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OO′=2OA=2a, |
OO″=2O′B−O′A=O′B+OB=2AB=2c, |
OO‴=2A′O″−OO″=AO″+AO=2OO″+2AO=2c+2a, |
⋮: |
OO′=2OB=b, |
OO″=2AO′−OO′=AO′+AO=2AB=2c, |
OO‴=2BO″−OO″=BO″+BO=OO″+2BO=2c+2b. |
⋮: |
The angular distances between the object and the images, that is, the angles OEO′, OEO″, &c. may be calculated by means of their tangents; thus, if EN be perpendicular to AB,
OEO′ | = | NEO′−NEO=tan−1NO′EN−tan−1NOEN, |
OEO″ | = | NEO″+NEO=tan−1NO″EN+tan−1NOEN, |
OEO‴ | = | NEO‴−NEO=tan−1NO‴EN−tan−1NOEN, &c. |
Thus supposing the distance AB is 5 inches,
that AO | = | 2, |
BO | = | 3, |
EN | = | 6, |
NO | = | 1. |
Then OO′=4; | OO″=10; | OO‴=14, &c. |
OO′=b; | OO″=10; | OO‴=16, &c. |
NOEN=16=.1666… | NEO=tan−1.1666…=9°27′12 nearly, |
NO′EN=56=.8333… | NEO′=tan−1.8333…=39°48′12; |
∴ OEO′=30°21′, |