0
$\begingroup$

I am trying to build an optical logic circuit using the horizontal or vertical orientations of light as the two binary states and laser as the source. Before I ask you my question I wanted to first share how I designed the NOT Gate. enter image description here

NOT Gate To explain the working let's assume I send in a horizontally oriented beam of light. This beam splits into two beams. Because it is horizontally oriented, the first split beam won't enter the V Polarizer and gets blocked. The second split beam enters the H Polarizer and consequently gets vertically oriented at the end and the lower ldr detects the light, notifying us that it is now in V state.

My Problem Now, I am not able to figure out another combination of polarizers for AND and OR Gate. Can anyone help me in giving me a lead?

$\endgroup$
  • $\begingroup$ You don't even say what the logical 0 and 1 are. How should one answer the question? $\endgroup$ – Norbert Schuch Aug 14 at 17:28
  • $\begingroup$ I have mentioned that in the second and third line. Read properly. $\endgroup$ – Bhavesh Aug 15 at 18:14
  • $\begingroup$ I see. (Note that "second and third line" depends on the screen!). But then, why don't you just use a lambda/2 plate for the not gate? Your setup seems unnecessarily complicated! $\endgroup$ – Norbert Schuch Aug 15 at 18:38
  • $\begingroup$ I don't have access to wave plates, only linear polarizers $\endgroup$ – Bhavesh Aug 16 at 14:28
2
$\begingroup$

You’ll need the state of one beam to affect the state of another. This is a nonlinear effect requiring a potentially complex arrangement of optics and materials to make it efficient enough to be practical. First, read about nonlinear optics, then try looking up all-optical switches, for example. You won’t be able to do it with a few polarizers since they are linear optical elements.

Also, you can improve your NOT gate by switching your design with a single half-waveplate oriented at 45$^\circ$. This new design would be 100% efficient.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.