The irradiance of light incident on a surface from a point source is 20.0 W m⁻² - Scottish Highers Physics - Question 15 - 2022

The irradiance ( I ) of light from a point source is inversely proportional to the square of the distance ( R ) from the source. The formula is:

I = rac{P}{4\\pi R^2}

where:

• P is the power (irradiance at 5m, which is 20.0 W m⁻²)

First, we find the power: For the initial distance (5.0 m):

$$P = I imes 4 \\pi R^2 = 20.0 \, ext{W m}^{-2} \times 4 \\pi (5.0)^2$$

Calculating:

$$P = 20.0 \, ext{W m}^{-2} \times 4 \\pi \times 25 = 20.0 \, ext{W m}^{-2} \times 100 ext{π} \\approx 6283.19 \, ext{W}$$

Now, we find the new irradiance at the new distance (25.0 m):

I_{new} = rac{P}{4 \\pi (25)^2} = \frac{6283.19 \, ext{W}}{4 \\pi (625)}

Calculating:

$$I_{new} = \frac{6283.19}{2500 \\pi} \approx 0.80 \, ext{W m}^{-2}$$

Thus, the irradiance at 25.0 m is approximately 0.80 W m⁻².