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33

$OO'$	$=$	$2 KO =2 θ'$ ,
$OO ″$	$=$	$2 HO' - OO' = HO' + HO =2 HK =2 ι$ ,
$OO ‴$	$=$	$2 KO ″ - OO ″ = KO ″ + KO = OO ″ +2 KO$
		2HO $=2 ι +2 θ'$ ,
&c.⁠&c.

The number of images is not unbounded in this case, as in that of two parallel mirrors, for when any one of the images, as $O (n)$ , or $O (n)$ gets between the lines $HI$ , $KI$ produced, no further reflexion can take place, as no rays proceeding from such a point could fall on the face of either of the mirrors.

In order to express this condition algebraically, we must observe, that of the first series of images, $O'$ , $O ‴$ , $O v.$ , &c. lie on one side of the mirrors, and $O ″$ , $O iv.$ , $O vi.$ , &c. on the other, and that if $O (2 n +1)$ , for instance, be the last, the distance $KO (2 n +1)$ , or $2 nι +2 θ + θ'$ , that is, $2 nι + ι + θ'$ , must be the first that is greater than $Kk$ or $π$ ,

that is, $(2 n +1) ι + θ > π$ ,

or $2 n +1> π - θ / ι$ .

If $O (2 n)$ be the last image, we must have $HO (2 n) < π$ ;

that is, $2 nι + θ > π$ , or $2 n > π - θ / ι$ ,

the same expression as before, $2 n +1$ being the number of images in the one case, and $2 n$ in the other.

In like manner we should find, that the number of images in the other series is the least whole number greater than $π - θ' / ι$ .

If $ι$ be a measure of $π$ , since $θ / ι$ and $θ' / ι$ are proper fractions, the number of images in each series must be $π / ι$ , and therefore the whole number of images $2 π / ι$ . In this case, however, two images