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Figure 6 shows a diagram of the Michelson-Morley interferometer that was used to try to detect the absolute motion of the Earth through the ether (either) - AQA - A-Level Physics - Question 4 - 2018 - Paper 7

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Figure 6 shows a diagram of the Michelson-Morley interferometer that was used to try to detect the absolute motion of the Earth through the ether (either). Light fr... show full transcript

Worked Solution & Example Answer:Figure 6 shows a diagram of the Michelson-Morley interferometer that was used to try to detect the absolute motion of the Earth through the ether (either) - AQA - A-Level Physics - Question 4 - 2018 - Paper 7

Step 1

Explain how the experiment provided a means of testing the idea that the Earth had an absolute motion relative to the ether.

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Answer

The Michelson-Morley experiment was designed to detect the presence of 'aether' by observing the interference patterns of light beams split and recombined after traveling different paths. It was predicted that if the Earth moved through the aether, the time taken for light to travel along the path of the motion would differ from the time taken in the opposite direction. This would lead to a shift in the interference pattern. When rotated 90 degrees, no such shift of 0.4 of the fringe spacing was observed, indicating that either the aether did not exist or that the Earth's motion relative to the aether could not be detected.

Step 2

The Michelson-Morley experiment provides evidence for one of the postulates of Einstein’s theory of special relativity. State this postulate.

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Answer

The laws of physics are the same in all inertial frames.

Step 3

State the other postulate of Einstein’s theory of special relativity.

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Answer

The speed of light is the same in all inertial frames of reference.

Step 4

Determine the percentage of muons that reach the detector.

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Answer

Given the muon's reference frame travel distance of 1310 m and a velocity of 0.990c, we use the time dilation formula. The time in the muon's frame is calculated as: t=distancevelocity=1310m0.990ct = \frac{distance}{velocity} = \frac{1310 m}{0.990c}, where c3.00×108m/sc \approx 3.00 \times 10^8 m/s. Thus: t13100.990×3.00×1084.41×106s.t \approx \frac{1310}{0.990 \times 3.00 \times 10^8} \approx 4.41 \times 10^{-6}s.

Using the graph from Figure 7, we can determine the percentage of muons that remain after this time. According to the graph, we find that approximately 85% of the muons remain after this duration.

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