Chapter 44: Problem 42

Question: Isaac Newton, having erroneously convineced himself that chromatic aberration was an inherent property of refracting telescopes, invented the reflecting telescope. He presented his second model of this telescope which has a magnifying power of 38 to the Royal Society. Incident light falls, closely paralle to the telscope axis, on the objective mirror M. Afet reflection from small mirror $M^\prime$, the rays form a real, inverted image in the focal palne through F. The image is then viewed through the eye piece. Show the angular magnification is $m_\theta~=~-f_{mirror}/f_{eye piece}$, where f_mirror is the focal length of the mirror and f_{eye piece} is the focal length of the lens. The $200~{\rm in}$ mirror in the reflecting telescope at Mt. Palomar in California has a focal length of $f~=~16.8~{\rm m}$. Estimate the size of the image formed in the focal plane of this mirror when the object is a meter stick $2.0~{\rm km}$ away. Assume parallel incident rays. The mirro of the reflecting astronomical telescope has an effective radius of curvature of $10~{\rm m}$. To give an angular magnification of 200 what must be the focal length of the eyepiece?

Solution: We basically proved that $m_\theta~=~-f_{mirror}/f_{eye piece}$ in class, by small modifications to our discussion of refracting telescopes. For a $1~{\rm m}$ object $2000~{\rm m}$ away from the mirror, the angle subtended between rays from either end of the stick is $\theta_1~\sim~1/2000$, and therefore the size of the image in the focal plane is 16.8/2000. We know that the angular magnification of the combined mirror-eye piece system is $m_\theta~=~-f_{mirror}/f_{eye piece}$, and so to get $\vert m_\theta\vert~=~200$with $f_{mirror}=10~{\rm m}$, we need $f_{eye piece}~=~0.05~{\rm m}~=~5~{\rm cm}$