JEE MAIN - Physics (2022 - 26th July Morning Shift - No. 2)
A parallel beam of light of wavelength $$900 \mathrm{~nm}$$ and intensity $$100 \,\mathrm{Wm}^{-2}$$ is incident on a surface perpendicular to the beam. The number of photons crossing $$1 \mathrm{~cm}^{2}$$ area perpendicular to the beam in one second is :
$$3 \times 10^{16}$$
$$4.5 \times 10^{16}$$
$$4.5 \times 10^{17}$$
$$4.5 \times 10^{20}$$
Explanation
$$\lambda$$ = 900 nm
I = 100 W/m2
A = 10$$-$$4
$$\Rightarrow$$ P = 10$$-$$2 W
$$\Rightarrow$$ Number of photons incident per second
$$ = {{{{10}^{ - 2}}\lambda } \over {hc}}$$
$$ = {{9 \times {{10}^{ - 11}} \times {{10}^2}} \over {6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}} \simeq 4.5 \times {10^{16}}$$
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