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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 index 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.

A ray through points P and Q.

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 to be satisfied by xAx_A and xBx_B such that the OPL is minimum. 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 sinθ1\sin \theta_1, sinθ2\sin \theta_2 and sinθ3\sin \theta_3 and derive that Snell’s Law holds for the angles θ1\theta_1 and θ3\theta_3:

n1sinθ1=n3sinθ3.\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.

From electromagnetic theory, the wave trains emitted by different atoms (point sources) in the source suffer random phase jumps due to e.g. collisions and therefore the fields emitted by different point sources in an extended classical light source can not interfere.

Problem 2.2 Perfect focusing by an ellipsoid and a hyperboloid.

Suppose that there are two media with refractive indices n1>n2n_1>n_2 and that point SS is at infinity in the medium with refractive index n2n_2. We will construct a surface (interface) between the two media such that all rays from SS are focused into the same point FF (see Figure 2a). Because SS is at very large distance, the rays entering from the right are parallel. Since all parallel rays have traveled the same distance when they hit the surface DDDD' perpendicular to the rays, all parallel rays have the same phase at their intersection points with the plane DDDD'.

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

n2cDA+n1cAF=constant,\begin{align*} \frac{n_2}{c} |DA| + \frac{n_1}{c}|AF| = \text{constant}, \end{align*}

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

(b) Show that by moving the plane DDDD' parallel to itself we can achieve that for the new plane DDDD' we get:

(c) Suppose next that n2>n1n_2>n_1. as shown at the right of Figure 2. Show that now, by the same argument as above, the interface is a hyperboloid with FF as one of its focal points.

(a) Ellipsoid (n_2<n_1) and (b) hyperboloid (n_2>n_1) to perfectly focus a parallel beam incident from the medium with refractive index n_2 into a point in a medium with refractive index n_1.

Figure 2:(a) Ellipsoid (n2<n1n_2<n_1) and (b) hyperboloid (n2>n1n_2>n_1) to perfectly focus a parallel beam incident from the medium with refractive index n2n_2 into a point in a medium with refractive index n1n_1.

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

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

Figure 7:Lens with hyperboloid surfaces for perfect imaging of a pair of points.

Problem 2.3 Perfect focussing 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 in 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 in 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. (3) 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 diretrix.

A paraboloid mirror.

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 converging incident rays to the left of the lens, that correspond to the virtual object, are shown in Figure 5. The lens has focal length ff.

An object is created by incoming rays.

Figure 5:An object is created by incoming rays.

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

Figure corresponding to Exercise “A convergent and divergent lens”.

Figure 6:Figure corresponding to Exercise “A convergent and divergent lens”.

(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 to the left (right) of the surface a medium with refractive index n1n_1 (n2n_2).

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

(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 PP' 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.

A spherical lens made of glass of index n_l in a medium of index n_m. The point S is imaged in 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:geo:single-surface-power-form to deduce that when SS is imaged by the first spherical surface as if the second spherical surface were absent, the image PP' has zz-coordinate si1s_{i1} with respect to the origin in VV, of SS, satisfies:

nmso1+nlsi1=nlnmR1,-\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 in VV.

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

so2=si1d.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=nmnlR2.-\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=(nlnm)(1R11R2)+nld(si1d)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 d0d\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 for 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 lens 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} is in contact to 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 which are surrounded by a medium with refractive index nn. Let the left and right lens have power P1{\cal P}_1 and P2{\cal P}_2, respectively and let the distance between their vertices be dd. 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

(1dnP1P1P2dnP1P2dn1dnP2).\left( \begin{array}{cc}1-\frac{d}{n}{\cal P}_1 & -{\cal P}_1-{\cal P}_2 - \frac{d}{n}{\cal P}_1{\cal P}_2 \\\frac{d}{n} & 1-\frac{d}{n}{\cal P}_2 \end{array}\right).

(b) Show that the coordinates of the image and object focal points are given by:

fi=P1+P2+dnP1P2n(1dn)P2f_i = \frac{P_1+P_2 + \frac{d}{n} {\cal P}_1{\cal P}_2}{n \left( 1 - \frac{d}{n}\right){\cal P}_2 }
fo=P1+P2+dnP1P2n(1dn)P1.f_o = \frac{P_1+P_2 + \frac{d}{n} {\cal P}_1{\cal P}_2}{n \left( 1 - \frac{d}{n}\right){\cal P}_1 }.

(c) Verify that these formulae are equivalent to Eq. (46),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 Lensmakers’ formula 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. The purpose of this problem is to determine the entrance and exit apertures of a system consisting of two thin lenses with a diaphragm using three methods: construction, applying the Lensmaker’s formula 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.

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.

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.