MURI teleconference 28 May Optical Antimatter. John Pendry and Sebastien Guenneau Imperial College London. 24 May 2004 page 1

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1 24 May 2004 page 1 MURI teleconference 28 May 2004 Optical Antimatter John Pendry and Sebastien Guenneau Imperial College London

2 05 March 2004 page 2 A Conventional Lens Contributions of the far field to the image.. exp( ik z cos θ+ ik x sin θ iω t) 0 0 θ.. are limited by the free space wavelength: θ= 90 gives maximum value of kx = k0 =ω c0 = 2π λ 0 the shortest wavelength component of the 2D image. Hence resolution is no better than, 2π 2πc = =λ0 k ω 0

3 05 March 2004 page 4 Negative Refractive Index and Focussing A negative refractive index medium bends light to a negative angle relative to the surface normal. Light formerly diverging from a point source is set in reverse and converges back to a point. Released from the medium the light reaches a focus for a second time.

4 05 March 2004 page 5 Negative Space A slab of n = 1 material thickness d, cancels the effect of an equivalent thickness of free space. i.e. objects are focussed a distance 2d away. An alternative pair of complementary media, each cancelling the effect of the other. The light does not necessarily follow a straight line path in each medium: d d General rule: two regions of space optically cancel if in each region ε, µ are reversed mirror images. The overall effect is as if a section of space thickness 2d were removed from the experiment.

5 05 March 2004 page 6 Complementary Media We can express our theorem in a graphical fashion: two complementary media have an optical sum of zero. We calculate the optical response of the rest of the system outside the lens region by cutting out the complementary media which comprise the lens and closing the gap between the two remaining halves of the systems. + = A graphical expression of our new theorem: complementary halves sum to zero. The optical properties of the rest of the system can be calculated by cutting out the media and closing the gap.

6 28 January 2004 page 1 A Negative Paradox ε 1 µ 11 ε=+ 1 µ=+1 1 The left and right media in this 2D system are negative mirror images and therefore optically annihilate one another. However a ray construction appears to contradict this result. Nevertheless the theorem is correct and the ray construction erroneous. Note the closed loop of rays indicating the presence of resonances. 2 1

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8 05 March 2004 page 8 Change the Shape: Change ε, µ see A J Ward & J B Pendry Journal of Modern Optics, (1996). step 1: make a coordinate transformation, q1( xyz,, ), q2( xyz,, ), q3( xyz,, ) step 2: rewrite Maxwell s equations in the new coordinate system the form of Maxwell s equations does not change, E = µµ H t, H = εε E t 0 0 step 3: Calculate the new values of ε, µ, QQ Q ε =ε, µ =µ Qi where, Q 2 i QQ Q i i 2 i i 2 Qi x y z = + + q q q i i i, E = Q E, H = Q H i i i i i i conclusion: We can transform the perfect lens solution into new geometries where the surfaces are curved, but we must change ε, µ.

9 24 May 2004 page 2 The Angular Resolution of Telescope θ = 1.22λ D θ D

10 Can we beat the aperture limit for angular resolution? Devices detect the direction of a wave by the oscillations on the surface of the detector. Obviously a larger detector senses more oscillations and is therefore more sensitive to direction. 19 November 2003 page 1

11 24 May 2004 page 3 Beating the Angular Resolution Limit Our strategy: capture all the light entering a circle radius r 1 and force it into a circle of radius r 3 r 3 r 1 r 1 D = 2r1 d = 2r3 the problem: any apparatus must be smaller than 1 r otherwise we do not win. We must capture without touching!

12 24 May 2004 page 4 Using Negative Space to Beat the Angular Resolution Limit r 3 r 1 3 r 1 r 2 In effect we wish to eliminate the space between the red and green spheres. we shall use optical antimatter located r 3 < r < r 2 to annihilate the region of empty space located r 2 < r < r 1. Provided that r2< r1 we have achieved our objective. Using a spherical coordinate transformation we arrive at the following prescription: x y z x y z , 0 3 ε =ε =ε = + r r < r< r 2 2 2, 3 2 ε =ε =ε r r r < r< r ε =ε =ε = + 1, r < r< x y z µ =ε, µ =ε, µ =ε x x y y z z 2

13 r 1 Our design objective. Left, to observers external to the red circle, r> r1, the system is invisible and appears to be transparent to incident radiation. Right: to observers internal to the green circle, r< r3, there is also no evidence of the boundary at r = r3 which is perfectly transparent to outgoing radiation. The material between r 1 and r 3 acts as a wavelength expander/contractor ε =+ 1, ε =+ 1, ε =+ 1, r < r and with identical values for, x y z ε = 1, ε = 1, ε = r r, r < r < r x y z ε =+ 1, ε =+ 1, ε =+ r r =+ r r, r < r x y z µ x =ε x, µ y =ε y, µ z =ε z 19 November 2003 page 4

14 4 4 2 r 3 r z () r () r ε =µ z 1 r 3 r = 0 r2 r 1 The variation of ( r) 2 2 ε z with radius r = x + y. Note that ε z is constant outside the active region r 3 < r< r 2 where the focussing takes place. 19 November 2003 page 5

15 r 1 r2 r 3 A suitably designed negative material (blue shading) placed in the cylindrical annulus between r 2 and r 3 will compress the wave field originally within the cylinder r 1 to fit inside the smallest cylinder radius r November 2003 page 3

16 R r φ r 1 r 2 r 3 A ray, represent by the red line, is incident on a cylinder with impact parameter R. A point along the trajectory is defined by the angle φ. 19 November 2003 page 6

17 08 March 2004 page 7 ` 1a) 1b) An optical turbine. A plane wave entering the red sphere from the left is captured and compressed inside the green sphere. a) A ray picture which shows only part of the rays being captured b) An exact solution of Maxwell s equations. The green sphere is filled with the compressed contents of the red sphere as predicted. The region outside the blue sphere is free space.

18 Three separate calculations of H z for a system containing a negatively refracting material between the cyan and green cylinders, and a high refractive index material inside the green cylinder. On the left is an overview showing all three cylinders: r 1 - red, r 2 - cyan, r 3 green. To the right an expanded scale showing just the inner cylinder. The top pair is calculated for low losses, δ= 0.001, the next pair for δ= 0.01, and the bottom pair for δ= 0.1. Note how as the loss is increased the fields are confined to the region occupied by the rays. 19 November 2003 page 9

19 05 March 2004 page 7 Conclusions Negative refraction provides a new way of focussing light Negative lenses have resolution limited only by loss A slab of negative index space optically annihilates an adjacent slab of positive index space Any two adjacent slabs obeying a negative mirror condition optically annihilate The result can be extended to spherical, cylindrical, and other geometries so that adjacent shells annihilate one another In principle we can use this result to remove the limitations on angular resolution imposed by aperture However the absolute limitations are now replaced by practical limitations due to losses