he triangles are equal in all respects andthe angle RPM = the angle RPM. But since RN and MP are parallel, the angle N£P = the angle RPM,and the angle NRQ= the angle RPM. Therefore the angle NRQ = ihe angle NRP. Hence RQ is in the same plane as the incident ray andthe normal, and makes with the… … all the reflected rays pass through P • hence P is the imageof P. 27. To trace the rays by which an eye sees aluminous point reflected in a mirror. Let AB, fig. 21,be the mirror, P the lumi-nous point, E the eye. Draw PM perpendicular to the p mirror and produce it to P, % making MP equal to MP.…
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