12.4 Optical Instruments: Magnifiers, Microscopes, Telescopes, Spectrometer & Fibre Optics

Optical Instruments: Magnifiers, Microscopes, Telescopes, Spectrometer & Fibre Optics

Optical instruments use lenses and mirrors to make small or distant objects appear larger, or to analyse light. Questions on this topic are mostly formula-based and fast, provided the sign conventions and the meaning of "magnifying power" are clear.

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1. Least Distance of Distinct Vision and Angular Magnification

The near point of a normal eye is about $d = 25\text{ cm}$; this is the least distance of distinct vision. How large an object looks depends on the angle it subtends at the eye, not on its actual size.

The magnifying power (angular magnification) of an instrument is

$M = \frac{\beta}{\alpha} = \frac{\text{angle subtended by the image at the eye}}{\text{angle subtended by the object at the near point (unaided)}}$

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2. The Simple Microscope (Magnifying Glass)

A convex lens of focal length $f$ is held close to the eye with the object inside its focal length. It forms a virtual, erect, enlarged image.

• Image at the near point (maximum magnification): $M = 1 + \frac{d}{f}$
• Image at infinity (relaxed eye): $M = \frac{d}{f}$

Example: a lens with $f = 5\text{ cm}$ gives $M = 1 + \frac{25}{5} = 6$ with the image at the near point. A shorter focal length (more powerful lens) gives larger magnification.

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3. The Compound Microscope

Two converging lenses are used:

• The objective (short focal length $f_o$) forms a real, inverted, magnified image of an object placed just beyond its focal point.
• The eyepiece (short focal length $f_e$) acts as a simple magnifier for that image.

The total magnifying power is the product of the two stages. With the final image at the near point,

$M = \frac{q}{p}\left(1 + \frac{d}{f_e}\right) \approx \frac{L}{f_o}\left(1 + \frac{d}{f_e}\right)$

where $L$ is approximately the distance between the lenses. The final image is virtual and inverted relative to the object.

Example: $f_o = 1\text{ cm}$, $f_e = 5\text{ cm}$, $L = 16\text{ cm}$: $M \approx \frac{16}{1}\left(1 + \frac{25}{5}\right) = 16 \times 6 = 96$.

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4. The Astronomical (Refracting) Telescope

The objective has a long focal length $f_o$ and forms a real, inverted image of a distant object at its focal plane. The eyepiece has a short focal length $f_e$ and magnifies that image.

In normal adjustment the final image is at infinity:

$M = \frac{f_o}{f_e}, \qquad \text{length of telescope} = f_o + f_e$

Example: $f_o = 100\text{ cm}$ and $f_e = 5\text{ cm}$ give $M = 20$ and a tube length of 105 cm. The final image is inverted, which does not matter for viewing stars; a terrestrial telescope adds an erecting lens and becomes longer by $4f$ of that lens.

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5. Resolving Power

Magnification is useless if two nearby points blur into one. By Rayleigh's criterion, two point sources are just resolved when the central maximum of one diffraction pattern falls on the first minimum of the other (see Section 12.3). For a circular aperture of diameter $D$:

$\theta_{\min} = 1.22\frac{\lambda}{D}$

• A telescope with a larger objective resolves finer detail; this is a main reason large telescopes are built.
• An electron microscope achieves far higher resolution than an optical one because electron wavelengths (de Broglie) are much shorter than visible wavelengths.

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6. The Spectrometer

A spectrometer measures refractive indices and wavelengths. Its three parts are:

1. Collimator: a slit at the focus of a convex lens, producing a parallel beam.
2. Turntable: holds a prism or diffraction grating and carries a circular scale with a vernier.
3. Telescope: focused for parallel rays, used to view and measure the deviated light.

With a prism set at minimum deviation $D_m$ and prism angle $A$, the refractive index is

$n = \frac{\sin\left(\frac{A + D_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$

With a grating, the wavelength follows from $d\sin\theta = m\lambda$.

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7. Michelson's Measurement of the Speed of Light

Michelson used a rotating eight-sided mirror. Light reflected from one face travelled to a distant mirror and back (a round trip of $2d$). The octagonal mirror was spun until the next face arrived in exactly the right position in the time the light took for the round trip, which is $\frac{1}{8}$ of a revolution. If $f$ is the rotation frequency,

$t = \frac{1}{8f} = \frac{2d}{c} \implies c = 16fd$

His results gave a value very close to the modern $c = 2.998 \times 10^8\text{ m/s}$.

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8. Fibre Optics

An optical fibre is a thin glass or plastic core of refractive index $n_1$ surrounded by cladding of lower index $n_2$. Light entering at a suitable angle strikes the core–cladding boundary at more than the critical angle and is trapped by total internal reflection:

$\sin\theta_c = \frac{n_2}{n_1}$

Example: $n_1 = 1.50$ and $n_2 = 1.48$ give $\sin\theta_c = 0.987$, so $\theta_c \approx 80.6^\circ$. Rays must strike the boundary at more than about $81^\circ$ to the normal.

Signal losses come from absorption by impurities, scattering by tiny variations in the glass, and bending. Signals are restored by repeaters or optical amplifiers along long links. Fibres are lighter than copper, immune to electromagnetic interference, carry far more information, and are used in telephone and internet links, endoscopes and sensors.