How was analog video converted to an electrical signal?
Before understanding the answer to this question, you need to understand analog video displays, usually called Cathode Ray Tube Televisions (CRT TVs), often shortened to just &quot;CRTs&quot; in casual speech. There are plenty of answers here on this site, and a more technical one on the CRT Wikipedia Page.
Now, before answering the question, there were a multitude of technological advances that led to the collection of 'standard' Video Camera Tubes, each with its own pros and cons, and advancements throughout the years. In this answer, I will assume a black and white system, will ignore the optics, gloss over the physics of electron beams and photo-cathodes (you did read and understand the CRT wiki page... didn't you? :D ),  and I will ignore the electronics and technology involved in the camera, as well as those in the encoding, transmission, reception, decoding, etc. as those items each are far beyond the scope of this answer. (Many have their own dedicated Stack Exchange!!) For a very simple overview, see here.
Here is the short answer:
Analog cameras were the reverse of CRT displays, and the incoming photons from the optics were converted to an output signal in broadly one of two ways.

Incoming photons hit a photo-cathode plate of some type, which emitted electrons. These electrons may or may not have hit a secondary plate with the ability to amplify the electron (not photon) image. This image was then scanned past an aperture (note that the entire image is scanned, not just one beam). Behind this aperture the signal was detected and converted to an electronic signal via a dissector tube (again, many technologies).

The incoming photon image hits another type of photo-cathode plate. In this second technology, an electron beam of constant strength is then shot towards the rear of this plate (scanned and focused in the way a display CRT is), and the plate in the middle creates a voltage relative to the strength of the photons on the front. This voltage is the output signal.
P.S. This answer is in response to an unanswered question asked here by J. Doe