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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).
Application Strategy
Worked Examples
Problem Statement
A converging lens of focal length 5.0 cm is used as a magnifying glass. The near point is 25 cm. Find (i) the object distance and angular magnification when the image is at the near point, and (ii) the magnification when the image is at infinity.
Lens FormulaAngular MagnificationNear Point vs Infinity
Strategic Toolkit
$$M = 1 + \frac{D}{f}, \quad M = \frac{D}{f}$$
Step 1
Case (i): Image at Near Point
Use $M = 1 + D/f$ and the lens formula to find $p$.
$M = 1 + \frac{25}{5} = 1 + 5 = 6$
From $\frac{1}{p} = \frac{1}{f} + \frac{1}{D} = \frac{1}{5} + \frac{1}{25} = \frac{6}{25}$
$p = \frac{25}{6} \approx 4.2$ cm
Step 2
Case (ii): Image at Infinity
Object placed at the focal point; eye views with relaxed accommodation.
$p = f = 5.0$ cm
$M = \frac{D}{f} = \frac{25}{5} = 5.0$
Step 3
Compare
Near-point viewing gives slightly higher magnification (6 vs 5) but requires eye strain.
Near point: $M = 6$, $p = 4.2$ cm
Infinity: $M = 5$, $p = 5.0$ cm
Problem Statement
A microscope has an objective of focal length 10 mm and an eyepiece of 25.0 mm focal length. The object is 10.5 mm from the objective. Find the distance between lenses and the total magnification (image at near point, D = 250 mm).
Lens FormulaLinear MagnificationAngular Magnification
Strategic Toolkit
$$M = \frac{q}{p} \left(1 + \frac{D}{f_e}\right)$$
Step 1
Objective Image Distance
Apply the lens formula to the objective.
$\frac{1}{q} = \frac{1}{f_o} - \frac{1}{p} = \frac{1}{10} - \frac{1}{10.5}$
$q = 210$ mm
Step 2
Eyepiece Object Distance
Apply the lens formula to the eyepiece with virtual image at $d = -250$ mm.
$\frac{1}{p_e} = \frac{1}{f_e} - \frac{1}{d} = \frac{1}{25} + \frac{1}{250} = \frac{11}{250}$
$p_e = 22.7$ mm
Step 3
Lens Separation and Total Magnification
Distance between lenses is $q + p_e$. Total magnification is $M_1 \times M_2$.
$L = 210 + 22.7 = 233$ mm
$M_1 = q/p = 210/10.5 = 20$, $M_2 = d/p_e = 250/22.7 = 11$
$M = 20 \times 11 = 220$
Problem Statement
An astronomical telescope in normal adjustment has magnifying power 5 and the two lenses are 24 cm apart. Find the focal lengths of the objective and eyepiece.
Normal AdjustmentTelescope MagnificationSimultaneous Equations
Strategic Toolkit
$$M = f_o / f_e, \quad L = f_o + f_e$$
Step 1
Set Up Equations
Two equations in two unknowns from the given data.
$f_o / f_e = 5 \implies f_o = 5f_e$
$f_o + f_e = 24$
Step 2
Solve
Substitute and solve.
$5f_e + f_e = 24 \implies 6f_e = 24 \implies f_e = 4$ cm
$f_o = 5 \times 4 = 20$ cm
Problem Statement
An optical fibre has a core refractive index of 1.50 and cladding refractive index of 1.48. Calculate the critical angle at the core-cladding interface and the maximum angle of entry from air.
Critical AngleSnell's LawTotal Internal Reflection
Strategic Toolkit
$$\sin\theta_c = n_2/n_1$$
Step 1
Find the Critical Angle
Apply Snell's law at the core-cladding boundary.
$\sin\theta_c = \frac{n_2}{n_1} = \frac{1.48}{1.50} = 0.9867$
$\theta_c = \sin^{-1}(0.9867) = 80.6°$
Step 2
Find Angle Inside Core
The ray must hit the core-cladding interface at the critical angle. The complementary angle gives the refraction angle inside the core at the entrance.
$\theta' = 90° - \theta_c = 90° - 80.6° = 9.4°$
Step 3
Find Maximum Entry Angle
Apply Snell's law at the air-core boundary.
$n_{air} \sin\theta = n_1 \sin\theta'$
$\sin\theta = 1.50 \times \sin(9.4°) = 1.50 \times 0.163 = 0.245$
$\theta = \sin^{-1}(0.245) = 14.2°$
Light entering at angles greater than 14.2° will not undergo total internal reflection.
Memorize formulas
Simple Microscope (Near Point)
Maximum magnification when image forms at the near point.
Formula
$M = 1 + \frac{D}{f}$
Simple Microscope (Infinity)
Magnification for relaxed eye viewing (image at infinity).
Formula
$M = \frac{D}{f}$
Compound Microscope
Objective linear magnification times eyepiece angular magnification.
Formula
$M = \frac{q}{p} \left( 1 + \frac{D}{f_e} \right)$
Telescope Magnification
Ratio of focal lengths in normal adjustment.
Formula
$M = \frac{f_o}{f_e}$
Telescope Tube Length
Sum of focal lengths in normal adjustment.
Formula
$L = f_o + f_e$
Rayleigh Limit of Resolution
Minimum resolvable angle through a circular aperture.
Formula
$\alpha_{min} = \frac{1.22\lambda}{D}$
Critical Angle (Fibre Optics)
Angle of incidence for total internal reflection at core-cladding boundary.
Formula
$\sin\theta_c = \frac{n_2}{n_1}$
Speed of Light (Michelson)
Measured using an eight-sided rotating mirror at distance d.
Formula
c = 16fd
Review
1
A converging lens of focal length 5.0 cm is used as a magnifying glass with the near point at 25 cm. When the image shifts from the near point to infinity, the magnification changes from:
Ans
6 to 5
Review
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Use the magnifying glass formulas
For a simple microscope, the angular magnification is different for a final image at the near point and for a final image at infinity. Substitute the given focal length and near-point distance into both expressions.
$$M_{near} = 1 + \frac{D}{f} = 1 + \frac{25}{5} = 6, \qquad M_{\infty} = \frac{D}{f} = \frac{25}{5} = 5$$
Read off the change in magnification
The shift from near point to infinity reduces the magnification by one unit, so the correct transition is from 6 to 5.
$$6 \to 5$$
2
A compound microscope has an objective of focal length 0.5 cm and an eyepiece of focal length 2.5 cm. The object is placed 0.6 cm from the objective and the final image is at infinity. The total magnification is approximately:
Ans
50
Review
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Find the objective image distance
Use the lens formula for the objective first. With the object just beyond the focal length, the objective forms a real enlarged intermediate image.
$$\frac{1}{v} = \frac{1}{0.5} - \frac{1}{0.6} = 2 - \frac{5}{3} = \frac{1}{3} \Rightarrow v = 3.0\text{ cm}$$
Compute objective magnification
The objective magnification is the ratio of image distance to object distance.
$$m_o = \frac{v}{u} = \frac{3.0}{0.6} = 5$$
Multiply by the eyepiece magnification
For final image at infinity, the eyepiece acts as a simple magnifier with angular magnification $D/f_e$.
$$M_e = \frac{D}{f_e} = \frac{25}{2.5} = 10, \qquad M = m_o M_e = 5 \times 10 = 50$$
3
Which of the following changes would increase the resolving power of a microscope?
Ans
Using oil immersion to increase the effective numerical aperture
Review
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Use the resolving-power trend
A microscope resolves finer detail when the minimum resolvable angle gets smaller. That happens when the effective numerical aperture increases or the wavelength decreases.
$$\alpha_{min} \propto \frac{\lambda}{\text{NA}}$$
Choose the option that increases NA
Oil immersion raises the refractive index between specimen and objective, increasing the effective numerical aperture. The other options reduce resolving power instead.
Using oil immersion to increase the effective numerical aperture
4
An astronomical telescope in normal adjustment has an objective of focal length 120 cm and an eyepiece of focal length 5 cm. The length of the telescope tube and the magnifying power are:
Ans
125 cm and 24
Review
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Apply normal-adjustment geometry
In an astronomical telescope adjusted for distant objects, the focal plane of the objective coincides with the focal plane of the eyepiece. So the tube length is the sum of the focal lengths.
$$L = f_o + f_e = 120 + 5 = 125\text{ cm}$$
Compute the magnifying power
For normal adjustment, angular magnification is the ratio of the objective focal length to the eyepiece focal length.
$$M = \frac{f_o}{f_e} = \frac{120}{5} = 24$$
5
A telescope objective has focal length 96 cm and diameter 12 cm. If the magnification is 24×, the minimum diameter of the eyepiece lens (so all light from a distant on-axis point falls on it) is:
Ans
0.50 cm
Review
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Use the exit-pupil relation
To collect all light from an on-axis distant object, the eyepiece aperture must at least match the exit pupil size. That size is the objective diameter divided by the magnification.
$$d_e = \frac{D}{M} = \frac{12}{24} = 0.50\text{ cm}$$
Select the minimum diameter
Any smaller eyepiece would clip the beam, so the minimum required diameter is 0.50 cm.
$$0.50\text{ cm}$$
6
In Michelson's experiment to measure the speed of light, an 8-sided mirror rotates at frequency $f$ and the fixed mirror is at distance $d$. If the distance $d$ is doubled while keeping the setup functional, the required rotation frequency must:
Ans
Be halved
Review
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Use Michelson's speed formula
For an 8-sided mirror, the measured speed of light is proportional to the product of rotation frequency and distance.
$$c = 16fd$$
Keep the speed of light fixed
Since $c$ is constant, the frequency must vary inversely with distance. Doubling the mirror separation therefore halves the required rotation frequency.
$$f \propto \frac{1}{d} \Rightarrow d \to 2d \; \Longrightarrow \; f \to \frac{f}{2}$$
7
The core and cladding of an optical fibre have refractive indices 1.6 and 1.4 respectively. The maximum angle of incidence in air for a ray to undergo total internal reflection inside the fibre is approximately:
Ans
51°
Review
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Find the critical angle at the core-cladding boundary
Total internal reflection starts when light inside the core hits the boundary at the critical angle.
$$\sin\theta_c = \frac{n_{cladding}}{n_{core}} = \frac{1.4}{1.6} = 0.875 \Rightarrow \theta_c \approx 61^\circ$$
Convert to the ray angle inside the core
The ray inside the fibre makes an angle of $90^\circ-\theta_c$ with the fibre axis.
$$\theta' = 90^\circ - 61^\circ = 29^\circ$$
Apply Snell's law at the air-core surface
Using the core index as the refracting medium gives the maximum launch angle in air.
$$\sin\theta = 1.6\sin 29^\circ \approx 1.6(0.485) = 0.776 \Rightarrow \theta \approx 51^\circ$$
8
In a multi-mode step-index fibre, signals are distorted because:
Ans
Rays taking different zigzag paths have different path lengths and arrive at different times
Review
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Identify modal dispersion
A multi-mode fibre supports many ray paths, not just one. Different modes do not travel the same geometric distance through the fibre.
Different modes arrive at different times
Link path length to arrival time
Rays that zigzag more strongly travel a longer path than rays closer to the axis, so they take longer to reach the end.
Longer zigzag path \Rightarrow later arrival
9
A graded-index fibre reduces dispersion compared to a step-index fibre because:
Ans
The gradually decreasing refractive index makes outer rays travel faster, compensating for longer paths
Review
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Use the graded-index idea
The refractive index is higher near the centre and lower near the edge of the core. That means outer rays move through a lower-index region while following longer paths.
Outer region: lower $n$, so higher speed
Show how dispersion is compensated
Because $v = c/n$, the outer rays travel faster and partially make up for their extra path length. This brings the arrival times closer together.
$$v = \frac{c}{n} \Rightarrow \text{lower } n \; \Rightarrow \; \text{higher } v$$
10
A glass rod in air has a critical angle of 39°. If the same rod is immersed in water ($n = 1.33$), the minimum angle for total internal reflection becomes approximately:
Ans
57°
Review
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Recover the glass refractive index
The given critical angle in air lets us solve for the glass index using the critical-angle relation.
$$\sin 39^\circ = \frac{1}{n_g} \Rightarrow n_g = \frac{1}{\sin 39^\circ} \approx 1.589$$
Recompute the critical angle in water
Replace air by water in the critical-angle formula. A higher surrounding refractive index makes the critical angle larger.
$$\sin\theta_c = \frac{n_{water}}{n_g} = \frac{1.33}{1.589} \approx 0.837 \Rightarrow \theta_c \approx 57^\circ$$
11
Sodium light of wavelength 589 nm is used with a microscope whose objective has an aperture of 0.90 cm. The limiting angle of resolution is approximately:
Ans
$8.0 \times 10^{-5}$ rad
Review
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Apply the diffraction limit
The limiting angle of resolution for a circular aperture follows the Rayleigh criterion.
$$\alpha_{min} = \frac{1.22\lambda}{D}$$
Substitute the given values
Use the sodium wavelength in meters and the aperture in meters, then evaluate.
$$\alpha_{min} = \frac{1.22(589\times 10^{-9})}{0.90\times 10^{-2}} \approx 7.99\times 10^{-5}\text{ rad}$$
Round to the nearest option
The computed value matches the first choice after rounding.
$$8.0 \times 10^{-5}\text{ rad}$$
Question 1 of 11
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1
A converging lens of focal length 5.0 cm is used as a magnifying glass with the near point at 25 cm. When the image shifts from the near point to infinity, the magnification changes from: