A student investigates how irradiance I varies with distance d from a point source of light - Scottish Highers Physics - Question 8 - 2015

From the data provided, we can observe the following:

• At d = 0 m, I = 134 W/m²
• At d = 0.3 m, I = 60.5 W/m²
• At d = 0.5 m, I = 33.6 W/m²
• At d = 0.6 m, I = 21.8 W/m²

To establish the relationship, we can use the inverse square law of light which states that irradiance is inversely proportional to the square of the distance:

$I \propto \frac{1}{d^2}$

This relationship can also be expressed as:

$I = \frac{k}{d^2}$

where k is a constant that can be determined using the data.

Using the relationship established earlier:

If we take the data at d = 0.6 m, we have:

$I = k \cdot (0.6)^2$

From the previous data at d = 0.5 m, using I = 33.6 W/m², we can find k:

$k = I \cdot d^2 = 33.6 \cdot (0.5)^2 = 8.4$

Now, we can find the irradiance at d = 0.6 m:

$I = \frac{8.4}{(0.6)^2} = 23.33 W/m²$

One way to improve the experiment is to use a smaller lamp. A smaller lamp will behave more like a point source of light, which will provide more accurate measurements for irradiance. This ensures that the light spreads uniformly from the source.

To find the irradiance from the lamp with a power output of 24 W at a distance of 2.0 m, we use:

$I = \frac{P}{4 \pi d^2}$

Substituting in the values:

$I = \frac{24}{4 \pi (2)^2} = \frac{24}{16 \pi} = \frac{24}{50.27} \approx 0.477 W/m²$