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A student carries out an experiment to verify the inverse square law for a point source of light - Scottish Highers Physics - Question 8 - 2023

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A student carries out an experiment to verify the inverse square law for a point source of light. (a) Describe an experiment to verify the inverse square law for a ... show full transcript

Worked Solution & Example Answer:A student carries out an experiment to verify the inverse square law for a point source of light - Scottish Highers Physics - Question 8 - 2023

Step 1

Describe an experiment to verify the inverse square law for a point source of light.

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Answer

To verify the inverse square law for a point source of light, set up an experiment where a light source is placed at a fixed position. Use a sensitive light detector (lux meter or photodiode) to measure the intensity of light (irradiance) at varying distances from the light source. The distances should be marked, for example, at 0.20 m, 0.30 m, 0.40 m, 0.50 m, and 0.60 m. Record the light intensity readings for each distance. Repeat the measurements to reduce errors and ensure accuracy. The data collected will show how the light intensity decreases as the distance from the source increases.

Step 2

State what is meant by the term irradiance.

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Irradiance is defined as the power per unit area incident on a surface, typically measured in watts per square meter (W/m²). It quantifies the intensity of light hitting a given area.

Step 3

Use all the data to establish the relationship between irradiance I and distance d.

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Answer

Using the recorded data, we can establish the inverse square relationship between irradiance (I) and distance (d) as follows:

  1. Calculate the product of irradiance and the square of the distance for each data point:

    • For d = 0.200 m: Iimesd2=142.0imes(0.200)2=5.68I imes d^2 = 142.0 imes (0.200)^2 = 5.68
    • For d = 0.300 m: Iimesd2=63.1imes(0.300)2=5.68I imes d^2 = 63.1 imes (0.300)^2 = 5.68
    • For d = 0.400 m: Iimesd2=35.5imes(0.400)2=5.68I imes d^2 = 35.5 imes (0.400)^2 = 5.68
    • For d = 0.500 m: Iimesd2=22.7imes(0.500)2=5.68I imes d^2 = 22.7 imes (0.500)^2 = 5.68
    • For d = 0.600 m: Iimesd2=15.8imes(0.600)2=5.69I imes d^2 = 15.8 imes (0.600)^2 = 5.69
  2. Notice that the product remains constant, verifying the relationship as: Iimesd2=constantI imes d^2 = constant or I = rac{constant}{d^2}.

Step 4

Explain why the irradiance decreases when the distance from a point source of light increases.

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Answer

The irradiance decreases with increasing distance from a point source of light due to the geometrical spreading of light. As the distance increases, the same amount of light energy is distributed over a larger area. The inverse square law explains this phenomenon, where the intensity of light (irradiance) is inversely proportional to the square of the distance from the source. Therefore, as distance (d) increases, the irradiance (I) decreases according to the formula: I = rac{P}{4 ext{π}d^2} where P represents the total power emitted by the light source.

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