Mirror focal distance:
F=2R=22.45=1.225m Because of u<F , where u - distance from mirror to object and v - distance from image to mirror, then:
u1−v1=F1 and
v=F−uFu=1.225−0.151.225×0.15=0.17m The height of the object is:
y′=yuv=7×0.150.17=7.98m Thus, the image will be imaginary, enlarged, with a height of 7.98 m, at a distance of 0.17 m behind the mirror.
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