Ok got something more from noticing that R3C7 being a light basically forces some lights in the bottom right which make it impossible to place a light in the L from R4C6-R5C6-R5C10. This basically means R3C5 is a light to illuminate R1C5. Then using the fact that one light will be needed in R4 to eliminate R4C6, along with the 2 clue in C8, it is possible to fix an area where 3 lights appear in C7, C8, C9. From here it is easy to finish.