Optics

A converging glass lens of focal length $20$ cm is dropped into water. Does it converge more strongly, less strongly, or not at all?

Less strongly — by a factor of about four. And a lens made of a material with a lower index than water would converge not at all: it would diverge.

The lens maker's formula uses the index of the lens relative to its surroundings:

$\frac{1}{f}=\left(\frac{n_{lens}}{n_{medium}}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

In air, $\left(1.5-1\right)=0.5$. In water, $\left(\frac{1.5}{1.33}-1\right)=0.128$, so $f$ grows from $20$ cm to about $78$ cm. Push $n_{lens}$ below $1.33$ and the bracket turns negative: the lens changes sign.

A lens has no focal length of its own. It has one only in a stated medium — which is why an air bubble in water, shaped like a convex lens, is a diverging lens.

1. The single refracting surface generates everything

Almost every ray-optics result at Advanced level descends from one relation. For refraction at a spherical surface of radius $R$ separating media $n_1$ and $n_2$, with the Cartesian convention (distances measured from the surface, positive along the incident light):

$\frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R}$

Apply it twice, once at each face of a thin lens, and the two intermediate images cancel to give the lens maker's formula. Set $n_2=-n_1$ and it becomes the mirror formula. Set $R\to \infty$ and it becomes the plane-surface result $\frac{v}{u}=\frac{n_2}{n_1}$, which is apparent depth.

Magnification at a single surface is $m=\frac{n_{1}v}{n_{2}u}$, reducing to $-v/u$ for a lens where the media match on both sides.

Illustration 1

A glass sphere of radius $10$ cm and index $1.5$ has a point object $30$ cm from its surface. Find where the first surface forms an image.

$\frac{1.5}{v}-\frac{1}{-30}=\frac{1.5-1}{10}$

$\frac{1.5}{v}=0.05-0.0333=0.0167\Rightarrow v=90$ cm

The image lies $90$ cm beyond the first surface, well outside the sphere — so it acts as a virtual object for the second surface.

Never stop at the first surface. A sphere or a thick lens needs the formula applied twice, with the first image becoming the object for the second stage.

Illustration 2

A fish is $2$ m below the surface of a pond. How deep does it appear to a bird looking straight down, and how high does the bird at $3$ m appear to the fish?

Looking down (from air into water, $n_1=1.33$, $n_2=1$): apparent depth $=\frac{2}{1.33}=1.5$ m.

Looking up (from water into air): apparent height $=3\times 1.33=4$ m.

The asymmetry is real. Each observer sees the other displaced away from the surface or towards it depending on which way the light is going, and the factor is $n$ one way and $1/n$ the other.

2. Combinations, silvering and the displacement method

Two thin lenses separated by $d$ combine as

$\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}-\frac{d}{f_1{f}_2},P=P_1+P_2-d{P}_1{P}_2$

Silvering one face of a lens makes it an equivalent mirror. Light passes the lens, reflects, and passes the lens again, so the powers add as

$P=2P_{lens}+P_{mirror},P_{mirror}=\frac{2}{R}$

Newton's formula measures distances from the foci rather than from the lens: if $x_1$ and $x_2$ are the object and image distances from the respective focal points, then $x_1{x}_2=f^2$.

The displacement method finds a focal length without ever locating the lens's centre. Fix an object and screen a distance $D>4f$ apart; there are exactly two lens positions giving a sharp image, separated by $d$, and

$f=\frac{D^2-d^2}{4D}$

Illustration 3

An object and screen are $100$ cm apart and a lens forms a sharp image at two positions $20$ cm apart. Find the focal length and the two magnifications.

$f=\frac{100^2-20^2}{4\times 100}=\frac{10000-400}{400}=24$ cm

The two magnifications are reciprocals: $m_1{m}_2=1$, with $m_1=\frac{D-d}{D+d}=\frac{80}{120}=0.667$ and $m_2=1.5$.

The two positions are the object and image distances swapped. That is why the magnifications multiply to one, and it gives a fast check on any answer.

Illustration 4

A thin equiconvex lens of index $1.5$ and radius of curvature $20$ cm has one face silvered. Find the focal length of the equivalent mirror.

$P_{lens}=\left(0.5\right)\left(\frac{1}{20}+\frac{1}{20}\right)=\frac{1}{20}$ per cm, so $f_{lens}=20$ cm.

$P_{mirror}=\frac{2}{20}=\frac{1}{10}$

$P=2\left(\frac{1}{20}\right)+\frac{1}{10}=\frac{1}{5}$, so the equivalent mirror has $f=5$ cm and is converging.

Silvering makes the system four times as powerful as the lens alone here. The light traverses the lens twice, and the reflecting surface contributes its own curvature on top.

3. Slabs, total internal reflection and fibres

A parallel slab of thickness $t$ shifts a ray sideways without changing its direction:

$\text{lateral shift}=t\frac{\sin \left(i-r\right)}{\cos r},\text{normal shift}=t\left(1-\frac{1}{n}\right)$

The normal shift is what makes an object under glass appear raised, and it is independent of where the object sits.

Beyond the critical angle $\sin C=n_2/n_1$, light is totally internally reflected. For an optical fibre with core $n_1$ and cladding $n_2$, the largest angle that can be launched and still be guided is fixed by the numerical aperture:

$\sin {\theta }_{max}=\sqrt{n_1^2-n_2^2}$

Illustration 5

A fibre has core index $1.50$ and cladding index $1.45$. Find the critical angle at the core-cladding boundary and the acceptance angle in air.

$\sin C=\frac{1.45}{1.50}=0.967\Rightarrow C=75.2^{\circ }$

$\sin {\theta }_{max}=\sqrt{1.50^2-1.45^2}=\sqrt{2.25-2.1025}=\sqrt{0.1475}=0.384$

${\theta }_{max}=22.6^{\circ }$

A small index difference gives a large critical angle but a modest acceptance cone. Fibres are deliberately made with a small step so that the range of ray paths is narrow, which keeps pulses from spreading.

4. Prisms: deviation, dispersion and correction

For any prism, $A+\delta =i+e$ and $r_1+r_2=A$. Deviation is minimum when the passage is symmetric, $i=e$, and then

$n=\frac{\sin \left(\frac{A+{\delta }_{min}}{2}\right)}{\sin \left(\frac{A}{2}\right)}$

For a thin prism this reduces to $\delta =\left(n-1\right)A$, and since $n$ depends on colour, the beam spreads. Dispersive power measures that spread relative to the mean deviation:

$\omega =\frac{n_v-n_r}{n_y-1}$

Two prisms of different glasses can then be combined to give deviation without dispersion (a direct-vision prism, requiring $\left(n_v-n_r\right)A+\left(n_v^{\prime }-n_r^{\prime }\right)A^{\prime }=0$) or dispersion without deviation. The lens analogue is an achromatic doublet, which needs

$\frac{{\omega }_1}{f_1}+\frac{{\omega }_2}{f_2}=0$

so the two lenses must have opposite signs of focal length.

Illustration 6

A prism of refracting angle $60^{\circ }$ gives a minimum deviation of $30^{\circ }$. Find the refractive index and the angle of incidence at minimum deviation.

$n=\frac{\sin 45^{\circ }}{\sin 30^{\circ }}=\frac{0.707}{0.5}=1.414$

At minimum deviation, $i=\frac{A+{\delta }_{min}}{2}=45^{\circ }$.

Minimum deviation is where the graph of $\delta$ against $i$ is flat, so small errors in setting $i$ barely change $\delta$ — which is exactly why the condition is used for measuring $n$.

Illustration 7

Design an achromatic doublet of net focal length $20$ cm from crown glass ($\omega =0.02$) and flint glass ($\omega =0.04$).

Achromatism needs $\frac{0.02}{f_1}+\frac{0.04}{f_2}=0$, so $f_2=-2f_1$.

Net power: $\frac{1}{f_1}+\frac{1}{f_2}=\frac{1}{f_1}-\frac{1}{2f_1}=\frac{1}{2f_1}=\frac{1}{20}$

$f_1=10$ cm (converging crown), $f_2=-20$ cm (diverging flint).

The diverging lens must be weaker than the converging one, or the doublet would not converge at all. Achromatism fixes the ratio; the required net power then fixes the magnitudes.

5. Velocity of images

Differentiate the lens or mirror equation and the image velocity follows. For a mirror, differentiating $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$ gives

$v_{image}=-m^2{v}_{object}\text{(along the axis)}$

while transverse velocities scale simply as $m$. The two are different powers of the magnification, which is why an approaching object's image appears to stretch.

Illustration 8

An object approaches a concave mirror of focal length $20$ cm at $5$ cm s${}^{-1}$ when it is $60$ cm away. Find the image speed.

$\frac{1}{v}=\frac{1}{20}-\frac{1}{60}=\frac{1}{30}$, so $v=30$ cm and $m=-\frac{v}{u}=-\frac{30}{60}$, giving $\left|m\right|=0.5$.

$v_{image}=m^2{v}_{object}=0.25\times 5=1.25$ cm s${}^{-1}$

The image moves four times slower here. As the object nears the focus, $m$ grows without limit and the image races away — which is why focusing a camera becomes so touchy at close range.

6. Huygens' principle and what a wavefront explains

Treat every point of a wavefront as a source of secondary spherical wavelets; the surface tangent to all of them, an instant later, is the new wavefront. Two laws fall straight out.

Reflection. Both ends of the wavefront travel in the same medium at the same speed, so the geometry is symmetric and the angle of reflection equals the angle of incidence.

Refraction. In a time $t$ the wavelet in medium 1 advances $v_{1}t$ while the one in medium 2 advances $v_{2}t$. Fitting them to the same interface gives

$\frac{\sin i}{\sin r}=\frac{v_1}{v_2}=\frac{n_2}{n_1}$

Note what this asserts: light travels slower in the denser medium. Newton's corpuscular theory predicted the opposite, and Foucault's direct measurement of the speed of light in water is what settled the argument in favour of waves.

The same picture explains why frequency cannot change at a boundary. Wavefronts arrive at the interface at a fixed rate and cannot pile up or vanish there, so they must leave at the same rate. Speed drops by $n$, so wavelength must too.

One further consequence is worth carrying into interference problems. Every ray from an object point to its image point takes the same optical path length $\int nds$ — the extra geometric length of an off-axis ray is exactly compensated by the thinner glass it passes through. That is what makes an image sharp rather than smeared.

Illustration 9

A plane wavefront in air strikes a glass surface at $60^{\circ }$. Given that the wavefront advances $30$ cm in air while advancing $20$ cm in glass over the same time, find the refractive index and the angle of refraction.

$n=\frac{v_1}{v_2}=\frac{30}{20}=1.5$

$\sin r=\frac{\sin 60^{\circ }}{1.5}=\frac{0.866}{1.5}=0.577\Rightarrow r=35.3^{\circ }$

The wavefront tilts because one end is held back. That is the entire physical content of refraction, and the ray picture is just a convenient shorthand for it.

Illustration 10

Light of vacuum wavelength $500$ nm enters glass of index $1.5$. Find the wavelength, frequency and speed inside, and the optical path length of $2$ cm of the glass.

Frequency is unchanged: $f=\frac{3\times 10^8}{5\times 10^{-7}}=6\times 10^{14}$ Hz

$v=\frac{3\times 10^8}{1.5}=2\times 10^8$ m s${}^{-1}$, and ${\lambda }^{\prime }=\frac{500}{1.5}=333$ nm

Optical path $=n\times$ geometric path $=1.5\times 2=3$ cm

Optical path is the currency of interference. Two cm of this glass delays light exactly as three cm of vacuum would, which is why a slab inserted in one arm shifts a fringe pattern.

7. Young's slits, past the fringe-width formula

The fringe width $\beta =\frac{\lambda D}{d}$ is the least of what Advanced asks. The intensity at a point of path difference $\Delta x$ is

$I=4I_0{\cos }^2\left(\frac{\pi \Delta x}{\lambda }\right)$

so the pattern is a raised cosine, not a set of sharp lines, and the maximum is four times one slit's intensity, not two — energy being redistributed, not created.

Three modifications recur.

A thin slab of index $n$ and thickness $t$ over one slit adds an extra path $\left(n-1\right)t$, shifting the whole pattern by

$\Delta y=\frac{\left(n-1\right)tD}{d}$

without changing the fringe width. Immersing the apparatus in a medium divides both $\lambda$ and $\beta$ by $n$. White light gives a white central fringe with coloured edges, because only the zeroth order has zero path difference for every wavelength at once.

Illustration 11

In a double-slit experiment with $\lambda =600$ nm, $d=0.5$ mm and $D=1.5$ m, a mica sheet of index $1.58$ and thickness $6\mu$m is placed over one slit. Find the fringe width and the shift.

$\beta =\frac{\left(6\times 10^{-7}\right)\left(1.5\right)}{5\times 10^{-4}}=1.8\times 10^{-3}$ m $=1.8$ mm

$\Delta y=\frac{\left(0.58\right)\left(6\times 10^{-6}\right)\left(1.5\right)}{5\times 10^{-4}}=1.04\times 10^{-2}$ m $=10.4$ mm

The shift is nearly six fringe widths, so the pattern moves bodily by about six fringes towards the covered slit while its spacing is untouched.

Illustration 12

Two slits have intensities in the ratio $4:1$. Find the ratio of maximum to minimum intensity in the pattern.

Amplitudes are in the ratio $2:1$.

$\frac{I_{max}}{I_{min}}=\frac{{\left(2+1\right)}^2}{{\left(2-1\right)}^2}=\frac{9}{1}$

Unequal slits never give complete darkness. Contrast is set by the amplitude ratio, and only equal amplitudes produce true zeros.

8. Thin films

A film of index $\mu$ and thickness $t$ viewed at refraction angle $r$ produces interference between the beams reflected from its two faces. The beam reflecting off the denser medium suffers a $\pi$ phase change, equivalent to half a wavelength, so for reflected light:

$2\mu t\cos r=\left(2n-1\right)\frac{\lambda }{2}\text{(bright)},2\mu t\cos r=n\lambda \text{(dark)}$

and the conditions are exactly interchanged for transmitted light. This is why a very thin soap film looks black just before it bursts: as $t\to 0$ the only surviving effect is the $\pi$ phase change, giving destructive interference at every wavelength.

Illustration 13

A soap film of index $1.33$ is viewed normally in white light and appears bright green at $550$ nm. Find its smallest possible thickness.

Normal incidence gives $\cos r=1$, and the first bright order is $n=1$:

$2\left(1.33\right)t=\frac{550}{2}\text{nm}\Rightarrow t=\frac{275}{2.66}=103$ nm

Antireflection coatings invert the same idea, choosing the thickness so that reflection is destructive at mid-spectrum, which is why coated lenses look faintly purple.

9. Diffraction and the limit of resolution

A single slit of width $a$ gives minima where

$a\sin \theta =n\lambda \left(n=1,2,3\dots \right)$

so the central maximum spans $2\lambda D/a$ — twice the width of every other maximum. Narrowing the slit widens the pattern, which is the reverse of the intuition borrowed from shadows.

Diffraction is also what limits every instrument. Two point sources are just resolved when the maximum of one falls on the first minimum of the other — the Rayleigh criterion:

${\theta }_{min}=\frac{1.22\lambda }{D}$

for a circular aperture of diameter $D$. Bigger apertures and shorter wavelengths both resolve finer detail, which is the entire justification for large telescopes and for electron microscopes.

Illustration 14

A slit of width $0.2$ mm is illuminated with $600$ nm light and the pattern observed on a screen $2$ m away. Find the width of the central maximum.

Width $=\frac{2\lambda D}{a}=\frac{2\left(6\times 10^{-7}\right)(2)}{2\times 10^{-4}}$

$=1.2\times 10^{-2}$ m $=12$ mm

Halving the slit width would double this to $24$ mm. Squeezing light spatially spreads it angularly, which is the optical face of the uncertainty principle.

Illustration 15

A telescope has an objective of diameter $10$ cm. Find its angular resolution at $550$ nm and the smallest crater it could resolve on the Moon, $3.8\times 10^8$ m away.

${\theta }_{min}=\frac{1.22\times 5.5\times 10^{-7}}{0.1}=6.7\times 10^{-6}$ rad

Smallest feature $={\theta }_{min}\times r=6.7\times 10^{-6}\times 3.8\times 10^8=2.5\times 10^3$ m

About two and a half kilometres. Magnification cannot improve on this: once the diffraction limit is reached, enlarging the image only makes a blur bigger.

Summary

• A lens has no focal length of its own: $\frac{1}{f}=\left(\frac{n_{lens}}{n_{medium}}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$, and the sign can flip in a dense medium.
• Master relation: $\frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R}$, with $m=\frac{n_{1}v}{n_{2}u}$. Apply it once per surface.
• Two lenses at separation $d$: $\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}-\frac{d}{f_1{f}_2}$.
• Silvered lens: $P=2P_{lens}+P_{mirror}$, with $P_{mirror}=2/R$.
• Displacement method: $f=\frac{D^2-d^2}{4D}$, and the two magnifications multiply to one.
• Slab: lateral shift $t\frac{\sin \left(i-r\right)}{\cos r}$, normal shift $t\left(1-\frac{1}{n}\right)$; fibre acceptance $\sin {\theta }_{max}=\sqrt{n_1^2-n_2^2}$.
• Prism: $A+\delta =i+e$; at minimum deviation $i=e$ and $n=\frac{\sin \left(\frac{A+{\delta }_{min}}{2}\right)}{\sin \left(\frac{A}{2}\right)}$.
• Achromatic doublet: $\frac{{\omega }_1}{f_1}+\frac{{\omega }_2}{f_2}=0$, so the two lenses must have opposite signs.
• Axial image velocity is $m^2$ times the object's; transverse velocity only $m$ times.
• Huygens gives both laws, and asserts light is slower in a denser medium — the point on which the wave theory was decided.
• Frequency is fixed at a boundary; $v$ and $\lambda$ both divide by $n$. Optical path $\int nds$ is what interference counts.
• $I=4I_0{\cos }^2\left(\frac{\pi \Delta x}{\lambda }\right)$: the maximum is four times a single slit, by redistribution.
• A slab over one slit shifts the pattern by $\frac{\left(n-1\right)tD}{d}$ without changing the fringe width; immersion divides $\beta$ by $n$.
• Unequal slits give $\frac{I_{max}}{I_{min}}={\left(\frac{A_1+A_2}{A_1-A_2}\right)}^2$ — never true darkness.
• Thin film in reflection: bright at $2\mu t\cos r=\left(2n-1\right)\frac{\lambda }{2}$; a vanishingly thin film looks black.
• Single slit minima at $a\sin \theta =n\lambda$; central maximum is twice as wide as the rest; Rayleigh limit $\theta =\frac{1.22\lambda }{D}$.