(a) At this intensity, what is the average power incident on a pupil of diameter

Question

(a) At this intensity, what is the average power incident on a pupil of diameter 8.0 mm?
W

(b) If this light is produced by an isotropic source 11.0 m away, what is the average power emitted by the source?
W

Under favorable conditions, the human eye can detect light waves with intensities as low as 2.5 x 10^-12 W/m^2 (a) At this intensity, what is the average power incident on a pupil of diameter 8.0 mm? W (b) If this light is produced by an isotropic source 11.0 m away, what is the average power emitted by the source? W

Explanation / Answer

Here ,

I = 2.5 *10^-12 W/m^2

a)

Power = area*I

Power = pi*(.004)^2 * 2.5 *10^-12

Power = 1.2564 *10^-16 W

b)

Here ,

as Intensity = P/(4pi*r^2)

2.5 *10^-12 = P/(4*pi*11^2)

P = 3.801 *10^-9 W

the average power emitted is 3.801 *10^-9 W

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