age 40, and further in later life.

Magnifying Glass (Simple Microscope)

A convex lens used as a simple magnifier has angular magnification:

$$M = 1 + \frac{D}{f} \quad \text{(image at near point)}$$ $$M = \frac{D}{f} \quad \text{(image at infinity, normal adjustment)}$$

Comparison Table — Optical Instruments

Instrument	Type of Image	Image Orientation	Final Image Location
Compound microscope	Virtual (via eyepiece)	Inverted	At near point or infinity
Astronomical telescope	Virtual (via eyepiece)	Inverted	At infinity
Terrestrial telescope	Virtual	Upright	At infinity
Camera	Real	Inverted	On film/sensor
Magnifying glass	Virtual	Upright	At near point or infinity

Spectacles and Corrective Lenses

Lens power is measured in dioptres (D): $P = \frac{1}{f}$ (with $f$ in metres). A concave lens has negative power; a convex lens has positive power. For myopia, the corrective lens focal length $f = -x$ metres where $x$ is the far point. For hypermetropia, $f = +y$ metres where $y$ is the near point of the person.

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Optical Instruments — Comprehensive WAEC Physics Notes

Derivation: Magnifying Power of a Compound Microscope

For the objective lens, the linear magnification is:

$$m_o = \frac{v_o}{u_o} = \frac{\text{image height}}{\text{object height}}$$

The image from the objective lies at distance $v_o$ from the objective, and this image is at distance $u_e$ from the eyepiece (since the eyepiece is a distance $L$ from the object, where $L$ is the tube length). For maximum magnification in normal adjustment, $u_e \approx f_e$ and the final image is at infinity, so the eyepiece angular magnification is $m_e = \frac{D}{f_e}$.

Thus total magnification: $M = m_o \times m_e = \frac{v_o}{u_o} \times \frac{D}{f_e}$.

If the intermediate image is not at the focal point of the eyepiece, the more general formula applies:

$$M = \frac{v_o}{u_o}\left(1 + \frac{D}{f_e}\right)$$

Derivation: Magnifying Power of an Astronomical Telescope

For a telescope in normal adjustment (final image at infinity), the object is effectively at infinity, so the objective forms its image at its focal point: $v_o = f_o$. This image is at the focal point of the eyepiece, so $u_e = f_e$. The angular magnification is the ratio of the angles subtended at the eye by the image and by the unassisted object:

$$\theta_o \approx \frac{h}{f_o} \quad \text{and} \quad \theta_e \approx \frac{h}{f_e}$$

Therefore:

$$M = \frac{\theta_e}{\theta_o} = \frac{f_o}{f_e}$$

Resolving Power of a Microscope

The resolving power is the ability to distinguish two close objects. The minimum separation $d$ that can be resolved is given by:

$$d = \frac{1.22\lambda}{2\mu\sin\alpha}$$

where $\lambda$ = wavelength of light, $\mu$ = refractive index of the medium between object and objective, and $\alpha$ = half-angle of the cone of light from the object (the aperture angle). The term $\mu\sin\alpha$ is the numerical aperture.

The Eye and Its Defects — Detailed Treatment

Defect	Cause	Symptoms	Correction	Lens Type
Myopia	Eyeball too long or lens too converging	Cannot see distant objects clearly	Diverging/concave lens	Negative power
Hypermetropia	Eyeball too short or lens too flat	Difficulty with near objects	Converging/convex lens	Positive power
Astigmatism	Cornea not spherical	Lines in one direction blurred	Cylindrical lens	Toric surface
Presbyopia	Loss of ciliary muscle elasticity	Near objects hard to read	Convex lens for reading	Bifocal/progressive

Lens Formula and Sign Convention (Cartesian Convention)

Using the Cartesian sign convention (light travels left to right):

• Object distance $u$: positive if object is on the incoming light side (real object, $u$ is negative in the formula below)
• Image distance $v$: positive if image is on the outgoing light side (real image)
• Focal length $f$: positive for converging lens, negative for diverging lens

$$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$$

WAEC Past Question Patterns

Typical WAEC questions on optical instruments include:

1. Calculating magnifying power given focal lengths and tube length
2. Identifying defects of vision and suggesting corrections
3. Drawing ray diagrams for compound microscope and telescope
4. Calculating the power of corrective lenses in dioptres
5. Comparing linear and angular magnification

⚡ WAEC Exam Tip: When drawing ray diagrams for a compound microscope, remember the intermediate image is real, inverted, and magnified. For the astronomical telescope, the intermediate image is also real and inverted. In normal adjustment, the final image is at infinity. Always state the sign convention clearly in numerical problems!

Numerical Worked Example (WAEC-style)

A compound microscope has an objective of focal length 1.0 cm and an eyepiece of focal length 5.0 cm, separated by a tube length of 20 cm. Find the magnifying power when the final image is at the near point (D = 25 cm).

Solution: For objective: $f_o = 1.0$ cm, $v_o = 20 - 5 = 15$ cm (since eyepiece focal length is 5 cm) Using $\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$ for the objective: $\frac{1}{1} = \frac{1}{u_o} + \frac{1}{15}$ → $u_o \approx 1.07$ cm $m_o = \frac{v_o}{u_o} = \frac{15}{1.07} \approx 14.0$

Eyepiece magnification: $m_e = 1 + \frac{D}{f_e} = 1 + \frac{25}{5} = 6$ Total $M = 14 \times 6 = 84$

Answer: Magnifying power ≈ 84

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