A-Level Physics May/June 2024 Q1(b)(i): The power of the light incident on the solar panel is 750W. Calculate the intensity of…

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  The relationship between intensity $I$, power $P$, and cross-sectional area $A$ is given by the formula: $I = \frac{P}{A}$

  The area of the square solar panel must be calculated first. The side length is 1300 mm, which must be converted to SI units (metres) before squaring. $A = \text{length}^2 = (1300 \times 10^{-3} \text{ m})^2$ $A = (1.3 \text{ m})^2 = 1.69 \text{ m}^2$

  Now, substitute the values for power and area into the intensity formula: $I = \frac{750 \text{ W}}{1.69 \text{ m}^2}$ $I = 443.786... \text{ W m}^{-2}$

  The input power (750 W) is given to two significant figures, so the answer should also be rounded to two significant figures. $I = 440 \text{ W m}^{-2}$

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  How the marks are awarded

  • C1 — Correctly stating the formula for intensity, $I = P/A$.
  • C1 — Correctly substituting values for power and area, including the conversion of the side length from mm to m and squaring it, as shown by the working $750 / (1300 \times 10^{-3})^2$.
  • A1 — Calculating the final correct answer of 440 W m⁻², which is correctly rounded to two significant figures.

  Common mistakes

  • Forgetting to square the side length to find the area, instead incorrectly calculating $750 / 1.3$.
  • Failing to convert the side length from millimetres to metres, leading to an erroneous calculation of $750 / 1300^2$.
  • Making a simple arithmetic error when squaring the side length, for example calculating $(1.3)^2$ as 2.6.
  • Providing the final answer to an inappropriate number of significant figures, such as 443.8 or 444, instead of rounding to 440 based on the input data.

  Examiner tip: Always ensure all quantities are converted to their base SI units before substituting them into a physics equation to avoid conversion errors in the final answer.

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