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And if
| m= | 32, r=r′=5, ∆=∞, v=.0006, |
| ∆′= | 15, ∆″=r=5, |
| A= | {−32·52152·10015+100(23.15−15)}v |
| = | −{5045+3409}v |
| = | −3509v |
| = | −.023 |
In this case also the aberration diminishes the focal length.
- For the double-convex lens r is negative,
A={m·∆″2∆′2·(∆′−r)2·(m∆−1∆′)+(∆″−r′)2·(1m∆′−1∆″)}·v.
And if
| m= | 32, r=r′=5, ∆=∞, v=.0006, |
| ∆′= | −15, ∆″=−r=−5, |
| A= | {32·52152·10015+100(−23.15+15)}v |
| = | {109+1409}v |
| = | 1509v |
| = | .01 |
Here the focal distance is negative, and the aberration positive, so that they are still contrary. Observe that the aberration is more than twice as great in the concave lens than in the convex with equal radii.
