A light detector (your eye) has an area of `2.00 xx 10^(-6) m^(2)` and absorbs `80%` of the incident light, which is at wavelength 500 nm. The detector faces an isotropic source, 3.00 m from the source. If the detector absorbs photons at the rate of exactly `4.000 s^(-1)`, at what power does the emitter emit light?

Explanation

To solve the problem step by step, we will follow the outlined approach:

### Step 1: Determine the area of the detector and the distance from the source
- The area of the detector (your eye) is given as \( A = 2.00 \times 10^{-6} \, \text{m}^2 \).
- The distance from the light source is \( r = 3.00 \, \text{m} \).

### Step 2: Calculate the total power incident on the detector
Since the source is isotropic, the power \( P \) emitted by the source can be related to the intensity \( I \) at the detector's location using the formula:
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