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Adopt positive object and real-image distances $u$ and $v$ measured from a converging thin lens, with the object on the left. A ray through the lens centre is undeviated in the thin-lens approximation. For object height $h$ and signed image height $h^{\prime }$, similar triangles give $h^{\prime }/h=-v/u$.

A second ray leaves the object parallel to the axis, reaches the lens at height $h$ and then passes through the far focal point. Its height at distance $v$ is $h^{\prime }=h(1-v/f)$. Equating the two descriptions givesThe signed transverse magnification is $\boxed{m=h^{\prime }/h=-v/u}$. Since a real image has $v>0$ for $u>0$, $m<0$: it is inverted. Its size relative to the object is $|m|=v/u$.

The derivation assumes the paraxial approximation, a negligible lens thickness and a focal length defined in the surrounding medium. The ray diagrams show actual paths as solid lines and backward extrapolations as dashed lines.

A real image is a point where the actual outgoing rays converge; a screen placed there can intercept the focused light. A virtual image is a point from which the outgoing rays appear to diverge: only their backward extensions intersect, so a screen at that apparent point does not receive a focused real image from the same rays.

For a converging thin lens with $f>0$ and a real object a distance $u>0$ away, the lens relation gives $v=fu/(u-f)$. If $u>f$, then $v>0$: the image is real, on the opposite side, and inverted. If $0<u<f$, then $v<0$: the image is virtual, on the same side as the object, upright and enlarged, with $m=-v/u=f/(f-u)>1$.

At $u=f$, the emergent rays are parallel and the image is at infinity; there is no finite focused image. For an object very far away, $v\to f$ from above. These are conditions for a real object in the stated lens convention, rather than claims about all systems with virtual incident objects.

For a lens between an object and a screen, $u+v=D$ and the thin lens relation requires $uv=fD$. Eliminating $v$ gives the quadratic equationFor $D>4f$, both roots lie between $0$ and $D$, and they exchange the object and image distances. Both therefore focus the image on the same screen, with reciprocal absolute magnifications. The distance between lens positions isThe existence qualification is essential: for $D=4f$ the two roots coincide and there is only one distinct position; for $D<4f$ there is none. For example, $D=3f$ makes the discriminant negative, providing a concrete counterexample to an unconditional assertion of two positions. This is the basis of the Bessel lens displacement method, with $f=(D^2-l^2)/(4D)$ when two distinct positions can be measured.

The requested source-image relation is the thin lens equation, using positive real-object distance $u$ and signed image distance $v$:Here $f>0$ for a converging lens, and $v<0$ denotes a virtual image. This is usually distinguished from Lensmaker's equation, $1/f=(n-1)(1/R_1-1/R_2)$ for a thin lens in air with the usual signed surface radii. The latter relates focal length to the lens material and curvature, rather than to object and image positions.^[ref]

For a magnifying glass held close to the eye, compare the angle $h/u$ with the unaided near-point angle $h/D_n$, where $D_n=25\mathrm{cm}$. Maximum usable angular magnification occurs with the virtual image at the near point, so $v=-D_n$. ThusThe object lies inside the front focal length. The emerging light rays diverge, while their backward extensions meet at an upright enlarged virtual image on the object's side. With the image at infinity instead, the relaxed-eye angular magnification would be $D_n/f=6.25$, which is smaller. A finite eye-to-lens distance $s$ changes the near-point result to $1+(D_n-s)/f$; the quoted maximum uses the conventional $s\simeq 0$ limit.^[ref]

The paraxial approximation restricts light rays to small slopes relative to the optical axis, so $\sin \beta \simeq \tan \beta \simeq \beta$ and geometric lengths can be expanded to second order in height. Treat the source's optical lens as a thin lens located at $x=0$, with $|y|\ll u,|v|$ and small surface slopes.

For positive real conjugate distances $u,v$, the collapsed-plane geometric path has lengthTo thin-element accuracy, a segment $s(y)=s_0-\alpha y^2$ of this path lies inside the optical lens, replacing the same length of air. The geometric air path is $S-s$, the material path is $s$, and the optical path length and travel time are thereforeThis distinguishes geometric length from optical path length; adding $ns$ to $S$ without subtracting the replaced air would count that segment twice.

For a nonaxial light ray, Fermat's principle gives $d\mathcal{L}/dy=0$ and henceThe focal length is obtained with the incident object at infinity:The constant thickness $s_0$ supplies a common delay but does not change the leading focal length. Since $\alpha$ has units of inverse length, this formula has the correct dimensions. These are thin lens results; paraxial approximation alone does not remove finite-thickness corrections in a general thick lens.

Substituting the stated refractive index and thickness coefficient, $\alpha =2{\mathrm{m}}^{-1}$, givesThe thin lens equation then givesThe positive image distance places a real image on the outgoing side of the optical lens, at $x=+1.75\mathrm{m}$. The object is beyond the focal point, so outgoing light rays actually converge there rather than only their backward extensions meeting.

In the paraxial approximation, apply refraction at each surface of the thin lens in air. With signed radii $R_1$ and $R_2$, the first surface obeys $n/v_1-1/u=(n-1)/R_1$, and the second obeys $1/v-n/v_1=(1-n)/R_2$. The separation of the surfaces is neglected, so the intermediate image coordinate $v_1$ cancels:This is Lensmaker's equation, with Cartesian coordinate system object coordinate $u<0$ for a real object. For an equiconvex thin lens and light travelling left to right, $R_1=R$ and $R_2=-R$. Hence

Set the red image at coordinate zero, the blue image at $\delta =0.010\mathrm{m}$, and the screen at $D=1.000\mathrm{m}$. Let the correcting thin lens be at $x$, with $\delta <x<D$. Both original images are then real objects for the correcting thin lens, with positive object distances $u_r=x$, $u_b=x-\delta$, and common positive image distance $v=D-x$. The thin lens equation givesThe refractive indices and Lensmaker's equation requireSubstitute the two focal lengths and cancel $D-x$:Thus the correcting thin lens is $100\mathrm{mm}$ beyond the red image and $900\mathrm{mm}$ before the screen. The required focal lengths and surface radius areThe blue focal length is shorter because the blue refractive index is larger; these two differing object distances are mapped to the same screen, removing the specified longitudinal chromatic aberration.

Use positive real object and image distances $u,v$ for a thin lens. The required absolute magnification is $M=1000/15=200/3$, and the specified LCD-to-screen separation gives $u+v=2000\mathrm{mm}$, not $v=2000\mathrm{mm}$. Since $v/u=M$,The thin lens equation yieldsPlace the projection lens $29.56\mathrm{mm}$ beyond the LCD towards the screen. Its real image is inverted; the signed magnification is $-200/3$.

Use a converging thin lens as a magnifying glass, with the screen at or just inside its front focal plane. It presents an upright virtual image at infinity or at the eye's near point, respectively. Let the near point distance be $D$; the conventional numerical assumption is $D=250\mathrm{mm}$, which is not specified in the paper.

Under the usual approximation that the eye is close to the eyepiece, interpreting $50\mathrm{mm}$ as the focal length $f$ gives the familiar angular magnificationsThe first is for a relaxed eye with the screen at the front focal plane. For the second, the virtual image is at signed distance $v=-D$ and the thin lens equation gives screen distance $u=fD/(D+f)=41.7\mathrm{mm}$. Thus the often intended maximum $6$ requires a screen-to-eye distance of approximately $41.7\mathrm{mm}$, rather than exactly $50\mathrm{mm}$.

If $50\mathrm{mm}$ instead fixes the physical screen-to-eye separation $s$, an eye adjacent to the optical lens sees the angular magnificationThis is achieved by $f=s$ for relaxed viewing, or by $f=sD/(D-s)=62.5\mathrm{mm}$ for a virtual image at the near point. The latter follows from $1/f=1/s-1/D$.

More generally, the source does not specify the eyepiece-to-eye distance either. If that distance is $e$, then $u=s-e$ and a virtual image at the near point has transverse magnification $(D-e)/u$. Its angular magnification relative to unaided viewing at $D$ isFor example, $s=50\mathrm{mm}$ and $e=25\mathrm{mm}$ give $M=9$, with $u=25\mathrm{mm}$ and $f=28.125\mathrm{mm}$. This is a concrete counterexample to a universal maximum of $5$ at fixed screen-to-eye separation. In ideal geometrical optics, $e\to s^{-}$ allows arbitrarily large $M$ with an arbitrarily short focal length; practical aperture and optical aberrations constrain this. A unique numerical maximum therefore needs the conventional near-eye eyepiece geometry and near point assumption. The conventional $f=50\mathrm{mm}$ answer is $6$, with those assumptions stated explicitly.

The phrase “image” does not specify whether a real image or a virtual image is required. The later mention of an image plane makes a projected real image a natural interpretation. Let $u>0$ be object distance and $v>0$ real-image distance from a converging thin lens. A transverse magnification of magnitude $15$ gives $v=15u$, and the object-to-image separation gives $u+v=200\mathrm{mm}$. HencePlace the optical lens $12.5\mathrm{mm}$ beyond the object along the illuminating beam. The real image is inverted and has transverse magnification $-15$.

For the alternative single-optical lens magnifying glass interpretation, the virtual image is upright and lies on the object's side, so signed $v=-15u$. Its separation from the object is $|v|-u=14u=200\mathrm{mm}$. The thin lens equation then givesThe source does not explicitly select between these two geometries; they should not be conflated with a $200\mathrm{mm}$ lens-to-image distance.

Use the principle of dark-field microscopy: block the unscattered beam where it is concentrated, while transmitting most light scattering at nonzero angles. All direct light rays of the collimated illumination are parallel, so the thin lens focuses them to one point in its back focal plane. Put a small opaque circular stop centred on that point, one focal length behind the optical lens. In the real image geometry above this is $11.719\mathrm{mm}$ behind the optical lens, well before the real image at $187.5\mathrm{mm}$.

A light ray leaving object coordinate $y$ with a small initial direction $\theta$ reaches the optical lens at $y+u\theta$. The thin lens changes its direction to $\theta -(y+u\theta )/f$. At the back focal plane its coordinate is thereforeThus direct illumination with $\theta =0$ is concentrated at the centre, whereas scattered light at different directions is distributed away from it, independently of $y$. A stop of radius $r_s$ removes approximately $|\theta |\lesssim r_s/f$; choose it just large enough to cover the direct beam's finite diffraction spot, illumination divergence and alignment errors. Scattered light inside that small angular cone is also lost, so negligible loss requires most useful scattered light to lie outside it. A stop in the object's real image plane instead removes object detail and does not generally separate scattered and direct light. For the virtual image interpretation, the same physical back focal plane filtering requires a downstream observation arrangement that actually passes through this plane.

Write $T$ for transit time and $\ell (y)=t_0-\alpha y^2$ for lens thickness to avoid using $t$ for both. In the thin lens approximation, the two geometrical distances to the lens plane are $\sqrt{a^2+d^2}$ and $\sqrt{b^2+d^2}$. Their sum counts the lens segment as vacuum. Replacing that segment by material of refractive index $n$ adds $(n-1)\ell (d)$ to the optical path length. Thus, to the retained thin-element accuracy,Using the Taylor expansion $\sqrt{1+u}=1+u/2+O(u^2)$ for $d\ll a,b$ givesThe omitted geometrical terms include $O(d^4/a^3+d^4/b^3)$ in optical path length, together with finite-thickness corrections discarded by the thin lens and paraxial approximations. The addition is $n-1$, because the underlying distance already includes propagation across the thickness once in vacuum.

Fermat's principle makes the optical path length, or transit time in a stationary medium, stationary under small variations of the light ray with fixed endpoints. It need not be a global minimum. In the result of part (a), the coefficient of $d^2$ isFor a family of paraxial light rays from the object to its image, stationarity with respect to $d$ requires this coefficient to vanish. Therefore the thin lens equation isFor a distant object, $a\to \infty$, and the image position $b=f$ defines the focal length. HenceThe axial member $d=0$ is stationary by symmetry for arbitrary $b$; requiring the neighbouring non-axial light rays to meet the same image is what determines $b$.

For the convex vacuum void, use internal refractive index $1$, external refractive index $n$, and the original $\alpha >0$ in the immersed thin lens formula:A concave glass thin lens of the same curvature magnitude in vacuum has ${\alpha }_{\mathrm{concave}}=-\alpha$ and thereforeBoth elements diverge, but their physical focal lengths differ by a factor $n$ for $n>1$. For example, $n=1.5$ makes the void's focal length magnitude $1.5$ times that of the concave glass element. Their reduced optical powers $n_{\mathrm{outside}}/f$ are equal; physical focal length also depends on the surrounding refractive index.

The Fourier expression is the scalar Fraunhofer diffraction approximation for coherent, effectively monochromatic illumination with a uniform incident plane wave over the aperture. At a distant screen the aperture-dependent quadratic propagation phase must be negligible, roughly $D^2/(\lambda z)\ll 1$ for transverse size $D$. Alternatively, a thin lens forms the angular Fourier transform in its back focal plane. That focal-plane construction is the appropriate one here; the aperture-to-lens distance need not itself be an enormous free-propagation distance. aperture features should be large compared with wavelength for this simple scalar transmission-mask model.

If complementary transmissions are $h$ and $1-h$, linearity gives $U_h+U_{1-h}=U_{\mathrm{free}}$. Away from the undiffracted forward beam, the uniform plane-wave contribution $U_{\mathrm{free}}$ vanishes in angular space, so $U_{1-h}=-U_h$ and their intensities agree. This is Babinet's principle. It equates the off-forward diffracted intensities, not the transmitted power or the on-axis intensity including the direct beam.

Let $X,Y$ be screen coordinates conjugate to aperture coordinates $x,y$, so $p\simeq kX/f$, $q\simeq kY/f$. With width $a=0.2\mathrm{mm}$ along $x$ and height $b=0.5\mathrm{mm}$ along $y$, define $\mathrm{sinc}u=\sin u/u$. The complementary rectangle has transformThus the obstruction's off-forward scattered component hasIts dark lines occur atThe central-lobe zero-to-zero size is $1.00\mathrm{mm}$ by $0.400\mathrm{mm}$; it is wider in the direction conjugate to the narrower obstruction. The central forward image of the illuminating beam must be distinguished from this scattered envelope in a real screen observation.

The source figure gives centre-to-centre separation $D=1.0\mathrm{mm}$. Choose the three rectangle centres at $x=-D,0,D$; an overall translation changes only the transform phase. Their scattered amplitude is proportional toThereforeAlong the screen's $x$ direction, the three-element interference factor has principal maxima separated by $\lambda f/D=0.100\mathrm{mm}$, under the single-rectangle envelope. Its zeros lie at $X=(m+1/3)(0.100\mathrm{mm})$ and $(m+2/3)(0.100\mathrm{mm})$. It also has weaker secondary maxima between principal orders. Every fifth principal order coincides with a single-rectangle zero and is missing; multiplication by the envelope slightly shifts maxima away from the nominal interference-factor positions. The figure plots the calculated scattered component and marks the separate direct-beam limitation.

Across a uniformly illuminated slit, the phase difference between its edges is $2\pi a\sin \theta /\lambda$. Integrating the complex amplitude across the slit gives the single-slit diffraction factorThus $\beta =\pi a/\lambda$ in the small-angle amplitude expression. The first noncentral zero has $q=\pi$, so ${\theta }_0\simeq \lambda /a$. At the stated approximate subsidiary maximum,The exact first side maximum solves $\tan q=q$, giving $q\simeq 4.493$ and intensity ratio $0.0472$; the requested approximation need not be mistaken for the exact maximum.

A virtual object is a point toward which incident rays are already converging and would meet if the lens were removed. With real-is-positive object/image distances it has $u<0$. The thin-lens equation gives $1/v=1/f-1/u$; as $u\to -\infty$, $v\to f$, so a distant virtual object images in the rear focal plane.

Without the lens, $y\simeq L\theta$ puts the central maximum at zero and the first dark fringes at $y=\pm L{\lambda }_0/a=\pm 0.100\mathrm{m}$. Subsequent zeros are integer multiples, and weaker subsidiary maxima lie between them.

For the finite-distance pattern compression, take the old screen as the prospective image plane at $L=6\mathrm{m}$, with $u=-L$, and the new screen at $d=0.10\mathrm{m}$. ThenThe plane surface has infinite radius of curvature. The transverse scale is $d/L=1/60$, yieldingThe finite-distance scaling can also be checked without treating an intensity pattern as a collection of geometrical point objects: the slit-to-screen Fresnel integral after a thin lens has quadratic coefficient $1/d-1/f=1/L$ and linear term $-y/d$. It therefore reproduces the old diffraction pattern at transverse coordinate $Y=Ly/d$, up to an overall amplitude and phase. If the old far-field plane is treated as infinitely distant, the equally useful leading approximation is $f=d$, $R=6.0\mathrm{cm}$; the finite-distance correction is only $1.7\%$.

The colour discussion below assumes a smooth white incident spectrum and neglects lens dispersion when comparing the ideal diffraction patterns.

For the real object and real image, take positive distances $u=f+u^{\prime }$ and $v=f+v^{\prime }$. The thin lens equation yieldsThis is Newton's lens formula. The distances $u^{\prime },v^{\prime }$ refer to the adjacent focal points, not the lens plane.

Use the thin lens equation with positive object and real-image distances $u,v$:The transverse magnification is $M=-v/u=-f/(u-f)=-4/296$. For an object dimension $500\mathrm{mm}$, the corresponding image dimension isThe minus sign of the magnification means that the image is inverted. This is the ideal thin-lens prediction; it assumes the relevant field is admitted by the camera and ignores lens distortion.

Keep the paper's $R_1,R_2>0$ as magnitudes for the two convex faces. With the standard signed-curvature convention measured to the right, the first surface has signed radius $+R_1$ and the second $-R_2$. This explains why the requested formula contains a sum of the two reciprocal magnitudes.

Let the axial lens thickness be $\ell$, and let $d_1$ be the distance from the first vertex to the intermediate focus that its refracted rays would reach without the second surface. The first refraction at a spherical surface givesAt the second face those rays are still converging toward that intermediate focus, so its object distance is negative, $u_2=-d_2$, with $d_2=d_1-\ell$. Applying the signed spherical-surface equation again givesIn the thin lens limit, $d_2\simeq d_1$. Adding cancels the intermediate-focus terms:The focal length is the image distance for parallel incident light rays, $u\to \infty$, hence the lensmaker's equation isThe same surrounding refractive index $n_1$ is assumed on both sides. For $n_2>n_1$ this biconvex thin lens has positive focal length. With signed radii $R_a=R_1$, $R_b=-R_2$, the equivalent familiar expression is $1/f=(n_2/n_1-1)(1/R_a-1/R_b)$.