(a) Intensity of \( S_2 \): \( I_{S_2} = 2 \times 10^{-6} = \frac{1}{500000} \) units. A1

[1 mark]

(b) Loudness of \( S_2 \): substitute \( I_{S_2} = 2 \times 10^{-6} \) into \( L = 10 \log_{10}(I \times 10^{12}) \). M1

\( L_{S_2} = 10 \log_{10}(2 \times 10^{-6} \times 10^{12}) = 10 \log_{10}(2 \times 10^6) \). A1

\( = 10 (\log_{10} 2 + 6) \approx 10 (0.3010 + 6) = 63.01 \approx 63.0 \) decibels. A1

[2 marks]

(c) Thunder’s loudness: \( 115 = 10 \log_{10}(I \times 10^{12}) \). M1

Solve: \( \log_{10}(I \times 10^{12}) = \frac{115}{10} = 11.5 \), so \( I \times 10^{12} = 10^{11.5} \). A1

\( I = \frac{10^{11.5}}{10^{12}} = 10^{-0.5} \approx 0.316227 \approx 0.316 \) units. A1

[3 marks]

Total [6 marks]

(a) Given: Intensity of \( S_1 \) is \( 10^{-6} \) units. Intensity of \( S_2 \) is twice that of \( S_1 \).

\( I_{S_2} = 2 \times 10^{-6} \) units.

(b) Loudness formula: \( L = 10 \log_{10}(I \times 10^{12}) \). For \( S_1 \), \( I_{S_1} = 10^{-6} \):

\( L_{S_1} = 10 \log_{10}(10^{-6} \times 10^{12}) = 10 \log_{10}(10^6) = 60 \) decibels, which matches.

For \( S_2 \), \( I_{S_2} = 2 \times 10^{-6} \):

\( L_{S_2} = 10 \log_{10}(2 \times 10^{-6} \times 10^{12}) = 10 \log_{10}(2 \times 10^6) \).

\( = 10 (\log_{10} 2 + 6) \approx 10 (0.3010 + 6) = 63.01 \approx 63.0 \) decibels.

(c) For thunder, \( L = 115 \) decibels:

\( 115 = 10 \log_{10}(I \times 10^{12}) \).

\( \log_{10}(I \times 10^{12}) = 11.5 \), so \( I \times 10^{12} = 10^{11.5} \).

\( I = 10^{11.5 – 12} = 10^{-0.5} \approx 0.316227 \approx 0.316 \) units.