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Problem 2.1 Principle of Fermat and Snell’s Law.

Consider a layer of thickness dd and refractive index n2n_2 which is sandwiched between two half spaces with refractive indices n1n_1 and n3n_3 as shown in Figure 1. A ray from point P=(xP,yP)P=(x_P,y_P) with yP>dy_P>d passes through point Q=(xQ,yQ)Q=(x_Q,y_Q) with yQ<0y_Q<0.

Ray from P above a slab of thickness d crosses at A, travels through the slab, exits at B, and reaches Q below.

Figure 1:A ray through points PP and QQ.

(a) Write a formula for the OPL of the ray from PP to QQ as shown in Figure 1.

(b) Find the equations satisfied by xAx_A and xBx_B at a stationary OPL. Hint: set the partial derivatives of the OPL with respect to xAx_A and xBx_B equal to zero.

(c) Express the equations derived under b) in terms of sin⁡θ1\sin \theta_1, sin⁡θ2\sin \theta_2 and sin⁡θ3\sin \theta_3 and derive that Snell’s Law holds for the angles θ1\theta_1 and θ3\theta_3:

n1sin⁡θ1=n3sin⁡θ3.\begin{align*} n_1 \sin \theta_1 = n_3 \sin \theta_3. \end{align*}

Note that the relationship between θ3\theta_3 and θ1\theta_1 is independent of the refractive index n2n_2 and the thickness dd of the middle layer.

(d) Does Snell’s Law Eq. (1) hold irrespective of the number of layers and their thicknesses in between the two half spaces? Explain your answer.

(e) Derive Eq. (1) by using the boundary conditions for the tangential components of the electromagnetic field, using Maxwell’s boundary conditions for electromagnetic fields.

Problem 2.2 Perfect focusing by an ellipsoid and a hyperboloid.

Suppose that two media have indices n1>n2n_1>n_2 and the source SS is far to the left in medium n2n_2. We seek an interface that focuses the parallel incident rays at FF in medium n1n_1, as in Figure 2. A plane DD′DD' perpendicular to the incident beam is a constant-phase reference plane.

(a) If point AA is on the interface sought, derive that

n2c∣DA∣+n1c∣AF∣=constant,\begin{align*} \frac{n_2}{c} |DA| + \frac{n_1}{c}|AF| = \text{constant}, \end{align*}

where the constant is the same for all points AA on the interface.

(b) Show that by moving the plane DD′DD' parallel to itself we can achieve that for the new plane DD′DD'

e∣DA∣=∣AF∣,\begin{align*} e|DA| = |AF|, \end{align*}

where e=n2/n1<1e=n_2/n_1<1. This is the focus–directrix definition of an ellipse in cross section, or an ellipsoid of revolution.

(c) Suppose next that n2>n1n_2>n_1. By the same phase-matching argument, show that the interface is a hyperboloid with FF as a focus.

Parallel rays in medium n2 enter a curved interface and converge at focus F in medium n1; an incident reference plane DD prime is marked.

Figure 2:Ellipsoidal interface (n2<n1n_2<n_1) focusing a parallel beam from medium n2n_2 at FF in medium n1n_1. The hyperboloidal case is derived separately.

(d) Use the previous results to describe a lens with refractive index n2>n1n_2>n_1 and having hyperboloid surfaces which perfectly image two given points S and P in the ambient medium with refractive index n1n_1.

Lens-shaped region n2 between lower-index surroundings n1 transforms rays diverging from S into rays converging at P.

Figure 3:Lens with hyperboloid surfaces for perfect imaging of a pair of points.

Problem 2.3 Perfect focusing by a parabolic mirror

Next we consider perfect focusing of parallel rays in air (n=1n=1) by a mirror. Let there be a parallel bundle of rays in air (n=1n=1) and suppose we want to focus all rays at point FF.

(a) We draw a plane Σ1\Sigma_1 perpendicular to the rays as shown in Figure 4. The rays that hit Σ1\Sigma_1 have traversed the same optical path length. Draw a second surface Σ2\Sigma_2 parallel to Σ1\Sigma_1. Consider rays hitting the mirror at A1A_1 and A2A_2. Derive that

OPL=∣W1A1∣+∣A1F∣=∣W2A2∣+∣A2F∣.\begin{align*} \text{OPL}=|W_1A_1| + |A_1F|= |W_2A_2| + |A_2F|. \end{align*}

(b) Derive that

∣W1A1∣+∣A1D1∣=∣W2A2∣+∣A2D2∣.\begin{align*} |W_1A_1| + |A_1D_1|= |W_2A_2| + |A_2D_2|. \end{align*}

(c) Show that Eq. (4) is satisfied for points AA for which ∣AF∣=∣AD∣|AF|=|AD|, and conclude that the mirror is a paraboloid with FF as focus and Σ2\Sigma_2 as directrix.

Parallel rays from reference plane Sigma1 reflect at two points of a parabolic mirror and reach focus F; a second reference plane Sigma2 is shown.

Figure 4:A paraboloid mirror.

Problem 2.4 Imaging of a virtual object.

A virtual object is at a distance d0d_0 behind a converging lens. The incoming rays would converge there if the lens were absent, as their dashed continuations in Figure 5 show. Construct the refracted rays for a lens of focal length ff.

Incoming rays converge toward a point behind a thin lens; dashed continuations locate the virtual object and a focal point on the axis.

Figure 5:Converging incoming rays define a virtual object to the right of the lens; the dashed lines show their continuations before the lens’s refraction is constructed.

(a) Construct the image when f=2f=2 cm, d0=4d_0=4 cm and the height of the object is 1 cm.

(b) Is the image real or virtual? Inverted or upright?

(c) Calculate the location of the image using the lens formula and compare it with your drawing. What is the magnification?

Problem 2.5 Suppose we have two thin lenses L1{\cal L}_1 and L2{\cal L}_2. Lens L1{\cal L}_1 is convergent (i.e. positive) with focal distance f1>0f_1>0 and lens L2{\cal L}_2 is divergent (i.e. negative) with focal distance f2<0f_2<0. The distance between the lenses is dd. Let there be an object at distance 2f12f_1 in front of lens L1{\cal L}_1, as shown in Figure 6.

Object before converging lens L1 and diverging lens L2; their focal points and lens separation d are labeled.

Figure 6:Converging lens L1L_1 followed by diverging lens L2L_2 at separation dd; their object and image focal points are labeled.

(a) Derive the condition on the distance dd between the lenses such that the final image is real.

(b) Let f1=3 cmf_1=3~\text{cm} and f2=−2 cmf_2=-2~\text{cm}. What should be the distance dd such that the final image is real and the magnification is 2?

Problem 2.6 System matrix for imaging by a spherical surface.

Consider a spherical surface with radius of curvature RR, with a medium of refractive index n1n_1 (n2n_2) to the left (right) of the surface).

(a) Derive the ray matrix between a plane to the left of the surface at a distance d1d_1 from the vertex and a plane to the right of the surface at a distance d2d_2 from the vertex. (As always, the rays are assumed to propagate from left to right.)

(b) Derive the conditions such that the plane at distance d2d_2 is the image of the plane at distance d1d_1. Express the formula in coordinates so=−d1s_o=-d_1, si=d2s_i=d_2 with respect to the vertex as origin and show that the result is identical to formula Eq. (26)

(c) Assume that n1=1n_1=1, n2=2n_2=2 and R=2R= 2 cm. Show by construction using the paraxial version of Snell’s Law that when the object is virtual with so=−4s_o=-4 cm, the coordinate of the image point is given by si=8s_i= 8 cm, in agreement with the formula derived in b).

(d) Construct again the image when so=−4s_o=-4 cm but now for the case that R=−2R=-2 cm and verify that the coordinate sis_i of the image point agrees with the formula derived in b).

Problem 2.7 The purpose of this problem is to derive the Lensmaker’s formula for a thin lens by applying the imaging formula of a spherical surface twice. The image PP of a point SS as shown in Figure 7 is computed in two steps. First the intermediate image P′P' of SS by the spherical surface with vertex V1V_1 is computed and then this intermediate image is imaged by the spherical surface with vertex V2V_2.

Two spherical lens surfaces with vertices V1 and V2 and centers C1 and C2; rays from S form an intermediate image P prime and final image P.

Figure 7:A spherical lens made of glass of index nln_l in a medium of index nmn_m. The point SS is imaged in PP.

(a) Use Eq. (26) to deduce that when SS is imaged by the first spherical surface as if the second spherical surface were absent, the zz-coordinate si1s_{i1} of the image P′P' with respect to the origin at V1V_1 satisfies:

−nmso1+nlsi1=nl−nmR1,-\frac{n_m}{s_{o1}}+\frac{n_l}{s_{i1}} =\frac{n_l-n_m}{R_1},

where so1s_{o1} is the zz-coordinate of SS with respect to the origin at V1V_1.

(b) Show that with respect to the origin at V2V_2 the zz-coordinate of P′P' is

so2=si1−d.s_{o2}= s_{i1}-d.

(c) Show that the zz-coordinate si2s_{i2} of P with respect to the origin at V2V_2 satisfies

−nlso2+nmsi2=nm−nlR2.-\frac{n_l}{s_{o2}}+\frac{n_m}{s_{i2}}=\frac{n_m-n_l}{R_2}.

(d) Add the results of a) and c) to derive

−nmso1+nmsi2=(nl−nm)(1R1−1R2)+nld(si1−d)si1.-\frac{n_m}{s_{o1}} + \frac{n_m}{s_{i2}} = (n_l-n_m)\left( \frac{1}{R_1}-\frac{1}{R_2}\right) + \frac{n_l d}{(s_{i1}-d) s_{i1}}.

(e) Derive the Lensmaker’s formula for the thin lens (see the Ray Matrix chapter) by taking the limit d→0d\rightarrow 0 in d).

Problem 2.8 System matrix for focusing.

Consider a ray transfer matrix

(ABCD)\begin{align*} \left( \begin{array}{cc}A & B \\C & D \end{array}\right) \end{align*}

between two planes.

(a) Suppose that any ray that is parallel to the optical axis in the first plane goes through a point on the optical axis in the second plane. This means that the second plane is the image focal plane of the system. What does this imply for the elements of the transfer matrix?

(b) Suppose that the first plane is the object focal plane, so that any ray emitted by the point on the optical axis in the first plane becomes collimated in the second plane. What does this imply for the elements of the transfer matrix?

(c) Consider two thin lenses with distance dd and focal distances f1if_{1i} and f2if_{2i}. Derive the transfer matrix linking the plane immediately before the first lens with the plane immediately behind the second lens. You may assume that the lenses are in air with refractive index n=1n=1.

(d) Use the condition that you found in a) to derive the image focal distance of a system consisting of two thin lenses with image focal distances f1if_{1i} and f2if_{2i} and distance dd. Verify that the result agrees with the distance for the image focal plane Eq. (45). Hint: let fif_i be the distance of the image focal point of the two-lens system to the second lens. Write the transfer matrix between the plane immediately before the first lens and the plane through the image focal point.

(e) Add a third thin lens with image focal distance f3if_{3i} in contact with the second lens. Derive the ray matrix of this system.

(f) Let fif_i be the distance from the image focal plane to the third lens. Use the condition derived in a) and the ray matrix derived in e) to derive fif_i.

Problem 2.9 Matrix for two thin lenses.

(a) Consider two thin lenses in a medium of refractive index nn. Their powers are P1\mathcal{P}_1 and P2\mathcal{P}_2, and the distance between them is dd. Use the ray vector (nα,y)T(n\alpha,y)^{T}, with angle α\alpha first and height yy second. Derive that the matrix between the planes immediately to the left of the first lens and the plane immediately to the right of the second lens is given by

(1−dnP2−P1−P2+dnP1P2dn1−dnP1).\left( \begin{array}{cc}1-\frac{d}{n}{\cal P}_2 & -{\cal P}_1-{\cal P}_2 + \frac{d}{n}{\cal P}_1{\cal P}_2 \\\frac{d}{n} & 1-\frac{d}{n}{\cal P}_1 \end{array}\right).

(b) Define the combined power Peff=P1+P2−(d/n)P1P2\mathcal{P}_{\mathrm{eff}}=\mathcal{P}_1+\mathcal{P}_2-(d/n)\mathcal{P}_1\mathcal{P}_2. Show that the image focal coordinate measured from the second lens and the object focal coordinate measured from the first lens are:

fi=n−dP1Peff.f_i = \frac{n-d\mathcal{P}_1}{\mathcal{P}_{\mathrm{eff}}}.
fo=−n−dP2Peff.f_o = -\frac{n-d\mathcal{P}_2}{\mathcal{P}_{\mathrm{eff}}}.

(c) With Pj=n/fji\mathcal{P}_j=n/f_{ji}, verify that these formulae agree with Eq. (46) and Eq. (45).

Problem 2.10 Entrance pupil of a system of two lenses. Consider a system of two lenses L1L_1 and L2L_2 with distance dd. The left lens L1L_1 has image focal distance f1if_{1i} and a1a_1, a2a_2 are the radii of the lens apertures of L1L_1 and L2L_2.

(a) Let lens L1L_1 be convergent with f1i=2 cmf_{1i}=2~\text{cm} and let the distance be d=1 cmd=1~\text{cm}. Furthermore, let a1=2 cma_1=2~\text{cm} and a2=1 cma_2=1~\text{cm}. Determine by construction with a ruler the entrance pupil. Compute also its position and radius using the thin-lens imaging equation and derive the tangent of the angle that the marginal ray makes with the optical axis for an object on the optical axis at 4 cm to the left of L1L_1.

(b) Same question when L1L_1 is a divergent lens with f1i=−6 cmf_{1i}=-6~\text{cm}, d=3 cmd=3~\text{cm} and a1=a2=1 cma_1=a_2=1~\text{cm}.

Problem 2.11 Diaphragm in a system of two thin lenses. Determine the entrance and exit pupils by ray construction, thin-lens imaging, and the matrix method. The situation is as shown in Figure 8. The focal distances of the two thin lenses are f1=10f_1=10 cm and f2=12f_2=12 cm and their distance d=6d = 6 cm. Suppose that the aperture stop is as shown in Figure 8. It is at a distance of 1.5 1.5 cm in front of the lens L1L_1 and has a diameter Da=5D_a=5 cm.

A diaphragm before two converging thin lenses L1 and L2; object and image focal points, lens spacing, and stop diameter are labeled.

Figure 8:Optical system with two thin lenses and an aperture stop. Lens L₁ (focal length f₁ = 10 cm) and lens L₂ (focal length f₂ = 12 cm) are separated by distance d = 6 cm, with an aperture diaphragm (diameter Dₐ = 5 cm) positioned 1.5 cm in front of L₁. This configuration is used to analyze entrance and exit pupils.

(a) Determine the position of the entrance pupil and its diameter DeD_{e} by the mentioned three methods.

(b) Determine the position of the exit pupil and its diameter DexitD_{exit} by the mentioned three methods.

(c) Compute the ff-number of the system.