## Nanosphere Lithography Geometry

All labeling refers to the figure above.

1. The length of the perpendicular bisector of the largest possible equilateral triangle that fits in the space between the circles, GI is:

GI = 3d/2 * (√3 -1 – 1/√3)

Proof: First, we note that the angle GHI is 60º and angle GHK is 30º, thus GI is 3 times longer than GK because tan(60º)/tan(30º) = 3.

GI = 3 GK.                                                                                                               [I]

Next we note that angle DAK is 30º, and calculate

AK = AD/cos(30º) = 2r/√3 = √3 r – r/√3,                                                                              [II]

GK = AK – r [III]

From III and II,

GK = √3 r – r/√3 – r = d/2 * (√3 -1 – 1/√3)                                                                           [IV]

where 2r = d (diameter)

Therefore from [I] and [IV],

GI = 3d/2 * (√3 -1 – 1/√3) QED.

2. The length of the perpendicular bisector of the entire cusp, GE is:

GE = AE – AG = √3 r – r.

Proof:

GE = AE – AG = AE – r.                                                                                                   [I]

AE = AB cos(30º) = √3 r.                                                                                                [II]

Therefore,

GE = √3 r – r. QED.

Note, the length produced during a real deposition depends on the deposition conditions and any subsequent annealing.

3. The percentage area occupied by the space between the nanosphere mask is: 1 – π/(√3*2) ≈ 9%

Proof: Since the nanosphere mask can be decomposed into identical unit triangles, ABC, we examine ABC.

Area of triangle ABC = ½ CB * AE                                                                                    [I]

AE = AB cos(30º) = √3d /2 = √3r                                                                                   [II]

Where d (sphere diameter) = 2r (radius).

Therefore from [I] and [II],

Area ABC = ½ 2r * √3r = √3 r^2                                                                                   [III]

This area includes 3 arcs of circles, each 1/6th of a circle in area, thus the area occupied by the circle arcs is

3 * π r^2 / 6 = π r^2 / 2                                                                                             [IV]

The area unoccupied by the circles is total area – circle area, from [III] and [IV], this is

r^2 (√3 – π/2).                                                                                                      [V]

As a percentage area, this is [V]/[III]

r^2(√3 – π/2)/ (√3 r^2) = 1 – π/(√3*2) ≈ 9% QED.

4. The % area occupied by the smallest equilateral triangle that fits in the space between the circles HIJ is: 7√3/3 –4 ≈ 4%

Proof: Since ABC and HIJ are both equilateral triangles, the area of HIJ is the square of the ratio between two similar sides.

% area = (GI/AE)2                                                                                                  [I]

AE = AB cos(30º) = √3d /2                                                                                      [II]

From above (see proof 1), GI = 3d/2 * (√3 -1 – 1/√3).                                                        [III]

Therefore, from [I], [II], [III], percentage area

(GI/AE)2 = 3 d/2 * (√3 -1 – 1/√3)2/(√3 d/2 )
= √3 *(√3 -1 – 1/√3)2 = 7√3/3 –4 ≈ 4%
QED.

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• ### Professor Richard Van Duyne

Discoverer of Surface-enhanced Raman Spectroscopy (1977)
Inventor of Nanosphere Lithography (1995) & Localized Surface Plasmon Resonance Spectroscopy (2000)

• ### Group Members

Professor Van Duyne has, in his career to date, advised a total of 87 graduate students and 47 postdoctoral fellows. Every year, Professor Van Duyne gives a talk to introduce new graduate students to our research. The 2017 seminar slides are available here.

• ### News

Professor Van Duyne was recently named a Vannevar Bush Faculty Fellow by the U.S. Department of Defense to conduct "high risk, high payoff" basic scientific research. Read more here