Question Video: Identifying the Effect of Coherent Light on a Resonant Cavity | Nagwa Question Video: Identifying the Effect of Coherent Light on a Resonant Cavity | Nagwa

# Question Video: Identifying the Effect of Coherent Light on a Resonant Cavity Physics • Third Year of Secondary School

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Consider two identical lasers. Energy is transferred to one of the lasers to make the laser emit coherent light. The coherent light is directed into the resonant cavity of the other laser. Which of the following most correctly describes the effect of the coherent light on atoms in the active medium of the laser that the light is directed into? [A] Electrons in the ground state transition to an excited state and then decay to the ground state. [B] Electrons in the ground state transition to an excited state, then decay to a metastable state, and then decay to the ground state. [C] Electrons in the ground state transition to a metastable state and then decay to the ground state.

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### Video Transcript

Consider two identical lasers. Energy is transferred to one of the lasers to make the laser emit coherent light. The coherent light is directed into the resonant cavity of the other laser. Which of the following most correctly describes the effect of the coherent light on atoms in the active medium of the laser that the light is directed into? (A) Electrons in the ground state transition to an excited state and then decay to the ground state. (B) Electrons in the ground state transition to an excited state, then decay to a metastable state, and then decay to the ground state. Or (C) electrons in the ground state transition to a metastable state and then decay to the ground state.

Before we start to answer this question, let’s make an outline of the scenario we’ve been given. We have two identical lasers. Coherent light from the first laser is directed into the resonant cavity of the other laser, meaning the first laser acts as an energy source. We need to work out how the coherent light from the first laser affects the electrons in the active medium of the second laser.

Let’s start by thinking about what’s going on in the active medium of the first laser. The active medium of a laser has three energy levels: a ground state, a metastable state, and an excited state. Let’s label the energies of each of these states. We’ll say that the ground state corresponds to an energy of 𝐸 g, the metastable state corresponds to an energy of 𝐸 m, and the excited state corresponds to an energy of 𝐸 e.

Most of the electrons will start out in the ground state. An external energy source can be used to excite the electrons by providing photons to the active medium. If an electron absorbs a photon of exactly the right energy, it can be excited to a higher state. Usually, the electrons in a laser transition from the ground state to the excited state, bypassing the metastable state. This occurs if the electron absorbs a photon with an energy equal to the difference between the excited state and the ground state, 𝐸 e minus 𝐸 g.

An electron in an excited state has a very short lifetime, meaning it will soon spontaneously decay down to the next state, the metastable state. This spontaneous decay causes a photon to be released, which has an energy equal to the difference between these two energy levels, 𝐸 e minus 𝐸 m.

However, the photons produced by spontaneous emission are not useful for producing laser light. The phase and direction of these photons is random, meaning that these photons don’t add up to produce a strong laser beam. However, once the electron is in the metastable state, it will stay there for a long time, long enough for stimulated emission to occur.

Stimulated emission is when a photon traveling through the active medium interacts with the electron and causes the electron to emit another photon and decay to a lower energy level. The emitted photon has the same energy, phase, and direction as the photon that caused the emission. Because this occurs when the electron moves from the metastable state back down to the ground state, the photons must have an energy equal to the difference between the energies of these two states, 𝐸 m minus 𝐸 g.

If lots of stimulated emissions occur in the medium, we end up with a large number of identical photons, which together can form a strong beam of laser light. All the photons in the beam produced by the first laser will have an energy of 𝐸 m minus 𝐸 g.

Now let’s think about what’s going on inside the second laser. We know this laser is identical to the first, so the active medium will have the same energy levels as before. Again, our electron is starting off in the ground state. This time, however, the beam from the first laser is being directed into the active medium. This means that lots of photons, all of which have an energy of 𝐸 m minus 𝐸 g, will travel through the active medium. What happens when an electron in the ground state absorbs one of these photons?

Well, initially, the electron has an energy 𝐸 g. When it absorbs the photon, its energy becomes 𝐸 g plus the energy of the photon, giving it a total energy of 𝐸 g plus 𝐸 m minus 𝐸 g. This simplifies to 𝐸 m, the energy of the metastable state. So, when an electron in the ground state absorbs a photon from the first laser, it will transition to the metastable state. The electron can then stay in the metastable state until another photon causes a stimulated emission, just like we discussed for the first laser. Just like before, this will cause the electron to emit a photon of energy 𝐸 m minus 𝐸 g and decay back down to the ground state.

So let’s summarize what happens to the electrons in the active medium of the second laser. An electron starts out in the ground state, absorbs a photon, transitions to the metastable state, then emits a photon, and moves back down to the ground state.

We’re now ready to answer this question. Let’s go through the answer options. We can rule out (A) and (B) because the electrons in the second laser never reach the excited state. They never come into contact with photons of a high enough energy for this to be possible. Instead, electrons in the ground state transition to a metastable state and then decay to the ground state, just as described in option (C). So option (C) is the correct answer to this question.

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