Two sources of light emit with a power of 200 | vedclass

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(D) The power $P$ of a light source is given by $P = n 	imes E_{photon}$,where $n$ is the number of photons emitted per second and $E_{photon} = \fra

Vedclass · Accepted Answer

$3: 5$

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(D) The power $P$ of a light source is given by $P = n 	imes E_{photon}$,where $n$ is the number of photons emitted per second and $E_{photon} = \frac{hc}{\lambda}$.For the first source: $P = n_1 	imes \frac{hc}{\lambda_1} = 200 \ W$.For the second source: $P = n_2 	imes \frac{hc}{\lambda_2} = 200 \ W$.Equating the two expressions: $n_1 	imes \frac{hc}{\lambda_1} = n_2 	imes \frac{hc}{\lambda_2}$.This simplifies to $\frac{n_1}{n_2} = \frac{\lambda_1}{\lambda_2}$.Given $\lambda_1 = 300 \ nm$ and $\lambda_2 = 500 \ nm$,we have $\frac{n_1}{n_2} = \frac{300}{500} = \frac{3}{5}$.

Two sources of light emit with a power of $200 \ W$. The ratio of the number of photons of visible light emitted per second by each source having wavelengths $300 \ nm$ and $500 \ nm$ respectively,will be:

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