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The following text is from Concepts of Physics by Dr. H.C.Verma, from chapter "Geometrical Optics", page 387, topic "Relation between $u$,$v$ and $R$ for Spherical Mirrors":

enter image description here

If the point $A$ is close to $P$, the angles $\alpha$,$\beta$ and $\gamma$ are small and we can write

$$\alpha\approx\frac{AP}{PO},\ \beta=\frac{AP}{PC}\ \ \ \text{and} \ \ \gamma \approx\frac{AP}{PI}$$

As $C$ is the centre of curvature, the equation for $\beta$ is exact whereas the remaining two are approximate.

The terms on the R.H.S. of the equations for the angles $\alpha$,$\beta$ and $\gamma$, are the tangents of the respective angles. We know that, when the angle $\theta$ is small, then $\tan\theta\approx\theta$. In the above case, this can be imagined as, when the angle becomes smaller, $AP$ becomes more and more perpendicular to the principal axis. And thus the formula for the tangent could be used.

But, how can this approximation result in a better accuracy for $\beta$ when compared to $\alpha$ and $\gamma$? I don't understand the reasoning behind the statement: "As $C$ is the centre of curvature, the equation for $\beta$ is exact whereas the remaining two are approximate." I can see the author has used "$=$" instead of "$\approx$" for $\beta$ and he supports this with that statement. But why is this so? Shouldn't the expression for $\beta$ be also an approximation over equality? Is the equation and the following statement really correct?

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I believe the author is assuming that you recognize that AP is an arc by the context of the image. The main point here then is that C is the center of curvature of the mirror while O and I are not. As C is the center, AC = PC, and the usual arc length formula can be applied exactly with angle $\beta$ even if $\beta$ isn't assumed small.

On the other hand, O and I are not centers of curvatures. Thus, OA $\neq$ PO and AI $\neq$ PI, so the arc length formula does not apply exactly. However, as you note, AP becomes more tangent to the horizontal axis as the angles become smaller. For small angles, arc AP is approximately a straight line perpendicular to the horizontal axis and the angles are approximately the tangents of the angles, so you can apply the tangent definition to relate AP to PO and AP to PI by the angles in this case.

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For the angle $\beta$ the author is apparantly thinking of AP as being the arc length along the circle rather than the length of the straight line joining A to P. Then $$ {\rm arclength}({\rm AP}) = \beta R $$ exactly.

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  • $\begingroup$ But then that is also the case with $\alpha$ and $\gamma$. $\endgroup$
    – user243267
    Dec 26, 2019 at 14:51
  • $\begingroup$ @FakeMod Yes. I think so. The book is not being very precise, and I agree with M. Guru Vishnu's complaint. $\endgroup$
    – mike stone
    Dec 26, 2019 at 14:56
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    $\begingroup$ @FakeMod Not so. As C is the center of curvature, AC = PC, so the arclength relation holds exactly as with a circle. For the other cases, OA is not equal to OP and AI is not equal to IP except approximately when the angles are small. $\endgroup$
    – Harjeq
    Dec 26, 2019 at 14:56
  • $\begingroup$ @JeqHar I definitely missed that point! Thanks for clarifying. $\endgroup$
    – user243267
    Dec 26, 2019 at 14:58

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