Optical Instruments

Visual Angle and the Simple Microscope

The Visual Angle determines the size of the image on the retina; a simple microscope uses a single convex lens to increase this angle by creating a Virtual Image from an object placed within its focal length.

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

Calculates the maximum angular magnification when the image is at the near point.

$M$=Magnifying Power(Dimensionless)

$D$=Least Distance of Distinct Vision(cm (typically 25 cm))

$f$=Focal length of the lens(cm)

Image at Infinity (relaxed eye)

→

$M = \frac{D}{f}$ — magnification drops by 1 compared to near-point case

$f \to D$

→

$M \to 2$ — the lens barely helps; useful magnification requires $f \ll D$

Near Point: The minimum distance ($D \approx 25$ cm) for sharp focus by the unaided human eye; increases with age.

Angular Magnification: The ratio of the angle subtended by the image (through the lens) to the angle subtended by the object placed at the near point (unaided eye).

Derivation Logic: From the lens formula with $q = D$, we get $1/p = 1/f + 1/D$, so $M = D/p = 1 + D/f$.

The Compound Microscope

A compound microscope uses two lenses: a short focal length Objective Lens to form a real intermediate image, and an Eyepiece to further magnify that image as a Virtual Image.

$$M = \frac{q}{p} \left( 1 + \frac{D}{f_e} \right)$$

Total magnification is the product of linear magnification from the objective ($q/p$) and angular magnification from the eyepiece ($1 + D/f_e$).

$q$=Objective image distance(cm)

$p$=Objective object distance(cm)

$f_e$=Eyepiece focal length(cm)

$D$=Least distance of distinct vision(cm (typically 25 cm))

$p \approx f_o$

→

Maximum magnification — the object is just beyond the objective's focus, making $q$ very large.

$p = f_o$

→

Image at infinity — $q \to \infty$, no real intermediate image forms.

Tube Length: The distance between the two lenses equals $q + p_e$, where $p_e$ is the eyepiece object distance.

Total Magnification: $M = m_o \times M_e$, combining linear magnification of the objective and angular magnification of the eyepiece.

Blue Light Advantage: Shorter wavelength reduces diffraction ($\alpha_{min} = 1.22\lambda/D$), allowing finer detail to be resolved.

The Astronomical Telescope

An astronomical telescope uses a large-aperture Objective Lens with long focal length to form a real image of a distant object, and a short focal length Eyepiece to view that image under Normal Adjustment.

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

Angular magnification in normal adjustment — the ratio of objective to eyepiece focal lengths.

$f_o$=Focal length of objective(cm)

$f_e$=Focal length of eyepiece(cm)

Normal Adjustment

→

Tube length $L = f_o + f_e$; final image at infinity (relaxed eye viewing).

Light Gathering Power: Proportional to the area of the objective ($\propto D^2$); a larger aperture captures more photons from faint celestial objects.

Inverted Image: Astronomical telescopes produce inverted images, acceptable for stars and planets but not for terrestrial use.

Design Trade-off: Increasing $f_o$ increases both magnification and tube length — very long focal lengths require reflecting telescopes using mirrors.

Simultaneous Equations: Given $M$ and $L$, solve $f_o/f_e = M$ and $f_o + f_e = L$ simultaneously to find both focal lengths.

Resolving Power and Diffraction

The Resolving Power is the ability to separate two close points, limited by Diffraction as described by the Rayleigh Criterion.

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

The minimum angular separation for two point sources to be just resolved. Resolving power $RP = 1/\alpha_{min} = D/(1.22\lambda)$.

$\alpha_{min}$=Minimum resolvable angle (limit of resolution)(rad)

$D$=Aperture diameter of the lens(m)

$\lambda$=Wavelength of light used(m)

Using blue light ($\lambda \approx 400$ nm) vs red light ($\lambda \approx 700$ nm)

→

Blue light reduces $\alpha_{min}$ by nearly half, so microscopes resolve finer detail with shorter wavelengths.

Airy Disk: Each point source produces a central bright disk with faint rings due to diffraction — two sources are resolved when the central maximum of one falls on the first minimum of the other.

Limit of Resolution: $\alpha_{min} = 1.22\lambda / D$ — the smaller this angle, the better the instrument resolves fine detail.

Grating Resolving Power: For a diffraction grating, $R = \lambda / \Delta\lambda = N \times m$, where $N$ is the number of rulings and $m$ is the diffraction order.

Dimensional Check: $\alpha_{min}$ has units of $[m]/[m] = $ rad (dimensionless angle), confirming correct usage.

The Spectrometer

A spectrometer is an optical device used to study spectra, measure the deviation of light by a prism, and determine wavelengths using a diffraction grating. It consists of three main components: a Collimator, a turn table, and a telescope.

Collimator: A tube with a convex lens and adjustable slit — when the slit is at the focal point of the lens, parallel rays emerge.

Turn Table: A rotating platform (graduated in half-degrees) on which a prism or grating is mounted.

Telescope: Rotatable around the same axis as the turn table, equipped with a vernier scale to measure angular positions precisely.

Setup Procedure: Adjust collimator for parallel rays, focus telescope at infinity, level the turn table so the prism's refracting edge is parallel to the rotation axis.

Michelson's Measurement of the Speed of Light

Michelson measured the speed of light using an eight-sided rotating mirror and a distant plane mirror separated by a known distance $d$.

$$c = 16fd$$

The speed of light equals 16 times the mirror rotation frequency times the one-way distance.

$c$=Speed of light(m/s)

$f$=Rotation frequency of the eight-sided mirror(Hz (rev/s))

$d$=Distance between the rotating mirror and distant plane mirror(m)

Principle: The time for light to travel from mirror M to distant mirror m and back ($2d/c$) must equal the time for one face of the octagonal mirror to rotate into the next face's position ($1/8f$).

Derivation: Setting $2d/c = 1/(8f)$ and solving gives $c = 16fd$.

Accepted Value: $c = 2.998 \times 10^8$ m/s $\approx 3.00 \times 10^8$ m/s in vacuum.

Key Insight: The speed of light in any material is always less than $c$; in air it is approximately equal to the vacuum value.

Optical Fibres and Total Internal Reflection

Optical fibres guide light over long distances using Total Internal Reflection. Light entering the fibre's core (refractive index $n_1$) reflects off the core-Cladding boundary ($n_2 < n_1$) whenever the angle of incidence exceeds the Critical Angle.

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

The critical angle for total internal reflection at the core-cladding interface.

$\theta_c$=Critical angle(degrees)

$n_1$=Refractive index of core (higher)(Dimensionless)

$n_2$=Refractive index of cladding (lower)(Dimensionless)

Glass-air boundary ($n_1 = 1.5$, $n_2 = 1.0$)

→

$\theta_c = \sin^{-1}(1.0/1.5) = 41.8°$

$n_1 = n_2$

→

$\theta_c = 90°$ — no total internal reflection possible.

Refractive Index: $n = c/v$ — the ratio of the speed of light in vacuum to the speed in the medium.

Acceptance Angle: The maximum angle of incidence at the fibre entrance for light to undergo TIR inside — found by combining Snell's law at the entrance face with the critical angle condition.

Advantages Over Copper: Much wider bandwidth, immunity from electromagnetic interference, lighter and thinner cables.

Optical fibres are classified into three types based on how they propagate light: single-mode step-index, multi-mode step-index, and multi-mode graded-index.

Single-Mode Step-Index: Very thin core (~5 µm), requires laser source, minimal dispersion — best for long-distance telecommunications.

Multi-Mode Step-Index: Larger core (~50 µm), constant $n_1$ in core that steps down to $n_2$ at cladding — suffers from modal dispersion (~33 ns/km).

Multi-Mode Graded-Index: Core refractive index decreases gradually from centre to edge — different paths take different times but travel at different speeds, reducing dispersion to ~1 ns/km.

Dispersion Problem: In step-index fibres, rays taking longer paths arrive later than axial rays, spreading the signal pulse. Graded-index fibres compensate because outer rays travel faster (lower $n$ means higher speed).