The irradiance on a surface 0-50m from a point source of light is I - Scottish Highers Physics - Question 12 - 2018

To determine the irradiance at a distance of 1.5 m from the point source of light, we can utilize the inverse square law of light, which states:

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

where:

• I is the irradiance
• P is the power of the source (assumed constant)
• r is the distance from the source.

Given that the irradiance at 50 m is I, we can re-arrange the formula for the distance 50 m: $I = \frac{P}{4\pi (50)^2}$

To find the irradiance at 1.5 m: $I' = \frac{P}{4\pi (1.5)^2}$

To express $I'$ in terms of $I$, we can set: $I' = I \times \left( \frac{(50)^2}{(1.5)^2} \right)$

Calculating the ratio: $\frac{(50)^2}{(1.5)^2} = \frac{2500}{2.25} \approx 1111.11$

Thus, the irradiance $I'$ at 1.5 m is roughly: $I' \approx 1111.11 \cdot I$

As calculated, substituting with the appropriate values leads us to answer Option A, indicating that the irradiance at 1.5 m is significantly greater compared to that at greater distance.