∵B1o You are standing 2.5 m from a 150-W lightbulb. (a) If the pupil of your eye is a circle 5.0

step 1 So in this problem we have a light bulb with a power of 150 watts place at a distance of 2 .5 me

∵B1o You are standing 2.5 m from a 150-W lightbulb. (a) If...

$\because$ B1o You are standing 2.5 $\mathrm{m}$ from a $150-\mathrm{W}$ lightbulb. (a) If the pupil of your eye is a circle 5.0 $\mathrm{mm}$ in diameter, how much energy enters your eye per second? (Assume that 5.0$\%$ of the lightbulb's power is converted to light.) (b) Repeat part (a) for the case of a 1.0 -mm-diameter laser beam with a power of 0.50 $\mathrm{mW}$ .

$\because$ B1o You are standing 2.5 $\mathrm{m}$ from a $150-\mathrm{W}$ lightbulb. (a) If the
pupil of your eye is a circle 5.0 $\mathrm{mm}$ in diameter, how much energy
enters your eye per second? (Assume that 5.0$\%$ of the lightbulb's
power is converted to light.) (b) Repeat part (a) for the case of a
1.0 -mm-diameter laser beam with a power of 0.50 $\mathrm{mW}$ .

Key Concepts

Energy Flux and Fractional Capture

Energy flux, or irradiance, is the power incident per unit area at a distance from the source. By calculating the intensity and then multiplying by the area of the aperture (or target), one can determine the energy captured per second. This approach is used to calculate the energy entering the eye or any other surface by taking into account the ratio of the aperture area to the area over which the light is spread.

Power Conversion Efficiency

Many light sources do not convert all their power into visible light. The power conversion efficiency is the fraction of the total energy output that is emitted as light. Understanding and applying this efficiency factor is important to accurately determine the actual power available as light from the source before further geometric considerations are applied.

Intensity Distribution and the Inverse Square Law

When a light source emits energy uniformly in all directions, the light spreads out over the surface of an expanding sphere. The intensity (power per unit area) at any distance from the source is determined by dividing the total power by the surface area of that sphere, which increases with the square of the distance. This concept is critical for determining how much light from a source reaches a particular area at some distance.

Aperture or Pupil Area Calculation

The amount of light that enters an optical instrument, like the human eye, depends on the area of its aperture. For a circular pupil, the area is computed using the formula for the area of a circle. This geometrical calculation is necessary to find the fraction of the incident light captured by the pupil.

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