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The diagram shows light from a point source, S, spreading out as it gets further from S - Edexcel - GCSE Physics - Question 5 - 2013 - Paper 1
Question 5
The diagram shows light from a point source, S, spreading out as it gets further from S. The intensity of light passing through the surface which is 1 m from S is 2... show full transcript
Worked Solution & Example Answer:The diagram shows light from a point source, S, spreading out as it gets further from S - Edexcel - GCSE Physics - Question 5 - 2013 - Paper 1
Step 1
Complete the sentence by putting a cross (×) in the box next to your answer.
Answer
To find the intensity of light at a distance of 2 m from the source, we apply the inverse square law, which states that intensity is inversely proportional to the square of the distance from the source. Therefore, the intensity at 2 m can be calculated as:
I = rac{P}{A} = rac{P}{4 ext{π}r^2}
Given that at 1 m, intensity is 2.5 W/m², we will find that:
- The intensity at 2 m is:
I_{2m} = rac{2.5}{(2)^2} = rac{2.5}{4} = 0.625 ext{ W/m²}
So, the options given correspond to:
- A = 2.5 + 2 = 4.5
- B = 2.5 + 4 = 6.5
- C = 2.5 × 2 = 5
- D = 2.5 × 4 = 10
We confirm that C (2.5 × 2) is not correct. Therefore, the best option to fill the sentence is D (2.5 × 4).
Step 2
Calculate the power of the light passing through the surface which is 1 m from S.
Answer
The power, , can be found from the intensity and area of the surface through which the light passes. Since we know the intensity at 1 m is 2.5 W/m² and the area (A) through which the light passes (assumed to be spherical at this distance) can be calculated based on the given information. Using the order of intensity and rearranging:
If the area at 1 m is given as 0.2 m²:
Therefore, the power of the light passing through the surface at 1 m from S is 0.5 W.