91Ó°ÊÓ

The phase delays of the light that will flow through the EO crystal must always be determined (that is, how far behind will it physically be, or, to put it another way, how far behind in time). The time it takes for light to pass through with a crystal having length $L$and refractive indices $n$is

$t_{EO}=\frac{L}{c/n}=n\frac{L}{c}$

This duration would be $t=L/c$if there was no EO crystalline. We understand that there will be a discrepancy.

$\mathrm{Δ}t=(n−1)\frac{L}{c}$

We will have interference patterns if the time difference is a large integer of the period; if it is of the type $(n+1/2)T$, we will have complete destructive interference. It's up to us to tell out which case we have.

localid="1650221016576" $$\begin{gathered} \mathrm{Δ}t=(1.522−1)\frac{6.7X{10}^{−6}m}{3X{10}^{8}m/s} \\ =1.1658X{10}^{−14}s \end{gathered}$$

The duration of the light will be

localid="1650221057416" $$\begin{gathered} T=\frac{1}{\mathrm{ν}}=\frac{\mathrm{λ}}{c}=\frac{1X{10}^{−6}m}{3X{10}^{8}m/s} \\ =3.\stackrel{}{3}X{10}^{−15}s \end{gathered}$$

The ratio will be close to perfect $3.5$, implying that the intervention will be completely damaging. As a result, the switch is closed when no force is applied.

Should go without saying that the first index of refraction to engage the switch is the one that achieves our ratio $4.0$. We could indicate this new index of refraction as $n_n$:

localid="1650221092100" $$\begin{gathered} \mathrm{Δ}t=\left(n_n−1\right)\frac{6.7X{10}^{−6}m}{3X{10}^{8}m/s} \\ =4X3.\stackrel{}{3}X{10}^{−15}s \end{gathered}$$

We could now calculate the new angle of refraction.

localid="1650221117693" $$\begin{gathered} n_n−1=\frac{4X3.\stackrel{}{3}X{10}^{−15}sX3X{10}^{8}m/s}{6.7X{10}^{−6}m} \\ =0.597 \end{gathered}$$

Ultimately

$n_n=1.597$