Consider an image projected onto a screen with parallel rays of light, as opposed to an image formed by the same rays that are passed through a cubic prism (assume that the amount of light that is reflected is negligible). The rays that pass through the prism will not be refracted since the angle of refraction = sin

A horizontal ray is required to undergo a 45º-angle change and this has to be achieved using a mirror. We need to find how the mirror should be oriented to achieve the desired change of angle.

sinIif we define the incoming and outgoing rays ray and the normal of the refractive surface as vectors and using a property of the cross-product we can say the following_{0}/n_{0}= sinI_{1}/n_{1}

Qand also_{0}xM_{1}= |Q_{0}||M_{1}| sinI_{0}= sinI_{0}

Qthus_{1}xM_{1}= |Q_{1}||M_{1}| sinI_{1}= sinI_{1}

NIf we introduce two new vectors S_{0}(Q_{0}xM_{1})= n_{1}(Q_{1}xM_{1})

Sor_{0}x M_{1}= S_{1}xM_{1}

(Sthis implies that (S_{1}-S_{0})xM_{1}= 0

S1 – SHow is useful for solving our problem? It turns out that reflection can be described using the same laws and as refraction and letting -n_{0}= ΓM_{1}

Γ =-2 nThis equation can be generalized to describe a ray reflecting off of a sequence of mirrors (see picture),_{1}cosI_{0}

Click on any image to get the PostScript source.and it is also convenient to introduce a new variable ρ which will equal Γ/-2. Thus using a property of the dot product we have:

ΓIn our case since we are only dealing with one mirror we can let i = 1_{i}= -2(S_{i-1}∙M_{i}) = -2ρ_{i}

Sand_{1}= S_{0}+ Γ_{1}M_{1}

ρSince the ray is horizontal and is a unit vector we will let S_{1}= S_{0}∙M_{1}

(√2/2, √2/2) = (1,0) – 2((1,0) ∙Mor_{1})M_{1}

(√2/2, √2/2) = (1,0) – 2(MIf we split the vector equation into two that involve only scalar values we will get_{x})M_{1}

√2/2 = 1 – 2M_{x}2

√2/2 = 0 – Mthus_{x}M_{y}

Mand_{x}= √((√2/2-1)/2) = .382

MWe can also find the angle of the mirror_{y}= -√2/(4M_{x}) = -.923

a = -tan-1(Mwhich is intuitively the right answer. //_{x}/M_{y}) = -tan^{-1}(.382/-.923) = 22.5º

SOf course 2d and 3d models are completely analogous, here the 3d one will be used. First we can see that if we split up the vectors into their individual components,_{i}= S_{i-1}+2(S_{i-1}∙M_{i})M_{i}

S_{ix}= S_{(i-1)x}+2M_{ix}(S_{(i-1)x}M_{ix}+S_{(i-1)x}M_{iy}+S_{(i-1)z}M_{iz})

S_{iy}= S_{(i-1)y}+2M_{iy}(S_{(i-1)x}M_{ix}+S_{(i-1)x}M_{iy}+S_{(i-1)z}M_{iz})

Sthe equation can be written in a matrix form. Specifically:_{iz}= S_{(i-1)z}+2M_{iz}(S_{(i-1)x}M_{ix}+S_{(i-1)x}M_{iy}+S_{(i-1)z}M_{iz})

| SWe can try to find the determinant of the matrix, for convenience, in the following calculations M_{ix}| | (1-M_{ix}^{2}) -2M_{ix}M_{iy}-2M_{ix}M_{iz}|| S_{(i-1)x}| | S_{iy}| = | -2M_{ix}M_{iy}(1-M_{iy}^{2}) -2M_{iy}M_{iz}|| S_{(i-1)y}| | S_{iz}| | -2M_{ix}M_{iz}-2M_{iy}M_{iz}(1-M_{iz}^{2}) || S_{(i-1)z}|

D = (1-2x^{2})((1-2y^{2})(1-2z^{2})-4y^{2}z^{2}) +2xy(-1(1-2z^{2})2xy - 4yz^{2}x) - 2xq((4xy^{2}z+2xz(1-2y^{2})

D = (1-2x^{2})(1-2y^{2}-2z^{2}+4y^{2}z^{2}-4y^{2}z^{2}) -2xy(2xy-4xyz^{2}+4yz^{2}x) -2xz(4xy^{2}z+2xz-4xzy^{2})

D = 1 - 2y2-2z^{2}-2x^{2}+4x^{2}y^{2}+4x^{2}z^{2}-4x^{2}y^{2}-4x^{2}z^{2}

D = 1 - 2(xsince M is a unit vector^{2}+y^{2}+z^{2})

xand^{2}+y^{2}+z^{2}= 1

D = -1If the system involves many mirrors then the transformation has to be applied to the vector multiple times. If we let the above mentioned matrix be R

(SThus we can see that the determinant of the matrix for the whole system is going to equal -1 if there are an odd number of matrices that are multiplied, i.e. and odd number of mirrors and it is going to equal 1 if the number of mirrors is even. The interpretation of this is that the image produced by an even number of mirrors is right handed i.e. "R" and an image produced by an even number of mirrors is left handed i.e. "Я". If the right-handed image is plain text, the left handed one is unreadable. This is a key principle in the design of prisms._{n}) = [R_{n}][R_{n-1}]...[R_{1}](S_{0})

If the front side is of length A then the C will be of length 1.414A. It can also be used to reverse an image i.e. turn it 180º(see picture).

In this case it will produce a left-handed image. Another example is the Dove prism. It is used to invert an image without changing its direction (see picture).

r = sinit's angle to the horizontal is^{-1}(sin(45º)/n)

a = sinnow we can find the length of side B. A is added to account for the offset of the top corner.^{-1}(sin(45º)/n) - 45º

B = A tan(sinif we let n equal 1.5170, then^{-1}(sin(45º)/n) - 45º)+A

B = 4.227Aand

C = 4.227A - 2A = 2.227ANote that the prisms symmetry implies that the rays hitting the bottom of the front side will hit the top of the back side. //

Is it possible to make a prism with the same effect as the Dove prism but more compact? In other words how can the size to aperture ration be reduced? If the prism is rotated around the central ray by 180 degrees it will produce the same image. Thus two Dove prism with their respective side B's placed together will produce an image twice the size, but it will retain it's original length. Note that the rays that hit the bottom and the top of the prism will take opposite paths. The only remaining concern is that the rays that are refracted by the bottom side of each prism will produce a double image. This is remedied by silvering the two bottom sides, which insures that no rays pass through.

Another prism that has the same effect as the Dove prism is the Pechan Prism (see Picture). It has an advantage of that it can be placed in diverging of converging light.

As noted above the right angle prism produces a left image. If however the same angle change is required, but a right handed image is needed, a Pentaprism can be used (see picture).

B = 1/cos(22.5º)A = 1.08239Awe can also find side c:

x = tan(22.5º)Aand

C = √(2(tan(22.5º)A)Notice that unlike the Dove prism, the dimensions of the Pentaprism are independent of the value of n.//^{2}) = .5857A

Rwhere R_{p}= 180º + R_{i}

RWhere R_{f}= R_{i}+ R_{p}= 2R_{i}+180

dRThis implies that the final image is rotated at twice the speed of the rotation of the prism. For this reason it is called a rotator. Let's see how this fact can be used._{f}= 2dR_{i}

A ray first passes through a right-angle prism, then a Dove prism, and then an Almici roof prism (a right-angle prism cannot be used at the bottom, because that would mean that the ray will be reflected 3 times and thus the final image will be left-handed). Now the top prism can be rotated while the bottom one remains fixed. How should the Dove prism behave? Since it rotates an image at twice the rate of the rotation of the incoming image, the Dove prism should rotate at half the angular speed of the right-angle prism. This is achieved using differential gears Keep in mind that normally a periscope would include several lenses to increase the field of view and add a magnification factor.

which produces exactly the same image in the viewfinder and on the film. Let's look at how this is achieved. After the rays pass through the lenses, the image is both upside down and has right and left reversed. This is inconsequential with respect to the film but must be corrected by the viewfinder. Until the moment that the picture is taken, the rays are reflected at 90º by a hinged mirror. Just before the shot, the mirror swings out of the way and the rays hit the film producing the photograph. Otherwise after being reflected off the mirror the rays are traveling vertically and need to be deflected 90º towards the observer. If a right-angled prism was used for this then the resulting image would appear upside down. This can be ascertained by bouncing an imaginary pencil off the surfaces of the mirror and the prism (see picture),

keeping in mind that the image is flipped vertically when it passes through the lens. If however, a pentaprism was used the image will appear oriented the right way vertically. However there is still a problem because the image was flipped horizontally by the lens, and the lens only. To get the right horizontal orientation a penta-roof prism needs to be used. It adds one more reflection and flips the image horizontally.

Michael Langford, Alfred A. Knopf,

Walter G. Driscoll, William Vaughan,

Rudolf Kingslake,