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# Lec25 - t = tOP O tOO =(tOP tOD(n1(tP O tDO(tV W(n2 tV W(n1...

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Δ t = t OPO - t OO = t OP + t PO - t OD ( n 1 ) - t DO - ( t V W ( n 2 ) - t V W ( n 1 )) = ( t OP - t OD ( n 1 )) + ( t PO - t DO ) - ( t V W ( n 2 ) - t V W ( n 1 )) = n 1 h 2 2 s + n 1 h 2 2 s - ( n 2 - n 1 ) V W and we still need to evaluate the distance V W . To evaluate this distance, notice that it is the distance on the axis between two spheres with radii R 1 and R 2 (non–concentric!) that intersect. From PDC 2 , R 2 - DC 2 = h 2 2 R 2 . But R 2 - DC 2 is also the distance WD . Therefore, WD = h 2 2 R 2 . Similarly, from PDC 1 , R 1 - DC 1 = h 2 2 R 1 = V D . Next, V W = V D - WD = h 2 2 R 1 - h 2 2 R 2 . We now require that Fermat’s Principle holds, or Δ t = 0. Therefore, n 1 h 2 2 s + n 1 h 2 2 s = ( n 2 - n 1 ) h 2 2 R 1 - h 2 2 R 2 n 1 s + n 1 s = ( n 2 - n 1 ) 1 R 1 - 1 R 2 . Now take the limit s → ∞ , which would give us an s value we identify with f . Therefore, n 1 f = ( n 2 - n 1 ) 1 R 1 - 1 R 2 , 58

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or 1 f = n 2 n 1 - 1 1 R 1 - 1 R 2 . We now define n := n 2 /n 1 , so that 1 f = ( n - 1) 1 R 1 - 1 R 2 . This last relation is sometimes called the lens maker’s formula . (Notice, it is a formula, not an equation.) If we now take s → ∞ we identify s with f . Then, n 1 f = ( n 2 - n 1 ) 1 R 1 - 1 R 2 1 f = ( n - 1) 1 R 1 - 1 R 2 which is exactly what we found for 1 /f . Therefore, 1 f = ( n - 1) 1 R 1 - 1 R 2 = 1 f Notice that we find that f =
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Lec25 - t = tOP O tOO =(tOP tOD(n1(tP O tDO(tV W(n2 tV W(n1...

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