Mirror Akari is a slight modification of a puzzle type made by Thomas Synder under the name irakAkari (though the inspiration comes to me without seeing that).
In an unsolved puzzle, there are four types of cells. The first two are identical to Akari, white cells and black cells. Some black cells are written with numbers. The other two are new to Mirror Akari, a mirror cell and a double mirror cell.
To solve a puzzle, one must mark some white cells to be “lights”, represented by a circle in the example above. A white cell can only contain one light. A mirror cell and a double mirror cell, even though they have white parts, may not have any light on them.
First, the number on a black cell indicates the number of lights which are orthogonally adjacent to the black cell. So a black cell with a zero has no light on immediate top, bottom, left, or right of it. At the opposite, a black cell with a four has lights in each of its immediate top, bottom, left, and right of it.
More, a light illuminates other cells. A light illuminates all the squares to each of the four directions until blocked with either the edge of the board or a black cell. So, in the solution above, R1C3 illuminates R1C2, R1C1, and R2C3. It doesn’t illuminate R4C3 since it’s blocked with a black cell on R3C3, and it doesn’t illuminate R2C4 either since the illumination path doesn’t travel diagonally. A light may not illuminate any other lights.
With the new mirrors, illumination paths can be bent. If an illumination path is going to a direction to the open side (the white part) of the mirror, it is reflected by the mirror and continues to the other open side. A double mirror acts as two mirrors placed in one; if an illumination hits a side of the mirror, it is reflected by that side of the mirror. So, in the example above, R2C2’s illumination to its right is reflected by the top-left part of the R2C3 mirror, reflecting the illumination upward to R1C3, and not downward to R3C3. A light cannot illuminate itself by being reflected by mirrors.
Some white cells must be marked as lights such that all white areas (white cells, the open side of a mirror cell, both sides of a double mirror cell) that don’t have any light on them must be illuminated.
Walkthrough of the example
This one is quite difficult for a first puzzle. Note that R1C3 and R2C2 are both adjacent to the 2, so because of the mirror on R2C3, only one of these can have a light. So R1C1 must have a light. Now, since the top (or left) side of R3C2 must be illuminated, a light must be placed on R2C2. The top part is finished.
The bottom part is a bit harder. Note that we must illuminate the bottom part of R2C3, so a light must be placed at R2C4 or R3C3. Besides that, the right part of R3C2 must be illuminated too, so a light must be placed at R3C3 or R3C4. If we choose R2C4 to be illuminated, R3C3 must not be illuminated, so R3C4 must be illuminated. However, this means R3C4 is illuminated by R2C4. So R2C4 is not illuminated, and thus R3C3 is illuminated. The only remaining cell not illuminated, R4C4, must have a light in order for R1C4 to be illuminated…clear.