<p>In an experiment designed to measure the speed of light, a laser is aimed at
ID: 1973353 • Letter: #
Question
<p>In an experiment designed to measure the speed of light, a laser is aimed at a mirror that is 60.0 km due north. A detector is placed 125 m due east of the laser. The mirror is to be aligned so that light from the laser refelects into the detector.When properly aligned, what angle should the normal to the surface of the mirror make with due south? Suppose the mirror is misaligned, so that the actual angle between the normal to the surface and due south is too large by 0.0040°. By how many meters (due east) will the reflected ray miss the detector?</p>Explanation / Answer
have solved a similar question already.. will paste them.. you could substitute your values
In an experiment designed to measure the speed of light, a laser is aimed at a mirror that is 49.0 km due north. A detector is placed 120 m due east of the laser. The mirror is to be aligned so that light from the laser reflects into the detector.
Suppose the mirror is misaligned, so that the actual angle between the normal to the surface and due south is too large by 0.005°. By how many meters (due east) will the reflected ray miss the detector?
ANSWER
The ideal bearing of the normal vector is ½·tan¹(120/49000) east of south. If this bearing is 0.005° too far east, then the reflected ray will have a bearing of
tan¹(120/49000)+2·0.005° east of south.
The reflected ray will cross the baseline joining the laser source and detector at a distance of
49000·tan(tan¹(120/49000)+0.01°) east of the source.
Calculate
49000·tan(tan¹(120/49000)+0.01°) = 128.552 m.
The ray will miss the detector by 8.5522 m (eastward).
As a check, the error can be approximated by r·,
where r = 49000 m is the radius and
= 0.01·/180 radians. This works out as
49000·0.01·/180 = 8.5521 m, which is within 0.1mm of the exact calculation.
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