Question

Optical computers require microscopic optical switches to turn signals on and off. One device for doing so, which can be implemented in an integrated circuit, is the Mach-Zender interferometer seen in FIGURE. Light from an on-chip infrared laser $(\lambda =1.000\mu \mathrm{m})$is split into two waves that travel equal distances around the arms of the interferometer. One arm passes through an electro-optic crystal, a transparent material that can change its index of refraction in response to an applied voltage. Suppose both arms are exactly the same length and the crystal’s index of refraction with no applied voltage is$1.522$.

a. With no voltage applied, is the output bright (switch closed, optical signal passing through) or dark (switch open, no signal)? Explain.

b. What is the first index of refraction of the electro-optic crystal larger than $1.522$that changes the optical switch to the state opposite the state you found in part a?

Short answer

(a) With no voltage applied. the output bright is closed.

(b) $1.597$is the first indexof refraction of the electro-optic crystal.

Step-by-step solution

Explanation (part a)

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.

$\mathrm{\Delta }t=(n-1)\frac{L}{c}$

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

The duration of the light will be

localid="1650221057416" $$\begin{gathered} T=\frac{1}{\nu }=\frac{\lambda }{c}=\frac{1X{10}^{-6}m}{3X{10}^{8}m/s} \\ =3.\stackrel{}{3}X{10}^{-15}s \end{gathered}$$

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.

Find first index of refraction (part b)

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

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

$n_n=1.597$