A source $S_{1}$ is producing $10^{15}$ photons per second of wavelength $5000\mathring{\mathrm{A}}.$ Another source $S_{2}$ is producing $1.02 imes 10^{15}$ photons per second of wavelength $5100\mathring{\mathrm{A}}.$ Then (power of $S_{2}$)/(power of $S_{1}$) is equal to

Question

A source $S_{1}$ is producing $10^{15}$ photons per second of wavelength $5000\mathring{\mathrm{A}}.$ Another source $S_{2}$ is producing $1.02 	imes 10^{15}$ photons per second of wavelength $5100\mathring{\mathrm{A}}.$ Then (power of $S_{2}$)/(power of $S_{1}$) is equal to

Dr. Abubacker · Accepted Answer

(b) $\frac{Power\ of\ S_{2}}{Power\ of\ S_{1}} = \frac{n_{2}\left( \frac{hc}{\lambda_{2}} \right)}{n_{1}\left( \frac{hc}{\lambda_{1}} \right)} = \frac{n_{2}\lambda_{1}}{n_{1}\lambda_{2}} = 1$

Q. A source $S_{1}$ is producing $10^{15}$ photons per second of wavelength $5000\mathring{\mathrm{A}}.$ Another source $S_{2}$ is producing $1.02 \times 10^{15}$ photons per second of wavelength $5100\mathring{\mathrm{A}}.$ Then (power of $S_{2}$)/(power of $S_{1}$) is equal to

Explanation

Reviewed by Dr. Abubacker

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