Q177 · Practice questionComplex unfamiliar7 marks
QUESTION 177 (7 marks)
A light clock has two mirrors separated vertically by in its own frame. A pulse travels from the lower mirror to the upper mirror and back to the lower mirror. The clock moves horizontally at relative to a laboratory. Its vertical separation is unchanged. Use . The diagram shows the laboratory-frame path of the upward pulse.
a)[2 marks]
Determine the proper time interval for the complete round trip.
b)[3 marks]
Use the path geometry to determine the laboratory interval for the complete round trip.
c)[2 marks]
Explain why the longer laboratory time is consistent with an unchanged speed of light.
WORKED SOLUTION
7 marksPractice marking scheme
ANSWER
(a) . (b) . (c) The laboratory path is longer; speed remains c, so transit time is longer.
Worked solution
(a) The emission and return occur at the same lower-mirror position in the clock rest frame. The round-trip path length is , so .
(b) Each half of the laboratory journey takes . The upward light path has length while the mirror moves horizontally. Thus , giving .
(c) The horizontal motion of the upper mirror makes the laboratory light path diagonal and longer than . Both observers measure light speed ; the extra path length requires extra time.
Equivalent physically justified methods and consistent equivalent units accepted.
Displayed decimals are model answers; accept appropriate significant figures and consistent rounding from stated constants.
Part a: Identify the complete round-trip length in the clock rest frame.
Part a: Calculate the proper interval between emission and return at the same mirror.
Part b: Construct the right-triangle relation for the light path.
Part b: Rearrange the relation without adding source speed to c.
Part b: Calculate the laboratory time.
Part c: Identify the longer diagonal path.
Part c: Relate extra travel time to invariant light speed.
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