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Physics Question on Ray optics and optical instruments

A parallel beam of light of wavelength 900 nm and intensity 100 Wm-2 is incident on a surface perpendicular to the beam. The number of photons crossing 1 cm-2 area perpendicular to the beam in one second is

A

3×10163 × 10^{16}

B

4.5×10164.5 × 10^{16}

C

4.5×10174.5 × 10^{17}

D

4.5×10204.5 × 10^{20}

Answer

4.5×10164.5 × 10^{16}

Explanation

Solution

Given,
λ=900 nmλ=900\ nm
I=100 W/m2I=100\ W/m^2
A=10−4A=10^{−4}
P=10−2 WP=10^{−2}\ W
Number of photons incident per second
=10−2λhc=\frac {10^{−2}λ}{hc}

=9×10−11×1026.63×10−34×3×108=\frac {9×10^{−11}×10^2}{6.63×10^{−34}×3×10^8}

≃4.5×1016≃4.5×10^{16}

So, the correct option is (B): 4.5×10164.5×10^{16}