Question #302959

1 decibel at 500 nm is 100 W, what is the number of watts at 550 nm?


1
Expert's answer
2022-02-28T10:07:00-0500

Answer

Power is inversely proportional to wavelength.

So

New power

P′=P∗λλ′=500nm∗100W550nm=90.91WP'=\frac{P*\lambda}{\lambda'}\\=\frac{500nm*100W}{550nm}\\=90.91W





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