Question

A 100 W light source emits uniformly in all directions. A photodetector having a circular active area whose diameter is 2 cm is placed 1 m away from the source, normal to the incident light. If the responsivity of the photodetector is 0.4 A/W, the photo-current generated in the detector, in units of mA, is
1. 1
2. 4
3. 100
4. 400

Answer 1

Correct Answer - Option 1 : 1

The whole power of 100 W will not fall on the active area

Intensity of light at a distance ‘d’ from an isotropic source is given by

\(I = \left( {\frac{{Power}}{{4\pi {d^2}}}} \right)\frac{W}{{{m^2}}}\)

Here the photodetector is at 1m distance

So, d = 1m

\(I = \frac{{100}}{{4\pi }}\frac{W}{{{m^2}}}\)

Power falling on photodetector = Intensity x active area

\(I = \frac{{100}}{{4\pi }} \times \pi {r^2} = \frac{{100}}{{4\pi }}\;\left( {\pi {{\left( {0.01} \right)}^2}} \right) = \frac{1}{{400}}W\)

Output current = Power on photodetector x Responsivity

\(Output\;current = \frac{1}{{400}} \times 0.4\; = \frac{1}{{1000}}mA = 1\;mA\)

Hence, the photo-current generated in the detector is 1 mA