Going through a medium?

$As\quad in\quad the\quad figure,\quad Mt\bot AB,\quad Mt\quad and\quad CD\quad intersect\quad at\quad H,\\ Mx\quad and\quad CD\quad intersect\quad at\quad I,\quad NK\bot MI.\\ In\quad the\quad right-angled\quad triagle\quad HMI,\quad \angle HMI=\angle xMt=90°-\angle AMx=45°.\\ So\quad HMI\quad is\quad an\quad isosceles\quad right-angled\quad triangle.\\ Hence\quad HI=HM=1cm.\\ In\quad the\quad right-angled\quad triangle\quad HMN,\quad \angle HMN=90°-60°=30°\\ So\quad MN=2HN\quad (I'll\quad prove\quad it\quad later)\\ By\quad Pythagorean\quad Theorem,\quad HM²+HN²=MN²\\ \qquad 1²+HN²=(2HN)²\\ \qquad HN²=\frac { 1 }{ 3 } \\ \qquad HN=\frac { 1 }{ \sqrt { 3 } } cm\quad (HN>0)\\ NI=HI-HN=\frac { 3-\sqrt { 3 } }{ 3 } cm\\ In\quad the\quad right-angled\quad triangle\quad NKI,\quad \angle NIK=45°\\ So\quad NKI\quad is\quad an\quad isosceles\quad right-angled\quad triangle.\\ Hence\quad NK=IK\\ By\quad Pythagorean\quad Theorem,\quad NK²+KI²=NI²\\ \qquad 2NK²={ \left( \frac { 3-\sqrt { 3 } }{ 3 } \right) }^{ 2 }\\ \qquad NK=\sqrt { \frac { 4-2\sqrt { 3 } }{ 6 } } cm\quad (NK>0)$ Do you realize that NK is the distance between Mx and Ny? :D

Using physics lateral shift =t sin(i-r)/cos r just use this formula of physics for lateral shift No need to do much calculation