PHALIN THAKKAR

More intermediate filters, more transmission?

What changes if several ideal projections share the rotation between crossed axes?

Note 12 · Research extension

Analytical result · Hypothetical examples are not physical measurements.

Let N be the number of angular steps after a fixed first 0° filter and before the final 90° filter. There are N − 1 intermediate filters.

For monotonic axes with gaps δj between 0 and π/2:

∑j=1Nδj=π2,T=∏j=1Ncos⁡2δj.\sum_{j=1}^{N}\delta_j=\frac{\pi}{2},\qquad T=\prod_{j=1}^{N}\cos^2\delta_j.

Because ln cosδ has second derivative −sec²δ, equal gaps maximize the product:

Tmax⁡=cos⁡2N(π2N).T_{\max}=\cos^{2N}\left(\frac{\pi}{2N}\right).

With one intermediate filter, N = 2 and Tmax = ¼. With two, N = 3: use 30° and 60°, giving Tmax = (3/4)³ = 27/64 ≈ 0.421875.

The fraction is relative to the intensity after the first filter. Add a factor ½ if the reference is an unpolarized source before that filter.

In the ideal limit, Tmax approaches 1 as N increases. Actual aligned transmission losses change this. If each of the N later filters additionally transmits an intensity fraction τ:

Tmax⁡,τ=τNcos⁡2N(π2N).T_{\max,\tau}=\tau^N\cos^{2N}\left(\frac{\pi}{2N}\right).

For the illustrative τ = 0.95, evaluating integer N gives a maximum near N = 7, with T ≈ 0.4894. Adding filters indefinitely eventually loses more intensity.

Continue the reasoning

The third filter

All twelve notes

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