A source emits laser beam with an average power of 6 mW. Find the number of photons emitted per second by the laser if 6328 $\mathring{A}$ is the $\lambda$ of emitted light.

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written 7.9 years ago by

neeta.vanage • 880

modified 3.9 years ago by

pedsangini276 • 4.8k

Subject :- Applied Physics 2

Topic :- Lasers

Difficulty :- Medium

engineering physics

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1 Answer

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written 7.8 years ago by

neeta.vanage • 880

modified 7.8 years ago by

saurabhsingh • 40

$$ $$ $ P =6* 10 -3 watts \\ T = 1 sec \\ λ = 6328*10^{-10} m \\ n = ? \\ $

$ \text{ E = hν is energy of one photon Therefore energy of n photons will be E = nhν } \\ $

$\text{ But ν = c /λ } \\ $

$ E = nhc /λ \\ P = E / t \\ 6 *10^{-3} = n* 6.634*10^{-34} * 3 *10^8 / 6328*10^{-10 } \\ n = 1.91*10^{16} \\ $

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